YOUR EXAM IS TODAY: Friday, March 13, 2026 | Ironwood 120 | Choose: 9-11 AM, 1-3 PM, or 6-8 PM
- ECE211_Weak_Areas_Focus.md - Deep-dive tutorials on your 8 weakest topics (based on your Canvas grades)
- ECE211_Video_Resources.md - All YouTube lecture & review videos, prioritized by your weak areas
- Formula_sheet_statics_reactions_2D_3D-1.pdf - Textbook support reaction tables (in .playwright-mcp folder)
| Priority | Topic | Your Score | Section in This Guide |
|---|---|---|---|
| 1 | 3D Particle Equilibrium | 0% (missing) | Chapter 3, Section 5.5 |
| 2 | 2D Rigid Body Equilibrium | 0% (missing) | Chapter 5, Section 7.5 |
| 3 | Moment About an Axis | 0% (missing) | Chapter 4, Section 6.3 |
| 4 | Springs & Normal Contact | 31% | Chapter 3, Section 5.3 |
| 5 | Dot Product | 50% | Chapter 2, Section 4.5 |
| 6 | Distributed Loads | 55% | Chapter 4, Section 6.6 |
| 7 | 2/3 Force Members | 56% | Chapter 5, Section 7.7 |
| 8 | 3D Rigid Body Equilibrium | 55% | Chapter 5, Section 7.6 |
- Exam Logistics
- Equation Sheet Template
- Chapter 1: General Principles
- Chapter 2: Force Vectors
- Chapter 3: Particle Equilibrium
- Chapter 4: Moments
- Chapter 5: Equilibrium of a Rigid Body
- Practice Problem Checklist
- Common Mistakes to Avoid
- Last-Minute Tips
- Date: Friday, March 13, 2026
- Location: CGCC Pecos Campus, Ironwood Building, Room 120
- Available Sessions (choose one):
- Morning: 9:00 AM - 11:00 AM
- Afternoon: 1:00 PM - 3:00 PM
- Evening: 6:00 PM - 8:00 PM
- Paper-based, in-person exam
- Duration: 2 hours
- Coverage: Modules 1-8 (Chapters 1-5 of the textbook)
- Structure: ~4 problems, similar to homework problems with different numbers
- Scratch paper will be provided (no need to bring your own)
- TI-83 or TI-84 calculator (no other calculator models permitted)
- One equation sheet: 8.5" x 11" paper, ONE SIDE ONLY
- Formulas and simple figures allowed
- NO worked-out problems (this means no full solutions, no example problems with numbers)
- Simple diagrams (support reaction diagrams, coordinate systems) ARE allowed
- Pencils/pens
- Student ID
- Cell phones (silence and put away)
- Notes beyond the single equation sheet
- Textbook
- Programmable calculators other than TI-83/84
- Worked-out example problems on your equation sheet
Your professor specifically mentioned having these programmed:
- Cross Product Program - Saves massive time on Ch 4 (moments) and Ch 5 (3D equilibrium)
- Video tutorial: https://www.youtube.com/watch?v=64HqdD7lr5E
- Dot Product Program - Helps with Ch 2 (angle between vectors, projections)
- Available on Canvas: "Calculator - Program the Dot Product Function" page
Make sure both programs are loaded and tested BEFORE the exam! Calculator memory is allowed. These programs alone can save 5-10 minutes on the exam.
Below is a recommended layout for your 8.5" x 11" one-sided equation sheet. Organize it in columns to maximize space. Use small but legible handwriting.
+---------------------------+---------------------------+---------------------------+
| COLUMN 1 (Left) | COLUMN 2 (Center) | COLUMN 3 (Right) |
| | | |
| GENERAL / CH.1 | CH.4: MOMENTS | CH.5: RIGID BODY EQ. |
| - W = mg | - M_O = r x F | - 2D: SFx=0, SFy=0, |
| - g = 9.81 m/s^2 | - M = Fd (scalar) | SM_O = 0 |
| - g = 32.2 ft/s^2 | - Cross product: | - 3D: SFx=SFy=SFz=0 |
| - 1 ft = 12 in | i j k | SMx=SMy=SMz=0 |
| - 1 mi = 5280 ft | rx ry rz | |
| - 1 kip = 1000 lb | Fx Fy Fz | SUPPORT REACTIONS (2D): |
| | = (ryFz-rzFy)i | - Roller: 1 (normal) |
| CH.2: VECTORS 2D | -(rxFz-rzFx)j | - Pin: 2 (Ax, Ay) |
| - Fx = F cos(th) | +(rxFy-ryFx)k | - Fixed: 3 (Ax,Ay,MA) |
| - Fy = F sin(th) | | - Cable: 1 (tension) |
| - F = sqrt(Fx^2+Fy^2) | MOMENT ABOUT AXIS: | - Smooth: 1 (normal) |
| - th = atan(Fy/Fx) | M_a = u_a . (r x F) | |
| - c^2=a^2+b^2-2ab*cosC | | SUPPORT REACTIONS (3D): |
| - a/sinA = b/sinB | COUPLE: | - Ball&socket: 3 |
| | M = r x F (free vector) | (Ax, Ay, Az) |
| CH.2: VECTORS 3D | - Same moment about | - Journal bearing: 4-5 |
| - r = (x2-x1)i + | ANY point | - Fixed: 6 |
| (y2-y1)j + (z2-z1)k | | (Ax,Ay,Az,MAx,MAy,MAz)|
| - |r| = sqrt(x^2+y^2+z^2)| EQUIVALENT SYSTEMS: | |
| - u_hat = r/|r| | Move F to new point: | [SIMPLE FIGURES:] |
| - F = F * u_hat | Add couple M = r x F | - Pin support diagram |
| - F = F(cos a i + | | - Roller diagram |
| cos b j + cos g k) | DISTRIBUTED LOAD: | - Fixed support diagram |
| - cos^2a+cos^2b+cos^2g=1 | F_R = integral w(x)dx | - Ball & socket diagram |
| | = area under curve | |
| DOT PRODUCT: | Location: at centroid | COMMON LOAD SHAPES: |
| - A.B = |A||B|cos(th) | of load area | - Rect: F=wL, at L/2 |
| - A.B = AxBx+AyBy+AzBz | | - Triangle: F=wL/2, |
| - th = acos(A.B/|A||B|) | Common shapes: | at L/3 from big end |
| - F_par = F . u_hat | - Rect: A=wL, x=L/2 | |
| - F_perp = sqrt( | - Tri: A=wL/2, x=L/3 | |
| F^2 - F_par^2) | (from larger end) | |
| | | |
| CH.3: PARTICLE EQ. | CH.3: SPRINGS | TWO-FORCE MEMBERS: |
| 2D: SFx=0, SFy=0 | F_s = k * delta_s | Force along the line |
| 3D: SFx=SFy=SFz=0 | delta_s = L - L_0 | between the 2 load pts |
| | Pulley: T same both sides| 3-FORCE: concurrent! |
+---------------------------+---------------------------+---------------------------+
- Write small but legible - Use a fine-point pen (0.5mm or smaller)
- Use color coding - Different colors for different chapters helps you find formulas fast
- Include simple support reaction diagrams - A quick sketch of each support type with its reactions is worth the space
- Include the cross product determinant layout - You will almost certainly need it
- Include the distributed load centroid locations - Rectangle at L/2, triangle at L/3 from the larger end
- Do NOT waste space on worked examples - They are not allowed and waste precious space
- Include unit conversion factors - These are easy to forget under pressure
- BASED ON YOUR GRADES, prioritize these on your sheet:
- Spring formula (F=k*delta, delta=L-L0) - you scored 31%
- Dot product formulas (both forms + projection) - you scored 50%
- Distributed load centroids (rectangle, triangle, trapezoid) - you scored 55%
- Two-force member rule + three-force concurrency - you scored 56%
- 3D support reactions table (ball&socket=3, fixed=6) - you scored 55%
- Moment about an axis formula (triple scalar product determinant) - you scored 0%
Statics is the study of bodies at rest or moving at constant velocity (zero acceleration). In statics, all forces and moments acting on a body must be in equilibrium -- they must sum to zero.
The fundamental requirement:
- Sum of all forces = 0 (translational equilibrium)
- Sum of all moments = 0 (rotational equilibrium)
Newton's First Law (Law of Inertia): A body at rest remains at rest, and a body in motion remains in uniform motion, unless acted upon by a net external force. In statics, bodies are at rest, so the net force is zero.
Newton's Second Law: F = ma. In statics, a = 0, therefore F_net = 0. This is the basis of all equilibrium equations.
Newton's Third Law: For every action, there is an equal and opposite reaction. This is critical for drawing free body diagrams -- if a support pushes up on a beam, the beam pushes down on the support.
Newton's Law of Gravitation: W = mg, where:
- W = weight (force due to gravity)
- m = mass
- g = gravitational acceleration
| System | g value | Weight unit | Mass unit |
|---|---|---|---|
| SI | 9.81 m/s^2 | Newtons (N) | kilograms (kg) |
| US Customary | 32.2 ft/s^2 | Pounds (lb) | slugs |
| Quantity | Unit | Symbol |
|---|---|---|
| Length | meter | m |
| Mass | kilogram | kg |
| Time | second | s |
| Force | Newton | N = kg*m/s^2 |
| Quantity | Unit | Symbol |
|---|---|---|
| Length | foot | ft |
| Mass | slug | slug = lb*s^2/ft |
| Time | second | s |
| Force | pound | lb |
- 1 ft = 12 in
- 1 mi = 5280 ft
- 1 kip = 1000 lb (kilopound)
- 1 kg = 1000 g
- 1 m = 100 cm = 1000 mm
- 1 kN = 1000 N
- 1 ft = 0.3048 m
- 1 lb = 4.448 N
- 1 slug = 14.59 kg
| Prefix | Symbol | Factor |
|---|---|---|
| Giga | G | 10^9 |
| Mega | M | 10^6 |
| Kilo | k | 10^3 |
| Milli | m | 10^-3 |
| Micro | mu | 10^-6 |
| Nano | n | 10^-9 |
In engineering statics, results are typically reported to 3 significant figures. When performing intermediate calculations, keep at least 4 significant figures to avoid rounding errors in the final answer.
Example:
- 1234 N = 1.23 kN (3 sig figs)
- 0.005678 m = 5.68 mm (3 sig figs)
For every statics problem, follow this systematic approach:
- Read the problem carefully - Identify what is given and what is asked
- Draw a diagram - Sketch the system, label all known quantities
- Draw a Free Body Diagram (FBD) - Isolate the body, show all forces
- Write equilibrium equations - Apply the relevant equilibrium conditions
- Solve the equations - Use algebra/trig to find unknowns
- Check your answer - Units correct? Magnitude reasonable? Signs make sense?
This is a large chapter covering Sections 2.1-2.9, split across Modules 1 and 2.
- Scalar: A quantity with magnitude only (mass, temperature, time, length)
- Vector: A quantity with both magnitude and direction (force, velocity, displacement)
Vectors are denoted with boldface (F) or an arrow above the letter. The magnitude is denoted |F| or simply F.
Two forces F1 and F2 acting at the same point can be replaced by a single resultant force FR using the parallelogram law:
- Place vectors tail-to-tail
- Complete the parallelogram
- The diagonal from the common tail is the resultant
An equivalent method:
- Place the tail of F2 at the head of F1
- The resultant FR goes from the tail of F1 to the head of F2
This extends to multiple forces: place them head-to-tail sequentially. The resultant goes from the first tail to the last head.
Any force can be resolved into components along two directions (typically x and y axes):
Given: Force F at angle theta from the positive x-axis
F_x = F cos(theta)
F_y = F sin(theta)
CRITICAL: The angle theta must be measured correctly. If the angle is measured from a different reference:
- From the y-axis: F_x = F sin(theta), F_y = F cos(theta)
- From the negative x-axis: F_x = -F cos(theta), F_y = F sin(theta)
Always sketch the force and its components to verify which trig function applies.
Given several forces, find the resultant by:
Step 1: Resolve each force into x and y components
Step 2: Sum all x-components and all y-components:
FR_x = SUM(F_ix) = F1x + F2x + F3x + ...
FR_y = SUM(F_iy) = F1y + F2y + F3y + ...
Step 3: Find the resultant magnitude and direction:
FR = sqrt(FR_x^2 + FR_y^2)
theta = tan^(-1)(FR_y / FR_x)
WATCH OUT for the quadrant! The arctangent function on your calculator only gives values in (-90, 90). You must check the signs of FR_x and FR_y to determine the correct quadrant:
| FR_x | FR_y | Quadrant | Angle from +x |
|---|---|---|---|
| + | + | I | theta (as calculated) |
| - | + | II | 180 - |
| - | - | III | 180 + |
| + | - | IV | 360 - |
Problem: Three forces act at point O. Find the resultant.
- F1 = 400 N at 30 degrees above the positive x-axis
- F2 = 250 N at 45 degrees above the negative x-axis (i.e., in quadrant II)
- F3 = 300 N pointing straight down (negative y-direction)
Solution:
Step 1: Resolve each force into components.
Force F1 (30 degrees from +x axis):
F1x = 400 cos(30) = 400(0.8660) = 346.4 N
F1y = 400 sin(30) = 400(0.5000) = 200.0 N
Force F2 (45 degrees from +x axis, in quadrant II, so 180 - 45 = 135 degrees from +x):
F2x = 250 cos(135) = 250(-0.7071) = -176.8 N
F2y = 250 sin(135) = 250(0.7071) = 176.8 N
Force F3 (straight down):
F3x = 0 N
F3y = -300 N
Step 2: Sum components.
FR_x = 346.4 + (-176.8) + 0 = 169.6 N
FR_y = 200.0 + 176.8 + (-300) = 76.8 N
Step 3: Resultant magnitude and direction.
FR = sqrt(169.6^2 + 76.8^2) = sqrt(28764 + 5898) = sqrt(34662) = 186.2 N
theta = tan^(-1)(76.8 / 169.6) = tan^(-1)(0.4528) = 24.4 degrees
Since FR_x > 0 and FR_y > 0, the resultant is in quadrant I. The angle is 24.4 degrees above the positive x-axis.
Answer: FR = 186 N at 24.4 degrees from the positive x-axis.
When you cannot easily use rectangular components (e.g., forces are not along standard axes), use the triangle method with:
Law of Cosines:
c^2 = a^2 + b^2 - 2ab cos(C)
where c is the side opposite angle C.
Law of Sines:
a / sin(A) = b / sin(B) = c / sin(C)
These are especially useful when two forces at an angle form a triangle with the resultant.
Example using Law of Cosines:
Two forces of 600 N and 400 N act at a point with an angle of 60 degrees between them. Find the resultant.
Using the parallelogram law, the triangle formed has an included angle of 180 - 60 = 120 degrees (the supplement, because in the triangle the angle between the two force vectors measured inside the triangle is 180 - 60).
FR^2 = 600^2 + 400^2 - 2(600)(400)cos(120)
FR^2 = 360000 + 160000 - 480000(-0.5)
FR^2 = 360000 + 160000 + 240000
FR^2 = 760000
FR = 872 N
Then use the Law of Sines to find the direction:
400 / sin(alpha) = 872 / sin(120)
sin(alpha) = 400 sin(120) / 872 = 400(0.8660) / 872 = 0.3972
alpha = 23.4 degrees
The resultant is 872 N at 23.4 degrees from the 600 N force.
In 3D, forces are expressed using unit vectors along the coordinate axes:
F = Fx*i + Fy*j + Fz*k
where i, j, k are unit vectors along x, y, z respectively.
The magnitude:
F = |F| = sqrt(Fx^2 + Fy^2 + Fz^2)
A force direction can be described by angles alpha, beta, gamma measured from the positive x, y, z axes respectively:
Fx = F cos(alpha)
Fy = F cos(beta)
Fz = F cos(gamma)
The direction cosines must satisfy:
cos^2(alpha) + cos^2(beta) + cos^2(gamma) = 1
The unit vector in the direction of F:
u_F = cos(alpha)*i + cos(beta)*j + cos(gamma)*k
A position vector from point A(x1, y1, z1) to point B(x2, y2, z2):
r_AB = (x2 - x1)*i + (y2 - y1)*j + (z2 - z1)*k
The magnitude (distance between A and B):
|r_AB| = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)
The unit vector from A toward B:
u_AB = r_AB / |r_AB|
This is extremely important! To express a force directed along a line from A to B:
F = F * u_AB = F * (r_AB / |r_AB|)
Problem: A cable exerts a 500 N force on point A(1, 2, 3) directed toward point B(4, -1, 6). Express this force as a Cartesian vector.
Solution:
Step 1: Position vector from A to B:
r_AB = (4-1)*i + (-1-2)*j + (6-3)*k
r_AB = 3i - 3j + 3k
Step 2: Magnitude of r_AB:
|r_AB| = sqrt(3^2 + (-3)^2 + 3^2) = sqrt(9 + 9 + 9) = sqrt(27) = 5.196 m
Step 3: Unit vector:
u_AB = r_AB / |r_AB| = (3i - 3j + 3k) / 5.196
u_AB = 0.5774i - 0.5774j + 0.5774k
Step 4: Force vector:
F = 500 * u_AB = 500(0.5774i - 0.5774j + 0.5774k)
F = 288.7i - 288.7j + 288.7k N
Verification: |F| = sqrt(288.7^2 + 288.7^2 + 288.7^2) = sqrt(3 * 83347) = 500 N. Checks out.
To add 3D vectors, simply add corresponding components:
FR = F1 + F2 + F3
FR_x = F1x + F2x + F3x
FR_y = F1y + F2y + F3y
FR_z = F1z + F2z + F3z
Then:
|FR| = sqrt(FR_x^2 + FR_y^2 + FR_z^2)
The dot product of two vectors A and B:
Geometric form:
A . B = |A| |B| cos(theta)
where theta is the angle between the two vectors (0 <= theta <= 180 degrees).
Component form:
A . B = Ax*Bx + Ay*By + Az*Bz
Combining both forms:
cos(theta) = (A . B) / (|A| |B|)
theta = cos^(-1)[(Ax*Bx + Ay*By + Az*Bz) / (|A| |B|)]
The scalar projection of F onto a line defined by unit vector u:
F_parallel = F . u = Fx*ux + Fy*uy + Fz*uz
Note: F_parallel can be negative! A negative result means the component points opposite to u.
The perpendicular component:
F_perpendicular = sqrt(F^2 - F_parallel^2)
Problem: Given F = 200i + 300j - 100k N and a line from origin O to point A(3, 4, 0). Find: (a) The angle between F and line OA (b) The component of F parallel to OA (c) The component of F perpendicular to OA
Solution:
Step 1: Find the unit vector along OA.
r_OA = 3i + 4j + 0k
|r_OA| = sqrt(9 + 16 + 0) = 5
u_OA = (3i + 4j) / 5 = 0.6i + 0.8j
Step 2 (a): Angle between F and OA.
F . u_OA = 200(0.6) + 300(0.8) + (-100)(0) = 120 + 240 + 0 = 360
|F| = sqrt(200^2 + 300^2 + 100^2) = sqrt(40000 + 90000 + 10000) = sqrt(140000) = 374.2 N
cos(theta) = 360 / (374.2 * 1) = 0.9620
theta = cos^(-1)(0.9620) = 15.8 degrees
Step 3 (b): Parallel component.
F_parallel = F . u_OA = 360 N
(The scalar projection IS the dot product with the unit vector.)
Step 4 (c): Perpendicular component.
F_perp = sqrt(374.2^2 - 360^2) = sqrt(140026 - 129600) = sqrt(10426) = 102.1 N
- Commutative: A . B = B . A
- Distributive: A . (B + C) = A . B + A . C
- Scalar multiplication: (cA) . B = c(A . B)
- Unit vector dot products: i . i = j . j = k . k = 1; i . j = i . k = j . k = 0
- Always draw a sketch showing the force and the coordinate system before resolving into components
- Double-check angle references - Is the angle from the x-axis or y-axis? Measured clockwise or counterclockwise?
- For 3D forces along a cable/line, the position vector approach (r/|r|) is almost always the cleanest method
- Verify your unit vector by checking that its magnitude equals 1
- When computing the resultant, keep track of signs meticulously. A force pointing left has a negative x-component
- Swapping sine and cosine when the angle is from the y-axis instead of x-axis
- Wrong quadrant for the arctangent - always check signs of components
- Forgetting the negative sign on components pointing in the negative direction
- Position vector direction wrong - r_AB goes FROM A TO B, meaning B's coordinates minus A's coordinates
- Not normalizing the unit vector - dividing r by |r| is essential
- Dot product gives a SCALAR, not a vector - do not put unit vectors on the result of a dot product
- Using the wrong angle in the Law of Cosines - the angle must be the included angle in the triangle, which is often the supplement of the given angle between forces
A "particle" in statics means we treat the object as a single point where all forces meet (concurrent forces). There are no moments to consider -- only force balance.
Equilibrium condition for a particle:
SUM(F) = 0
In component form:
2D: SFx = 0 and SFy = 0 (2 equations, can solve for 2 unknowns)
3D: SFx = 0, SFy = 0, SFz = 0 (3 equations, can solve for 3 unknowns)
A Free Body Diagram (FBD) is the most important tool in statics. For a particle:
- Draw the particle as a point
- Show ALL forces acting on the particle:
- Applied forces (given loads)
- Weight (if applicable): W = mg, acting downward
- Cable/rope tensions: along the cable, pulling away from the particle
- Spring forces: F = ks (Hooke's Law)
- Normal forces: perpendicular to the contact surface
- Reaction forces at supports
- Label all forces with magnitudes (if known) or variables (if unknown)
- Label all angles with respect to a reference axis
- Always in tension (cables cannot push)
- Force acts along the cable, directed away from the particle
- A cable over a frictionless pulley has the same tension on both sides
- Force: F = k * s
- k = spring constant (stiffness), units: N/m or lb/ft
- s = deformation (stretch or compression) = |l - l_0|
- l = deformed length, l_0 = natural (unstretched) length
- Spring can be in tension (stretched) or compression (compressed)
- Draw the FBD of the particle
- Establish an x-y coordinate system (usually horizontal/vertical, but can be rotated for convenience)
- Resolve all forces into x and y components
- Apply equilibrium equations:
SFx = 0: (sum of all x-components) = 0 SFy = 0: (sum of all y-components) = 0 - Solve the system of two equations for the unknowns (maximum 2 unknowns)
Problem (similar to Problem 3-21): A 50 kg crate is supported by two cables as shown. Cable AB makes a 30-degree angle with the horizontal, and cable AC makes a 45-degree angle with the horizontal. Find the tension in each cable.
B C
\ /
30 \ / 45
\ /
\ /
A
|
| W = 50(9.81) = 490.5 N
v
Solution:
Step 1: FBD at point A. Three forces act on point A:
- T_AB: tension in cable AB, directed from A toward B (up-left at 30 deg from horizontal)
- T_AC: tension in cable AC, directed from A toward C (up-right at 45 deg from horizontal)
- W: weight, pointing straight down = 50(9.81) = 490.5 N
Step 2: Set up the coordinate system with x horizontal (right +) and y vertical (up +).
Step 3: Resolve forces into components.
Cable AB (up and to the left, 30 deg from horizontal):
T_AB_x = -T_AB cos(30) (negative because pointing LEFT)
T_AB_y = +T_AB sin(30) (positive because pointing UP)
Cable AC (up and to the right, 45 deg from horizontal):
T_AC_x = +T_AC cos(45)
T_AC_y = +T_AC sin(45)
Weight:
W_x = 0
W_y = -490.5 N
Step 4: Apply equilibrium.
SFx = 0:
-T_AB cos(30) + T_AC cos(45) = 0
-0.8660 T_AB + 0.7071 T_AC = 0
T_AC = (0.8660 / 0.7071) T_AB = 1.2247 T_AB ... (Equation 1)
SFy = 0:
T_AB sin(30) + T_AC sin(45) - 490.5 = 0
0.5 T_AB + 0.7071 T_AC = 490.5 ... (Equation 2)
Step 5: Substitute Equation 1 into Equation 2.
0.5 T_AB + 0.7071(1.2247 T_AB) = 490.5
0.5 T_AB + 0.8660 T_AB = 490.5
1.366 T_AB = 490.5
T_AB = 359.0 N
Then: T_AC = 1.2247(359.0) = 439.7 N
Answer: T_AB = 359 N, T_AC = 440 N
Step 6: Verification -- check SFy: 359 sin(30) + 440 sin(45) - 490.5 = 179.5 + 311.1 - 490.5 = 0.1 (approximately zero, rounding).
Key insight: If you have 2 unknowns, you need exactly 2 independent equations. The two equilibrium equations (SFx = 0 and SFy = 0) give you exactly that.
Tip: If one equation has only one unknown, solve it first. This often happens when:
- A force is purely horizontal (then SFy might have only one unknown from other forces)
- A force is purely vertical (then SFx might have only one unknown)
Tip: You can choose any orientation for your x-y axes! Sometimes rotating the axes to align with a surface or cable simplifies the algebra. For example, on an inclined plane, aligning x along the slope means the normal force has no x-component.
- Draw the 3D FBD
- Express all forces as Cartesian vectors: F = Fxi + Fyj + Fz*k
- For forces along cables/lines, use the position vector method:
F = F * u = F * (r / |r|) - Apply equilibrium:
SFx = 0, SFy = 0, SFz = 0 - Solve the system of three equations for up to 3 unknowns
Problem (similar to Problem 3-43): A 200 N crate is supported by three cables meeting at point A located at (0, 3, 0). The cables connect to:
- Point B at (2, 0, 0)
- Point C at (-1, 0, 2)
- Point D at (-1, 0, -1)
Find the tension in each cable.
Solution:
Step 1: The weight acts downward at A: W = -200j N
Step 2: Find unit vectors from A to each anchor point.
Cable AB: A(0,3,0) to B(2,0,0)
r_AB = (2-0)i + (0-3)j + (0-0)k = 2i - 3j
|r_AB| = sqrt(4 + 9) = sqrt(13) = 3.606
u_AB = (2i - 3j) / 3.606 = 0.5547i - 0.8321j
Cable AC: A(0,3,0) to C(-1,0,2)
r_AC = (-1)i + (-3)j + (2)k
|r_AC| = sqrt(1 + 9 + 4) = sqrt(14) = 3.742
u_AC = (-0.2673i - 0.8018j + 0.5345k)
Cable AD: A(0,3,0) to D(-1,0,-1)
r_AD = (-1)i + (-3)j + (-1)k
|r_AD| = sqrt(1 + 9 + 1) = sqrt(11) = 3.317
u_AD = (-0.3015i - 0.9045j - 0.3015k)
Step 3: Express forces as Cartesian vectors.
F_AB = T_AB * u_AB = T_AB(0.5547i - 0.8321j)
F_AC = T_AC * u_AC = T_AC(-0.2673i - 0.8018j + 0.5345k)
F_AD = T_AD * u_AD = T_AD(-0.3015i - 0.9045j - 0.3015k)
W = -200j
Step 4: Equilibrium equations.
SFx = 0 (i-components):
0.5547 T_AB - 0.2673 T_AC - 0.3015 T_AD = 0 ... (1)
SFy = 0 (j-components):
-0.8321 T_AB - 0.8018 T_AC - 0.9045 T_AD - 200 = 0 ... (2)
SFz = 0 (k-components):
0.5345 T_AC - 0.3015 T_AD = 0 ... (3)
Step 5: Solve the system.
From (3): T_AD = (0.5345/0.3015) T_AC = 1.773 T_AC
Substitute into (1):
0.5547 T_AB - 0.2673 T_AC - 0.3015(1.773 T_AC) = 0
0.5547 T_AB - 0.2673 T_AC - 0.5345 T_AC = 0
0.5547 T_AB = 0.8018 T_AC
T_AB = 1.4455 T_AC
Substitute both into (2):
-0.8321(1.4455 T_AC) - 0.8018 T_AC - 0.9045(1.773 T_AC) = 200
-1.2029 T_AC - 0.8018 T_AC - 1.6037 T_AC = 200
-3.6084 T_AC = 200
T_AC = -55.42 N
The negative sign indicates our assumed direction was wrong -- but in this constructed example, the geometry may not support equilibrium with all positive tensions. In actual exam problems, the geometry is set up so all tensions are positive. The solution method is the same regardless.
Key takeaway: The systematic process is always:
- Position vectors -> unit vectors -> force expressions
- Sum components in each direction = 0
- Solve the 3x3 system of equations
When a spring is part of a particle equilibrium problem:
Spring force: F_spring = k * s
Where s is the deformation:
s = l - l_0 (positive if stretched, negative if compressed)
- l = current (deformed) length
- l_0 = natural (undeformed) length
- k = spring constant
The spring force direction:
- If stretched: pulls toward the spring (tension)
- If compressed: pushes away from the spring (compression)
Common problem type: Find the deformed position of a particle attached to springs and cables under a load.
- Always start with an FBD - Never skip this step
- Assume tension directions - Assume all cables are in tension (pulling away from particle). If you get a negative answer, the assumption was wrong
- In 3D, use the position vector method - It is systematic and avoids angle measurement errors
- Count unknowns vs. equations - In 2D you have 2 equations, so you can find 2 unknowns. In 3D you have 3 equations for 3 unknowns
- For pulleys, the tension is the same on both sides of a frictionless pulley
- Missing a force on the FBD - Especially weight! Always check if weight is acting
- Wrong direction of cable tension - Tension always pulls AWAY from the particle along the cable
- Sign errors - Carefully assign positive/negative based on your coordinate system
- Angle from wrong reference - Double-check if angle is from horizontal or vertical
- Forgetting to convert units - If weight is in kg, multiply by g to get Newtons. If given in lb, it is already a force
- In 3D, using wrong point order in position vector - r_AB = B - A, not A - B
This is the largest chapter, covering Modules 4, 5, and 6 (Sections 4.1-4.9).
The moment of a force about a point O is the tendency of the force to cause rotation about that point.
Scalar definition:
M_O = F * d
where:
- F = magnitude of the force
- d = perpendicular distance from point O to the line of action of the force
IMPORTANT: d is the PERPENDICULAR distance, not the distance from O to the point of application.
Direction (2D): The moment is either clockwise (CW) or counterclockwise (CCW). By convention:
- CCW = positive (follows right-hand rule about z-axis)
- CW = negative
Varignon's Theorem states that the moment of a force about a point equals the sum of the moments of the force's components about that point.
This is incredibly useful because it avoids finding the perpendicular distance d:
M_O = Fx * dy - Fy * dx
where dx, dy are the x and y distances from O to the point where the force is applied.
Sign convention for Varignon's Theorem (CRITICAL):
- A component that would cause CCW rotation about O is positive
- A component that would cause CW rotation about O is negative
- Determine this by inspection for each component separately
Problem: A 500 N force acts at point A located 3 m to the right and 2 m above point O. The force is directed at 40 degrees above the horizontal (to the right). Find the moment about O.
Solution using Varignon's Theorem:
Force components:
Fx = 500 cos(40) = 383.0 N (to the right)
Fy = 500 sin(40) = 321.4 N (upward)
Position of A relative to O: dx = 3 m (right), dy = 2 m (up)
Moment contributions (check rotation direction of each component about O):
-
Fx (rightward) at a point 2 m above O: This would cause clockwise rotation about O
- Contribution: -Fx * dy = -(383.0)(2) = -766.0 N*m
-
Fy (upward) at a point 3 m right of O: This would cause counterclockwise rotation about O
- Contribution: +Fy * dx = +(321.4)(3) = +964.3 N*m
Total moment:
M_O = -766.0 + 964.3 = +198.3 N*m = 198 N*m (CCW)
The moment about point O using the cross product:
M_O = r x F
where:
- r = position vector from O to ANY point on the line of action of F
- F = the force vector
Cross product formula (determinant method):
M_O = | i j k |
| rx ry rz |
| Fx Fy Fz |
M_O = (ry*Fz - rz*Fy)i - (rx*Fz - rz*Fx)j + (rx*Fy - ry*Fx)k
Memory aid for the cross product:
i component: ry*Fz - rz*Fy (cover the i column, take determinant)
j component: -(rx*Fz - rz*Fx) (cover the j column, NEGATE the determinant)
k component: rx*Fy - ry*Fx (cover the k column, take determinant)
WATCH THE NEGATIVE SIGN ON THE j COMPONENT! This is the most common error.
Problem: Force F = 50i + 80j - 30k N acts at point A(2, -1, 3) m. Find the moment about the origin O.
Solution:
Position vector from O to A:
r = 2i - 1j + 3k m
Cross product M_O = r x F:
M_O = | i j k |
| 2 -1 3 |
| 50 80 -30 |
i-component: (-1)(-30) - (3)(80) = 30 - 240 = -210 j-component: -[(2)(-30) - (3)(50)] = -[-60 - 150] = -(-210) = +210 k-component: (2)(80) - (-1)(50) = 160 + 50 = 210
M_O = -210i + 210j + 210k N*m
Magnitude: |M_O| = sqrt(210^2 + 210^2 + 210^2) = 210sqrt(3) = 363.7 Nm
If you have programmed your TI-84 with the cross product program from the instructor's video, you can enter the six components (rx, ry, rz, Fx, Fy, Fz) and get the result directly. This saves significant time on exams.
Manual verification method: Even if you use the program, do a quick sanity check on at least one component to make sure you entered the values correctly.
Sometimes you need the moment about a specific axis (not just a point). The moment about an axis a is the component of M_O that acts along that axis.
Formula (Triple Scalar Product):
M_a = u_a . (r x F)
where:
- u_a = unit vector along the axis
- r = position vector from any point on the axis to any point on the line of action of F
- F = the force
Determinant form:
M_a = | u_ax u_ay u_az |
| rx ry rz |
| Fx Fy Fz |
This evaluates to a scalar (the magnitude of the moment about the axis).
Problem: Force F = 100i + 200j - 50k N acts at point A(1, 2, -1). Find the moment about the y-axis.
Solution:
The y-axis has unit vector u_y = j = 0i + 1j + 0k.
Choose any point on the y-axis (e.g., the origin O) for the position vector:
r_OA = 1i + 2j - 1k
Triple scalar product:
M_y = | 0 1 0 |
| 1 2 -1 |
| 100 200 -50|
Expanding along the first row:
M_y = 0[(2)(-50) - (-1)(200)] - 1[(1)(-50) - (-1)(100)] + 0[(1)(200) - (2)(100)]
M_y = 0 - 1[-50 + 100] + 0
M_y = -1[50]
M_y = -50 N*m
The negative sign means the moment acts in the -j direction (i.e., opposite to the positive y-axis direction).
When the axis is one of the coordinate axes, the moment about that axis is simply the corresponding component of M_O:
- M about x-axis = (M_O)_x = the i-component of r x F
- M about y-axis = (M_O)_y = the j-component of r x F
- M about z-axis = (M_O)_z = the k-component of r x F
This is because u_x = i, u_y = j, u_z = k, and the dot product with M_O simply extracts that component.
A couple consists of two forces that are:
- Equal in magnitude
- Opposite in direction
- Separated by a perpendicular distance d
The moment of a couple:
M = F * d (scalar, 2D)
M = r x F (vector, 3D)
where r is the position vector from any point on one force to any point on the other force.
- A couple produces ROTATION ONLY, no translation (the net force is zero: F + (-F) = 0)
- A couple moment is a FREE VECTOR -- it has the same moment about EVERY point in space
- Couples can be moved anywhere without changing their effect
- Couples can be added by vector addition of the couple moments
- A couple can be represented as a moment vector (curved arrow or double-headed arrow)
Couples are essential for:
- Representing the effect of moving a force to a different point (equivalent systems)
- Describing torques applied to shafts
- Simplifying force systems
Two force systems are equivalent if they produce the same external effect (same resultant force and same resultant moment about any point).
When you move a force F from point A to point B, you must add a couple moment to maintain equivalence:
Original: Force F at point A
Equivalent: Force F at point B + Couple moment M = r_BA x F
where r_BA is the position vector from B to A (from the NEW point to the OLD point).
Think of it this way: Moving the force to B changes the moment about B. The couple M compensates for that change.
Any system of forces and couples can be reduced to a single resultant force and a single resultant couple moment at a chosen point O:
F_R = SUM(F_i) (vector sum of all forces)
M_R_O = SUM(M_i) + SUM(r_i x F_i) (sum of all couples plus moments of forces about O)
Problem (similar to 4-120): Three forces act on a beam:
- F1 = 200 N upward at x = 1 m from point A
- F2 = 500 N at 60 degrees below horizontal at x = 3 m from A
- F3 = 100 N to the right at x = 5 m from A
Find the equivalent resultant force and its location (distance from A) such that the system can be replaced by a single force.
Solution:
Step 1: Resolve all forces.
F1: F1x = 0, F1y = +200 N
F2: F2x = 500 cos(60) = 250 N, F2y = -500 sin(60) = -433.0 N (below horizontal)
F3: F3x = 100 N, F3y = 0
Step 2: Resultant force.
FR_x = 0 + 250 + 100 = 350 N
FR_y = 200 + (-433.0) + 0 = -233.0 N
FR = sqrt(350^2 + 233^2) = sqrt(122500 + 54289) = sqrt(176789) = 420.5 N
theta = tan^(-1)(-233 / 350) = -33.6 degrees (below horizontal, quadrant IV)
Step 3: Find the location d from A where the single resultant force produces the same moment about A.
Moment about A from the original system:
M_A = F1y(1) + F2x(0) + F2y(3) + F3x(0) + F3y(5)
Wait -- we need to be more careful. Using Varignon's theorem about point A, considering each force component's moment:
Point A is at x = 0, y = 0 (left end of beam, assuming beam is along x-axis):
From F1 at x=1: F1y(+200) acts at 1 m from A. 200 N upward at x=1 causes CCW rotation.
M_A from F1 = +200(1) = +200 N*m
From F2 at x=3: F2y(-433) at x=3 causes CW rotation. F2x(250) at x=3 has no moment arm (if beam is horizontal and F2x is along beam).
Actually, if the beam is along the x-axis and all forces act on the beam, then:
- Vertical forces (Fy) produce moments about A: M = Fy * x_distance
- Horizontal forces (Fx) at height y=0 on the beam produce no moment about A (moment arm = 0)
M_A from F2y = (-433.0)(3) = -1299.0 N*m
M_A from F2x = 250(0) = 0 (force is along the beam, no perpendicular distance)
M_A from F3 = 0 (horizontal force on horizontal beam at same height)
Total M_A = 200 - 1299 = -1099 N*m
Step 4: For the single resultant force at distance d from A:
M_A = FR_y * d (only vertical component creates moment about A on horizontal beam)
-1099 = (-233.0) * d
d = 1099 / 233.0 = 4.72 m from A
Note: The actual problem 4-120 has answer FR = 356 N at theta = -51.8 degrees, d = 3.32 m. The approach is identical; the specific numbers differ.
A distributed load is a force spread over a length (in 2D) or area (in 3D). It is described by a load intensity function w(x) with units of force per length (N/m or lb/ft).
The distributed load can be replaced by a single resultant force:
Magnitude:
F_R = integral of w(x) dx = area under the load curve
Location (x-bar):
x_bar = [integral of x * w(x) dx] / [integral of w(x) dx]
= centroid of the load area
Uniform (rectangular) load:
w(x) = w_0 (constant)
F_R = w_0 * L (area of rectangle)
x_bar = L/2 (centroid at midpoint)
Triangular load (zero at one end, maximum at other):
w(x) = w_0 * x / L (linearly increasing from 0 to w_0)
F_R = (1/2) * w_0 * L (area of triangle)
x_bar = (2/3) * L (from the zero end, or L/3 from the maximum end)
CRITICAL: The centroid of a triangle is at 1/3 of the base from the larger end (or 2/3 from the smaller end / vertex).
Trapezoidal load (different values at each end): Split into a rectangle + triangle, or use the trapezoidal formulas:
w(x) = w_1 + (w_2 - w_1)(x/L) (from w_1 at x=0 to w_2 at x=L)
Split into:
- Rectangle: w_1 * L, centered at L/2
- Triangle: (w_2 - w_1) * L / 2, centered at 2L/3 from the w_1 end
F_R = w_1*L + (w_2 - w_1)*L/2 = L(w_1 + w_2)/2
x_bar = [w_1*L*(L/2) + (w_2-w_1)*L/2*(2L/3)] / F_R
Problem: A beam 6 m long has a triangular distributed load that increases from 0 at the left end (A) to 900 N/m at the right end (B). Find the resultant force and its location from A.
Solution:
Step 1: Resultant magnitude.
F_R = (1/2) * w_max * L = (1/2)(900)(6) = 2700 N
Step 2: Location from A (the zero-load end). The centroid of a triangle is at 2/3 of the base from the vertex (zero end):
x_bar = (2/3) * L = (2/3)(6) = 4.0 m from A
Equivalently, x_bar = L/3 from the maximum end: (1/3)(6) = 2.0 m from B, which is 4.0 m from A. Consistent.
Answer: F_R = 2700 N acting at 4.0 m from A (or 2.0 m from B).
For complex load distributions, break them into simpler shapes (rectangles + triangles), find the resultant of each shape, then combine:
- Find F_R1, x_bar1 for shape 1
- Find F_R2, x_bar2 for shape 2
- Total resultant: F_R = F_R1 + F_R2
- Location: x_bar = (F_R1 * x_bar1 + F_R2 * x_bar2) / F_R
- For 2D moments, use Varignon's Theorem -- it avoids the perpendicular distance calculation
- For 3D moments, use the cross product -- it is systematic and handles any geometry
- The cross product is NOT commutative: r x F is NOT equal to F x r (they are negatives of each other)
- For moment about an axis, the triple scalar product gives a SCALAR result
- Use your TI-84 cross product program to save time and reduce errors
- For distributed loads, always identify the shape (rectangle, triangle, or composite)
- The couple moment is a free vector -- it can be moved to any point without additional couples
- WRONG SIGN ON THE j-COMPONENT of the cross product -- the j-component has a NEGATIVE sign in front of the 2x2 determinant. This is the single most common error
- Wrong r vector for moment -- r must go FROM the moment point TO a point on the line of action of F. The direction matters!
- Confusing moment of a force with a couple -- A single force about a point creates a moment. A couple is two equal/opposite forces
- Wrong centroid location for triangular loads -- It is at 1/3 from the MAXIMUM end, NOT from the minimum end
- Forgetting the couple when moving a force -- If you move F from A to B, you MUST add M = r_BA x F
- Using d = straight-line distance instead of perpendicular distance in the scalar moment formula M = Fd
- In equivalent systems, forgetting to include existing couple moments -- When reducing a system, include ALL couples (both given couples and those from force moments)
This chapter covers Modules 7 and 8 (Sections 5.1-5.6) and builds on everything from Chapters 2-4.
In Chapter 3, we studied particle equilibrium (all forces concurrent at a point, no moments).
In Chapter 5, we study rigid body equilibrium. Forces are NOT all concurrent, so we must consider moments as well as forces.
Equilibrium of a rigid body requires:
2D: SFx = 0, SFy = 0, SM_O = 0 (3 equations --> 3 unknowns)
3D: SFx = 0, SFy = 0, SFz = 0, (6 equations --> 6 unknowns)
SMx = 0, SMy = 0, SMz = 0
This is the most critical skill in statics. A wrong FBD means a wrong answer every time.
- Draw the outline of the body SEPARATE from its supports and connections
- Do not include supports in the drawing -- only the body itself
Three categories:
- Applied forces/loads -- given in the problem (concentrated forces, distributed loads, couples)
- Weight -- W = mg, acting at the center of gravity (usually the geometric center for uniform bodies)
- Support reactions -- forces and moments exerted by supports (see tables below)
- Applied forces: draw in the given direction
- Weight: always downward
- Reactions: draw in the assumed direction (if you guess wrong, the answer will be negative, which is fine)
- Label all forces with magnitudes (if known) or variables (Ax, Ay, MA, etc.)
- Label all distances and dimensions
- Label all angles
This table is ESSENTIAL. You must know these for the exam.
+-------------------+---+--------------------------------------------------+
| Support Type | # | Reactions |
+-------------------+---+--------------------------------------------------+
| Roller/Rocker | 1 | One force NORMAL to the support surface |
| /\ | | (perpendicular to the surface the roller sits on)|
| / \ | | |
| ------ | | |
+-------------------+---+--------------------------------------------------+
| Pin (Hinge) | 2 | Two force components: Ax (horizontal) |
| o | | and Ay (vertical) |
| /|\ | | |
+-------------------+---+--------------------------------------------------+
| Fixed Support | 3 | Two force components: Ax, Ay |
| (Built-in/Wall) | | Plus one moment reaction: MA |
| |=== | | |
+-------------------+---+--------------------------------------------------+
| Smooth Surface | 1 | One force NORMAL to the surface at the |
| | | point of contact |
+-------------------+---+--------------------------------------------------+
| Cable/Rope | 1 | Tension along the cable (AWAY from the body) |
| ~~~~ | | Cable CANNOT push, only pull |
+-------------------+---+--------------------------------------------------+
| Link/Strut | 1 | Force along the line of the link |
| (two-force member)| | (can be tension or compression) |
+-------------------+---+--------------------------------------------------+
| Smooth Pin in | 1 | Force perpendicular to the slot |
| Slot | | |
+-------------------+---+--------------------------------------------------+
| Pin on Collar | 1 | Force perpendicular to the rod |
| (on smooth rod) | | |
+-------------------+---+--------------------------------------------------+
| Fixed Collar | 2 | Force perpendicular to rod + couple moment |
| (on smooth rod) | | (prevents rotation and transverse movement) |
+-------------------+---+--------------------------------------------------+
Textbook Reference: These are from Table 5-1, Hibbeler Statics 14th Ed (pp. 210-211). The downloadable PDF with diagrams is on Canvas: Formula_sheet_statics_reactions_2D_3D.pdf
- A roller allows movement parallel to the surface, so it can only push perpendicular to the surface
- A pin prevents all translation but allows rotation, so it has two force reactions but NO moment reaction
- A fixed support prevents all translation AND rotation, so it has two forces AND one moment
- Each constrained degree of freedom = one reaction
- 2D has 3 DOF (x-translation, y-translation, rotation)
- Roller constrains 1 DOF -> 1 reaction
- Pin constrains 2 DOF -> 2 reactions
- Fixed constrains 3 DOF -> 3 reactions
+---------------------+---+--------------------------------------------------+
| Support Type | # | Reactions |
+---------------------+---+--------------------------------------------------+
| Ball and Socket | 3 | Three force components: Ax, Ay, Az |
| | | (prevents translation in all 3 directions) |
| | | (allows rotation about all 3 axes) |
+---------------------+---+--------------------------------------------------+
| Single Journal |4-5| Depends on configuration: |
| Bearing | | 2 forces + 2 moments (if axial thrust free) |
| | | 3 forces + 2 moments (if axial thrust present) |
+---------------------+---+--------------------------------------------------+
| Thrust Bearing | 5 | Three force components + two moment components |
| | | (Ax, Ay, Az, MAx, MAz - no moment about shaft |
| | | axis since bearing allows shaft rotation) |
+---------------------+---+--------------------------------------------------+
| Fixed Support | 6 | Three forces: Ax, Ay, Az |
| (Built-in) | | Three moments: MAx, MAy, MAz |
| | | (prevents ALL translation and ALL rotation) |
+---------------------+---+--------------------------------------------------+
| Ball on Smooth | 1 | One force normal to the surface |
| Surface | | |
+---------------------+---+--------------------------------------------------+
| Cable | 1 | One tension force along the cable |
+---------------------+---+--------------------------------------------------+
| Single Hinge | 5 | Three forces + two moments |
| (3D) | | (allows rotation about hinge axis) |
+---------------------+---+--------------------------------------------------+
| Single Smooth Pin | 5 | Three forces + two moments |
| (3D) | | (couple moments not applied if supported |
| | | elsewhere - see textbook examples) |
+---------------------+---+--------------------------------------------------+
| Journal Bearing | 5 | Two forces + three couple-moment components |
| (square shaft) | | (prevents rotation that round shaft allows) |
+---------------------+---+--------------------------------------------------+
Textbook Reference: Table 5-2, Hibbeler Statics 14th Ed (pp. 246-247). KEY NOTE from textbook: For journal bearings, thrust bearings, smooth pins, and hinges: "The couple moments are generally NOT applied if the body is supported elsewhere." This is a common exam detail! The downloadable PDF with diagrams is on Canvas: Formula_sheet_statics_reactions_2D_3D.pdf
- 3D has 6 DOF (3 translations + 3 rotations)
- Fixed support constrains all 6 -> 6 reactions
- Ball and socket constrains 3 translations -> 3 reactions (all rotations free)
- Each constrained DOF = one reaction
- Draw the FBD (4-step method from Section 7.2)
- Check: Number of unknowns should be <= 3 (for a single body in 2D)
- Write equilibrium equations:
SFx = 0: Sum of all horizontal forces = 0 SFy = 0: Sum of all vertical forces = 0 SM_O = 0: Sum of all moments about point O = 0 - Choose point O wisely! Pick a point where the most unknowns pass through (so they drop out of the moment equation)
- Solve the equations for the unknowns
- Check your answer -- substitute back, verify all three equations are satisfied
This is a strategic decision that can make the algebra much easier:
- Choose a point where unknown forces intersect -- forces passing through the moment point create no moment, so they disappear from the moment equation
- Pin supports are great moment points -- both Ax and Ay pass through the pin
- You can write moment equations about MULTIPLE points if it helps avoid simultaneous equations
Alternative sets of equilibrium equations (all valid for 2D):
- SFx = 0, SFy = 0, SM_A = 0 (standard)
- SFx = 0, SM_A = 0, SM_B = 0 (where A and B don't share a line perpendicular to x)
- SM_A = 0, SM_B = 0, SM_C = 0 (where A, B, C are not collinear)
Problem (similar to 5-22): A horizontal beam AB is 8 m long. It has a pin support at A and a roller support at B. A 10 kN downward force acts at 3 m from A, and a 6 kN downward force acts at 6 m from A. Find the support reactions.
10 kN 6 kN
| |
v v
A--+--+--+--+--+--+--+--B
^ 3m 6m ^
| |
Pin Roller
(Ax,Ay) (By)
Solution:
Step 1: FBD shows:
- At A: Ax (horizontal), Ay (vertical) -- pin support
- At B: By (vertical, upward) -- roller support
- Applied: 10 kN down at x=3m, 6 kN down at x=6m
Step 2: Three unknowns (Ax, Ay, By) and three equations. Solvable!
Step 3: Write equations.
SFx = 0:
Ax = 0
(No horizontal applied forces, so the horizontal reaction is zero.)
SM_A = 0 (choosing A eliminates Ax and Ay from the moment equation):
-10(3) - 6(6) + By(8) = 0
-30 - 36 + 8By = 0
8By = 66
By = 8.25 kN
SFy = 0:
Ay - 10 - 6 + By = 0
Ay - 10 - 6 + 8.25 = 0
Ay = 7.75 kN
Answer: Ax = 0, Ay = 7.75 kN (up), By = 8.25 kN (up)
Verification -- SM_B = 0:
Ay(8) - 10(8-3) - 6(8-6) = 7.75(8) - 10(5) - 6(2) = 62 - 50 - 12 = 0 ✓
Problem: A cantilever beam (fixed at A, free at B) has length L = 4 m. A uniform distributed load w = 3 kN/m acts over the entire length. Find the reactions at A.
Solution:
Step 1: FBD at the fixed support A: Ax, Ay, MA (three reactions).
Replace the distributed load with its resultant:
F_R = w * L = 3(4) = 12 kN (downward)
Acting at x = L/2 = 2 m from A (centroid of rectangle)
Step 2: Equilibrium equations.
SFx = 0: Ax = 0
SFy = 0: Ay - 12 = 0, so Ay = 12 kN (upward)
SM_A = 0: MA - 12(2) = 0, so MA = 24 kN*m (counterclockwise)
Answer: Ax = 0, Ay = 12 kN (up), MA = 24 kN*m (CCW)
Problem (similar to 5-32): A beam AC is supported by a pin at A and a cable BC. A 5 kip force acts at the midpoint of AC at 30 degrees below horizontal. AC is 10 ft long and horizontal. Cable BC connects from C to point B located 6 ft directly above A.
B
|\
| \ cable
6 | \
| \
A----+----C
^ 5m ^
Pin 5 kip at 30 deg below horizontal
Solution:
Step 1: FBD
- At A: Ax, Ay (pin)
- Cable BC: tension T_BC along cable from C toward B
- Applied: 5 kip at 30 degrees below horizontal at midpoint (x = 5 ft from A)
Step 2: Find the direction of cable BC.
B is at (0, 6), C is at (10, 0)
r_CB = (0-10)i + (6-0)j = -10i + 6j
|r_CB| = sqrt(100 + 36) = sqrt(136) = 11.662 ft
u_CB = -0.8575i + 0.5145j
So T_BC has components:
T_BC_x = -0.8575 * T_BC (to the left)
T_BC_y = +0.5145 * T_BC (upward)
Applied force at midpoint (30 degrees below horizontal):
Fx_applied = 5 cos(30) = 4.330 kip (assume to the right)
Fy_applied = -5 sin(30) = -2.500 kip (downward)
Step 3: Equilibrium.
SM_A = 0 (to eliminate Ax and Ay):
Fy_applied(5) + T_BC_y(10) + T_BC_x(0) + Fx_applied(0) = 0
Wait, let me be more careful. Taking moments about A (counterclockwise positive):
The applied force components at (5, 0):
- Fx_applied = 4.330 at height y=0: no moment about A (force through A's elevation) Actually, Fx (horizontal) at point (5,0) relative to A at (0,0): The moment = Fx * (y-distance) - Fy * (x-distance) Moment from Fx_applied = -4.330 * 0 = 0 (horizontal force at same height as A) Moment from Fy_applied = -(-2.500) * 5 ... no, let me use Varignon's:
For moment about A = (0,0):
- Fy_applied = -2.500 kip at x = 5: creates CW moment = (-2.500)(5) = -12.5 kip*ft Wait: -2.500 (downward) at x = 5 to the right of A creates CLOCKWISE rotation. CW is negative: contribution = -2.500 * 5 = -12.5 ... Actually:
Let me use the sign convention carefully. For Varignon's theorem about A:
- A force component Fy at horizontal distance x from A: M = Fy * x (positive Fy at positive x gives CCW)
- A force component Fx at vertical distance y from A: M = -Fx * y (positive Fx at positive y gives CW)
Applied force at (5, 0):
M from Fy_applied = (-2.500)(5) = -12.5 kip*ft
M from Fx_applied = -(4.330)(0) = 0
Cable force at C(10, 0):
M from T_BC_y = (0.5145 T_BC)(10) = 5.145 T_BC kip*ft
M from T_BC_x = -(-0.8575 T_BC)(0) = 0
SM_A = 0:
-12.5 + 5.145 T_BC = 0
T_BC = 12.5 / 5.145 = 2.429 kip
SFx = 0:
Ax + 4.330 + (-0.8575)(2.429) = 0
Ax + 4.330 - 2.083 = 0
Ax = -2.247 kip (i.e., 2.247 kip to the left)
SFy = 0:
Ay + (-2.500) + (0.5145)(2.429) = 0
Ay - 2.500 + 1.250 = 0
Ay = 1.250 kip (upward)
The actual Problem 5-32 answers are F_BC = 1.82 kip, F_A = 2.06 kip, which would come from different geometry/loading. The method is the same.
A two-force member is a body with forces applied at ONLY two points (no distributed loads, no couples, no other forces).
For a two-force member in equilibrium, the two forces must be:
- Equal in magnitude
- Opposite in direction
- Collinear (both act along the line connecting the two application points)
Recognizing a two-force member simplifies the problem because you know the force direction immediately -- it must be along the line connecting the two points.
Examples of two-force members:
- A straight link pin-connected at both ends with no other loads
- A cable (always a two-force member)
- A hydraulic cylinder connecting two pin joints
How to identify:
- The member has forces (including reactions) at exactly 2 points
- No distributed loads act on it
- No couples act on it
- Its weight is either negligible or not considered
A three-force member is a body with forces applied at exactly three points.
For a three-force member in equilibrium, the three forces must be either:
- Concurrent (all three lines of action intersect at a single point), OR
- All parallel
This property can help determine the direction of unknown forces. If you know two of the three force lines of action, their intersection point tells you where the third force must also pass through.
Six independent equilibrium equations:
SFx = 0, SFy = 0, SFz = 0
SMx = 0, SMy = 0, SMz = 0
This means you can solve for up to 6 unknowns.
- Draw the 3D FBD with all support reactions
- Express all forces as Cartesian vectors
- Choose a moment point (often a support with many reactions to eliminate unknowns)
- Write force equilibrium (3 scalar equations)
- Write moment equilibrium (3 scalar equations from the vector moment equation)
- Solve the system of 6 equations
Problem (similar to 5-67): A horizontal plate is supported by three vertical cables at points A, B, and C. A force of 500 N acts downward at a point on the plate. The plate weighs 200 N. Find the tension in each cable.
This type of problem often involves:
- SFz = 0 (vertical equilibrium): T_A + T_B + T_C = 500 + 200 = 700 N
- SMx = 0 (moment about x-axis): involves T_A, T_B, T_C at their y-coordinates
- SMy = 0 (moment about y-axis): involves T_A, T_B, T_C at their x-coordinates
The system of 3 equations (from the 3 cable tensions as unknowns) is solved simultaneously.
From the given answers to Problem 5-67: T_A = 7.27 kN, T_B = 16.5 kN, T_C = 14.8 kN.
- Use the moment equation about a point where multiple unknowns intersect - this reduces the number of unknowns in that equation
- For plates supported by cables, take moments about an axis connecting two cable attachment points - this eliminates two tensions from that moment equation
- Express each cable tension using the unit vector method from Chapter 2
- Use the cross product for computing moments in 3D
- Be systematic - write out all six equations even if some are trivial
A problem is statically determinate when the number of unknowns equals the number of independent equilibrium equations:
- 2D: 3 unknowns with 3 equations
- 3D: 6 unknowns with 6 equations
If there are more unknowns than equations, the problem is statically indeterminate. You cannot solve it with equilibrium alone (need additional equations from material properties/deformation). These will NOT be on the ECE 211 midterm.
A body is improperly constrained if:
- The support reactions are concurrent (all pass through one point) -- body can rotate about that point
- The support reactions are all parallel -- body can translate perpendicular to the forces
- There are fewer reactions than equations (partial constraints)
Improper constraints lead to an unstable body that cannot be in equilibrium under general loading.
- The FBD is EVERYTHING -- spend extra time making it correct. An error here cascades through all calculations
- Choose moment points strategically -- a good choice can turn a system of 3 equations into sequential single-unknown equations
- For a beam with a pin and roller, take moments about the pin to find the roller reaction directly, then use force equilibrium for the pin reactions
- For distributed loads, replace with the resultant force at the centroid BEFORE writing equilibrium equations
- Always check your answer by verifying that an unused equilibrium equation is satisfied
- Watch for two-force members in trusses and frames -- they simplify the analysis
- In 3D, be extra careful with signs -- draw the coordinate system clearly and stick with it
- Forgetting a reaction force on the FBD -- especially the moment reaction at a fixed support
- Wrong number of reactions -- using 2 reactions for a fixed support (it needs 3 in 2D)
- Putting the roller reaction in the wrong direction -- it must be PERPENDICULAR to the surface
- Taking moments about the wrong point -- make sure you know the distance from the moment point to each force
- Wrong moment arm -- the moment arm is the PERPENDICULAR distance from the point to the LINE OF ACTION
- Sign errors in moment equations -- always determine CW vs CCW for each force
- Not replacing distributed loads before writing equilibrium equations
- Forgetting the weight of the body -- if the problem gives the mass, include W = mg
- In 3D, forgetting a moment reaction -- a fixed support in 3D has 6 reactions
Use this checklist to verify you have worked through all review problems. The answers are provided for self-checking.
| Problem | Topic | Key Answer(s) | Completed? |
|---|---|---|---|
| 3-21 | 2D particle equilibrium | (See video for full solution) | [ ] |
| 3-27 | 2D particle equilibrium | m = 26.7 kg | [ ] |
| 3-43 | 3D particle equilibrium | F_AD = 763 N, F_AC = 392 N, F_AB = 523 N | [ ] |
| Problem | Topic | Key Answer(s) | Completed? |
|---|---|---|---|
| 4-120 | Equivalent Systems | FR = 356 N, theta = -51.8 deg, d = 3.32 m (up from A) | [ ] |
| 4-156 | Distributed Load | (See solution video) | [ ] |
| Problem | Topic | Key Answer(s) | Completed? |
|---|---|---|---|
| 5-22 | 2D rigid body equilibrium | Bx = 1.86 kN, By = 8.78 kN, N_A = 3.71 kN | [ ] |
| 5-32 | 2D with cable/two-force member | F_BC = 1.82 kip, F_A = 2.06 kip | [ ] |
| 5-67 | 3D rigid body equilibrium | T_A = 7.27 kN, T_B = 16.5 kN, T_C = 14.8 kN | [ ] |
-
Instructor Review Videos (Fanny Silvestri on YouTube):
- Ch2: Adding 3D Vectors, Position Vector and Unit Vector, Dot Product
- Ch3: Statics 3-21, 3-27, 3-43
- Ch4: Moments in 3D using Cross Product, TI-84 Cross Product Programming, Equivalent Systems, Distributed Loads (4-156)
- Ch5: Statics 5-22, 5-32, 5-67
-
EngineeringStatics.org -- Chapters 3, 4, 5 for additional lectures and worked examples
For each problem:
- Attempt the problem without looking at the answer
- Compare your final answer to the values above
- If wrong, identify where you went wrong
- Redo the problem from scratch until you get the correct answer
- If you cannot get the correct answer, watch the corresponding instructor video
This section compiles all common mistakes from each chapter into a single reference.
-
Unit errors -- mixing SI and US Customary units in the same equation. Always check that all quantities are in consistent units before computing.
-
Calculator in wrong mode -- make sure your TI-84 is in DEGREE mode (not radian) if your angles are in degrees. Press [MODE] and check. This causes silently wrong answers.
-
Premature rounding -- keep at least 4 significant figures in intermediate calculations. Only round to 3 significant figures for the final answer.
-
Not reading the problem carefully -- missing a given force, using the wrong dimension, or answering the wrong question.
-
Skipping the FBD -- this is the number one cause of errors. ALWAYS draw the FBD, even if the problem seems simple.
-
Swapping sin and cos -- If the angle is from the x-axis: Fx = F cos(theta), Fy = F sin(theta). If from the y-axis, they swap.
-
Wrong quadrant for arctan -- The calculator's tan^(-1) gives values in (-90, 90). You must adjust for the correct quadrant based on the signs of the components.
-
Position vector direction -- r_AB = B_coordinates - A_coordinates (FROM A, TO B).
-
Unit vector not normalized -- Always divide by the magnitude to get a true unit vector.
-
Dot product gives scalar, not vector -- Do not attach i, j, k to a dot product result.
-
Missing weight -- If the problem gives mass, you must calculate W = mg and include it in the FBD.
-
Wrong tension direction -- Cable tension acts ALONG the cable, AWAY from the particle (it pulls, never pushes).
-
Sign errors in equilibrium -- Forces in the negative direction must have negative signs in the equations.
-
More unknowns than equations -- In 2D, you can only solve for 2 unknowns. In 3D, only 3. If you have more, you may be missing a constraint or need to isolate a different particle.
-
NEGATIVE SIGN ON j-COMPONENT OF CROSS PRODUCT -- The j-component of r x F has a negative sign in the determinant expansion. This is the most common computational error. The formula is: M = (ryFz - rzFy)i - (rxFz - rzFx)j + (rxFy - ryFx)k
-
Wrong r vector -- r must go FROM the moment point TO a point on the force's line of action.
-
Using straight-line distance instead of perpendicular distance -- In M = Fd, d must be perpendicular to F's line of action.
-
Centroid of triangular load -- The centroid is at 1/3 from the LARGE end (maximum load), or 2/3 from the small end (zero load). Confusing these is very common.
-
Forgetting the couple when moving a force -- When translating a force to a new point, you MUST add a compensating couple moment.
-
Not including given couple moments in the system -- When reducing a force system, do not forget any couples that were part of the original system.
-
Wrong number of reactions at supports -- Roller = 1, Pin = 2, Fixed = 3 (in 2D). Memorize these.
-
Roller reaction in wrong direction -- The roller reaction is ALWAYS perpendicular to the surface the roller sits on. If the surface is inclined, the reaction is inclined too.
-
Forgetting the moment reaction at a fixed support -- Fixed supports resist rotation, so they have a moment reaction MA in addition to Ax and Ay.
-
Wrong moment arm in SM = 0 -- The moment of a force about a point depends on the perpendicular distance from the point to the force's line of action. Use Varignon's theorem to avoid mistakes.
-
Not replacing distributed loads with resultant -- Before writing equilibrium equations, replace any distributed load with its resultant force at the appropriate location.
-
Two-force member not recognized -- If a member has forces at only two points and no other loads, it is a two-force member. The force must act along the line connecting the two points.
-
In 3D, missing a reaction -- Ball and socket = 3, fixed = 6. Count them carefully.
-
Improper moment point choice -- While any point works, a poor choice means more simultaneous equations. Choose points where unknown forces pass through.
Below are fully worked practice problems targeting your weakest topics. Work through each one with pencil and paper before reading the solution.
Problem Statement: A 900-N crate is supported by three cables meeting at point A. The cables are attached to the ceiling at points B, C, and D with coordinates (in meters):
- A = (0, 0, 0) (the ring where cables meet)
- B = (2, 3, 0)
- C = (-1, 3, 2)
- D = (-1, 3, -1.5)
The weight W = 900 N acts downward at point A. Find the tension in each cable.
Solution:
Step 1: Find position vectors from A to each attachment point.
r_AB = B - A = (2, 3, 0) m
r_AC = C - A = (-1, 3, 2) m
r_AD = D - A = (-1, 3, -1.5) m
Step 2: Find the magnitude of each position vector.
|r_AB| = sqrt(2^2 + 3^2 + 0^2) = sqrt(4 + 9 + 0) = sqrt(13) = 3.606 m
|r_AC| = sqrt((-1)^2 + 3^2 + 2^2) = sqrt(1 + 9 + 4) = sqrt(14) = 3.742 m
|r_AD| = sqrt((-1)^2 + 3^2 + (-1.5)^2) = sqrt(1 + 9 + 2.25) = sqrt(12.25) = 3.500 m
Step 3: Find unit vectors along each cable.
u_AB = r_AB / |r_AB| = (2/3.606, 3/3.606, 0/3.606) = (0.5547, 0.8321, 0)
u_AC = r_AC / |r_AC| = (-1/3.742, 3/3.742, 2/3.742) = (-0.2673, 0.8018, 0.5345)
u_AD = r_AD / |r_AD| = (-1/3.500, 3/3.500, -1.5/3.500) = (-0.2857, 0.8571, -0.4286)
Step 4: Write force vectors in terms of unknowns T_B, T_C, T_D.
F_B = T_B * u_AB = T_B (0.5547i + 0.8321j + 0k)
F_C = T_C * u_AC = T_C (-0.2673i + 0.8018j + 0.5345k)
F_D = T_D * u_AD = T_D (-0.2857i + 0.8571j - 0.4286k)
W = -900j
Step 5: Apply equilibrium (SF = 0). Sum components in x, y, z.
SFx = 0: 0.5547*T_B - 0.2673*T_C - 0.2857*T_D = 0 ... (1)
SFy = 0: 0.8321*T_B + 0.8018*T_C + 0.8571*T_D - 900 = 0 ... (2)
SFz = 0: 0*T_B + 0.5345*T_C - 0.4286*T_D = 0 ... (3)
Step 6: Solve the system of 3 equations, 3 unknowns.
From Eq (3):
0.5345*T_C = 0.4286*T_D
T_C = (0.4286/0.5345)*T_D = 0.8019*T_D ... (*)
Substitute (*) into Eq (1):
0.5547*T_B - 0.2673*(0.8019*T_D) - 0.2857*T_D = 0
0.5547*T_B - 0.2143*T_D - 0.2857*T_D = 0
0.5547*T_B - 0.5000*T_D = 0
T_B = 0.5000/0.5547 * T_D = 0.9014*T_D ... (**)
Substitute (*) and (**) into Eq (2):
0.8321*(0.9014*T_D) + 0.8018*(0.8019*T_D) + 0.8571*T_D = 900
0.7501*T_D + 0.6428*T_D + 0.8571*T_D = 900
2.2500*T_D = 900
T_D = 400.0 N
Back-substitute:
T_C = 0.8019 * 400.0 = 320.8 N
T_B = 0.9014 * 400.0 = 360.6 N
>>> FINAL ANSWERS: T_B = 360.6 N, T_C = 320.8 N, T_D = 400.0 N <<<
Check your understanding: In 3D particle equilibrium you always get exactly 3 equations (SFx=0, SFy=0, SFz=0) so you can solve for at most 3 unknowns. Always start by finding unit vectors along each cable/rope.
Problem Statement: A 500-N lamp is supported at point A by cables AB, AC, and strut AD. The coordinates are:
- A = (0, 0, 0)
- B = (1, 2, 2)
- C = (-2, 2, 1)
- D = (0, -3, 0)
Strut AD can only push (compression). Find all forces.
Solution:
Step 1: Position vectors and magnitudes.
r_AB = (1, 2, 2) |r_AB| = sqrt(1 + 4 + 4) = sqrt(9) = 3.000 m
r_AC = (-2, 2, 1) |r_AC| = sqrt(4 + 4 + 1) = sqrt(9) = 3.000 m
r_AD = (0, -3, 0) |r_AD| = sqrt(0 + 9 + 0) = 3.000 m
Step 2: Unit vectors.
u_AB = (1/3, 2/3, 2/3) = (0.3333, 0.6667, 0.6667)
u_AC = (-2/3, 2/3, 1/3) = (-0.6667, 0.6667, 0.3333)
u_AD = (0, -1, 0)
Note: The strut pushes point A, so force from strut on A is directed FROM D toward A. Since D is at (0,-3,0), the direction from D to A is (0,3,0), unit vector = (0,1,0). So F_AD acts in +j direction on point A.
u_DA = (0, 1, 0) (strut pushes A upward)
Step 3: Equilibrium equations.
F_B = T_B(0.3333i + 0.6667j + 0.6667k)
F_C = T_C(-0.6667i + 0.6667j + 0.3333k)
F_D = F_D(0i + 1j + 0k)
W = -500j
SFx = 0: 0.3333*T_B - 0.6667*T_C = 0 ... (1)
SFy = 0: 0.6667*T_B + 0.6667*T_C + F_D - 500 = 0 ... (2)
SFz = 0: 0.6667*T_B + 0.3333*T_C = 0 ... (3)
Step 4: Solve.
From Eq (3):
0.6667*T_B = -0.3333*T_C
T_B = -0.5*T_C
Since tension cannot be negative, let us re-examine. A negative T_B would mean that cable AB is in compression, which is impossible for a cable. This means the geometry given would cause cable AB to go slack. However, for exam practice, let us proceed assuming these are rods (not cables) that can take tension or compression.
From Eq (1):
0.3333*T_B = 0.6667*T_C
T_B = 2*T_C ... (i)
From Eq (3):
0.6667*T_B + 0.3333*T_C = 0
0.6667*(2*T_C) + 0.3333*T_C = 0
1.3333*T_C + 0.3333*T_C = 0
1.6667*T_C = 0
T_C = 0
Then T_B = 0 from (i), and from Eq (2): F_D = 500 N.
>>> FINAL ANSWERS: T_B = 0 N, T_C = 0 N, F_D = 500.0 N <<<
The strut carries the entire load because it is aligned vertically.
Check your understanding: Always pay attention to the direction a strut pushes. A strut in compression pushes the particle away from the strut's other end. If your answer gives a negative cable tension, the cable would be slack (T=0) in reality.
Problem Statement: A ring at the origin supports a weight of 750 N. Three cables pull on the ring with the following unit vector directions:
- Cable 1: u_1 = (0.6, 0.8, 0) -- lies in the xy-plane
- Cable 2: u_2 = (-0.4, 0.7, 0.5916)
- Cable 3: u_3 = (-0.3, 0.6, -0.7416)
Find the tension in each cable.
Solution:
Step 1: Write equilibrium equations directly.
SFx = 0: 0.6*T_1 - 0.4*T_2 - 0.3*T_3 = 0 ... (1)
SFy = 0: 0.8*T_1 + 0.7*T_2 + 0.6*T_3 - 750 = 0 ... (2)
SFz = 0: 0*T_1 + 0.5916*T_2 - 0.7416*T_3 = 0 ... (3)
Step 2: From Eq (3):
0.5916*T_2 = 0.7416*T_3
T_2 = (0.7416/0.5916)*T_3 = 1.2536*T_3 ... (*)
Step 3: Substitute into Eq (1):
0.6*T_1 - 0.4*(1.2536*T_3) - 0.3*T_3 = 0
0.6*T_1 - 0.5014*T_3 - 0.3*T_3 = 0
0.6*T_1 = 0.8014*T_3
T_1 = 1.3357*T_3 ... (**)
Step 4: Substitute into Eq (2):
0.8*(1.3357*T_3) + 0.7*(1.2536*T_3) + 0.6*T_3 = 750
1.0686*T_3 + 0.8775*T_3 + 0.6*T_3 = 750
2.5461*T_3 = 750
T_3 = 294.6 N
Step 5: Back-substitute.
T_2 = 1.2536 * 294.6 = 369.3 N
T_1 = 1.3357 * 294.6 = 393.5 N
>>> FINAL ANSWERS: T_1 = 393.5 N, T_2 = 369.3 N, T_3 = 294.6 N <<<
Verification (SFy): 0.8(393.5) + 0.7(369.3) + 0.6(294.6) = 314.8 + 258.5 + 176.8 = 750.1 ~ 750 N. Checks out.
Check your understanding: When unit vectors are given directly, you skip the position-vector step. Always verify your answer by plugging back into at least one equilibrium equation.
Problem Statement: A 6-m horizontal beam AB is supported by a pin at A and a roller at B. The beam carries:
- A 4 kN downward force at 2 m from A
- A 6 kN downward force at 4.5 m from A
- A 2 kN*m clockwise couple at the midpoint (3 m from A)
Find the support reactions at A and B.
Solution:
Step 1: Draw the Free Body Diagram.
4 kN 2 kN*m (CW) 6 kN
| | |
v v v
A-----+------+-----+------+-----+----B
^ 2m 1m 0m 1.5m 1.5m ^
| |
Ax-> By
Ay
Pin at A provides: Ax (horizontal) and Ay (vertical). Roller at B provides: By (vertical only).
Step 2: Apply equilibrium equations.
SFx = 0:
Ax = 0
SM_A = 0 (take moments about A, CCW positive):
-4(2) - 6(4.5) - 2 + By(6) = 0
-8 - 27 - 2 + 6*By = 0
-37 + 6*By = 0
By = 37/6
By = 6.167 kN (upward)
Note: The 2 kN*m clockwise couple contributes -2 to the moment sum (it is clockwise = negative in our CCW+ convention). A couple has the same moment about ANY point, so its position does not matter.
SFy = 0:
Ay + By - 4 - 6 = 0
Ay + 6.167 - 10 = 0
Ay = 3.833 kN (upward)
>>> FINAL ANSWERS: Ax = 0, Ay = 3.833 kN (up), By = 6.167 kN (up) <<<
Verification (SM_B = 0):
-Ay(6) + 4(6-2) + 6(6-4.5) + 2 = ?
-3.833(6) + 4(4) + 6(1.5) + 2
= -23.0 + 16 + 9 + 2 = 4.0 ... should be 0
Let me recheck: taking moments about B, CCW positive:
Ay(6) - 4(6-2) - 6(6-4.5) - 2 = 0
Ay(6) - 4(4) - 6(1.5) - 2 = 0
6*Ay - 16 - 9 - 2 = 0
6*Ay = 27
Ay = 4.5 kN ...
Let me redo the moment about A more carefully. Taking CCW as positive about A:
SM_A = 0:
Moments about A:
4 kN at 2 m: force is downward, creates CW moment = -4(2) = -8
6 kN at 4.5 m: force is downward, creates CW moment = -6(4.5) = -27
2 kN*m CW couple: = -2
By at 6 m: force is upward, creates CCW moment = +By(6)
-8 - 27 - 2 + 6*By = 0
6*By = 37
By = 6.167 kN
Now verify with SM_B:
SM_B = 0 (CCW positive):
Ay at 6m to the left: Ay is upward at A, which is 6m left of B.
Upward force to the left of point creates CCW = +Ay(6)
4 kN at (6-2)=4m to the left of B: downward force left of B creates CW = -4(4) = -16...
Wait -- downward force to the LEFT of B: position is to the left, force is down. r cross F: using right hand rule, r points left (-x), F points down (-y). (-x) cross (-y) = +z = CCW. So +4(4).
Actually let me just use the sign convention carefully:
r from B to where Ay acts = -6 m (leftward)
Ay acts upward (+)
Moment = (-6)(+Ay) = -6*Ay ...
No, moment = r x F in 2D is just: M = x*Fy - y*Fx where (x,y) is position relative to moment point.
Let me place origin at A, x to the right, y up.
Re-doing SM_B (moments about B at x=6):
For each force, moment about B = (x_force - 6) * Fy_force ...
No: M_B = (x - x_B)*Fy for vertical forces (taking CCW positive).
Ay at x=0: M = (0 - 6)*(+Ay) = -6*Ay
4 kN down at x=2: M = (2 - 6)*(-4) = (-4)(-4) = +16
6 kN down at x=4.5: M = (4.5 - 6)*(-6) = (-1.5)(-6) = +9
Couple: -2 (clockwise, independent of point)
SM_B = -6*Ay + 16 + 9 - 2 = 0
-6*Ay + 23 = 0
Ay = 23/6 = 3.833 kN ✓
Both methods give Ay = 3.833 kN. The original answers are confirmed correct.
>>> FINAL ANSWERS (confirmed): Ax = 0, Ay = 3.833 kN (up), By = 6.167 kN (up) <<<
Check your understanding: A couple moment is the same about every point -- you do not multiply it by a distance. Always verify your answer using a moment equation about a different point than the one you solved with.
Problem Statement: A horizontal beam AC is 5 m long with a fixed support at A. A 3 kN force acts at point B (2 m from A) directed at 40 degrees below the positive x-axis (down and to the right). A 2 kN upward force acts at C (the free end, 5 m from A). Find the reactions at the fixed support A.
Solution:
Step 1: Resolve the 3 kN force at B into components.
F_Bx = 3 cos(40°) = 3 * 0.7660 = 2.298 kN (to the right, +x)
F_By = -3 sin(40°) = -3 * 0.6428 = -1.928 kN (downward, -y)
Step 2: Draw FBD. Fixed support at A gives three reactions: Ax, Ay, and M_A.
M_A (CCW)
↺
A=====B==============C
^ | |
| 3kN at 40° below 2 kN (up)
Ax→ horizontal
Ay↑
Step 3: Equilibrium equations.
SFx = 0:
Ax + 2.298 = 0
Ax = -2.298 kN
Ax = 2.298 kN to the LEFT.
SFy = 0:
Ay - 1.928 + 2 = 0
Ay + 0.072 = 0
Ay = -0.072 kN
Ay = 0.072 kN DOWNWARD (very small, nearly zero).
SM_A = 0 (CCW positive):
M_A + (-1.928)(2) + (2)(5) = 0
M_A - 3.856 + 10 = 0
M_A + 6.144 = 0
M_A = -6.144 kN*m
Note: The horizontal component of the 3 kN force (F_Bx) passes through the beam axis (assuming beam is along x-axis), so it creates no moment about A (its moment arm about A is zero since it acts along the line of the beam).
M_A = 6.144 kN*m CLOCKWISE.
>>> FINAL ANSWERS: Ax = 2.298 kN (left), Ay = 0.072 kN (down), M_A = 6.144 kN*m (CW) <<<
Verification (SM_C = 0):
M_A + Ay(5) + Ax(0) + 1.928(5-2) = 0
Wait, let me be careful. Taking moments about C:
M_A + Ay*(5) + (-1.928)*(5-2) + Ax*(0) = 0
But Ax is horizontal along the beam -- moment arm from C is zero (along the beam axis).
Actually Ax is horizontal. If the beam is horizontal, Ax's line of action is along the beam. The perpendicular distance from C to Ax's line is 0 (they're collinear). So Ax contributes zero moment about C. What about Ay?
Ay acts at A (x=0). About C (x=5):
Moment from Ay = Ay * 5 (Ay is downward = -0.072, so moment = (-0.072)(5)...
Let me use the formula: M_C from a force at position x relative to C:
M_A (couple, free vector) = -6.144
Ay at x = 0: relative to C at x=5, position = -5. Ay = -0.072 (down). M = (-5)*(-0.072) = +0.36
Ax at A: horizontal force, perpendicular distance = 0 (beam is horizontal). M = 0.
1.928 kN down at x=2: relative to C, position = -3. M = (-3)*(-1.928) = +5.784
2.298 kN right at x=2: along beam, moment arm = 0. M = 0.
2 kN up at C: at C itself, moment = 0.
SM_C = -6.144 + 0.36 + 5.784 = 0.00 ✓
Check your understanding: At a fixed support in 2D, you always have three unknowns: Ax, Ay, and a moment M_A. Always resolve angled forces into x and y components before writing equilibrium equations.
Problem Statement: An L-shaped bracket is pinned at A at the bottom-left corner. The bracket extends 3 m to the right to point B, then 2 m upward to point C. A cable connects C to a wall anchor at point D, which is directly to the left of C (so the cable is horizontal). A 500-N downward load is applied at B. Find the tension in cable CD and the pin reactions at A.
Solution:
Step 1: Draw FBD.
D -------T------ C
|
| 2 m
|
A----------------B
^ |
| 3 m 500 N (down)
Ax→
Ay↑
Cable CD is horizontal (to the left), so T acts horizontally to the left at C.
Step 2: Equilibrium equations.
SM_A = 0 (CCW positive, eliminates Ax and Ay):
Moment from 500 N at B: The 500 N force is downward at B, which is 3 m to the right of A.
M_500 = -(500)(3) = -1500 N*m (CW)
Moment from T at C: T acts horizontally to the left at C. C is located at (3, 2) relative to A. For a horizontal force F_x at position (x, y), the moment about the origin = -F_x * y (from the cross product). T acts in the -x direction, so F_x = -T:
M_T = -(-T)(2) = +2T (CCW)
SM_A = 0:
-1500 + 2T = 0
T = 750 N
SFx = 0:
Ax - T = 0
Ax - 750 = 0
Ax = 750 N (to the right)
SFy = 0:
Ay - 500 = 0
Ay = 500 N (upward)
>>> FINAL ANSWERS: T = 750 N, Ax = 750 N (right), Ay = 500 N (up) <<<
Verification (SM_C = 0):
Ax acts at A = (0,0) relative to C = (3,2). Position of A relative to C = (-3,-2).
Moment from Ax (rightward = +x): (-2)*(750) = -1500...
Using M = x*Fy - y*Fx where (x,y) is position relative to C:
From Ay at A: position relative to C = (-3, -2). Fy = +500. M = (-3)(500) = -1500.
Wait, M = x*Fy - y*Fx.
From Ay: x=-3, Fx=0, Fy=+500. M = (-3)(500) - (-2)(0) = -1500
From Ax: x=-3, Fx=+750, Fy=0. M = (-3)(0) - (-2)(750) = +1500
From 500N at B: position of B relative to C = (0, -2). Fx=0, Fy=-500. M = (0)(-500) - (-2)(0) = 0
SM_C = -1500 + 1500 + 0 = 0 ✓
Check your understanding: When taking moments, choose a point where the most unknowns pass through it. A pin at A eliminates both Ax and Ay from the moment equation, letting you solve for the remaining unknown directly.
Problem Statement: A force F = (40i - 20j + 30k) N is applied at point P = (3, 1, -2) m. Find the moment of this force about the axis that passes through the origin and has direction u = (0, 0.6, 0.8).
Solution:
The moment about an axis is: M_a = u . (r x F)
where u is the unit vector along the axis, r is the position vector from any point on the axis to the point of force application, and the dot product projects the moment vector onto the axis.
Step 1: Verify u is a unit vector.
|u| = sqrt(0^2 + 0.6^2 + 0.8^2) = sqrt(0 + 0.36 + 0.64) = sqrt(1.00) = 1.0 ✓
Step 2: Choose r from a point on the axis to P.
The axis passes through the origin, so use r = position of P - origin:
r = (3, 1, -2) m
Step 3: Compute r x F using the determinant method.
r x F = | i j k |
| 3 1 -2 |
| 40 -20 30 |
i component: (1)(30) - (-2)(-20) = 30 - 40 = -10
j component: -[(3)(30) - (-2)(40)] = -[90 - (-80)] = -[90 + 80] = -170
k component: (3)(-20) - (1)(40) = -60 - 40 = -100
r x F = (-10i - 170j - 100k) N*m
Step 4: Take the dot product with the unit vector.
M_a = u . (r x F)
= (0)(-10) + (0.6)(-170) + (0.8)(-100)
= 0 - 102 - 80
= -182 N*m
The magnitude is 182 N*m. The negative sign means the moment is in the -u direction (opposite to the specified axis direction).
>>> FINAL ANSWER: M_a = -182 Nm (or 182 Nm in the -u direction) <<<
Check your understanding: The formula M_a = u . (r x F) is also called the scalar triple product. You can also compute it as a single 3x3 determinant with rows: u, r, F. The sign tells you the rotation direction relative to u (positive = right-hand rule around u).
Problem Statement: A pipe wrench applies a force F = (20i + 10j - 15k) N at point A = (0.5, 0.3, 0) m. Find the moment about the axis defined by the line from B = (0, 0, 0) to C = (1, 0, 0) -- i.e., the x-axis.
Solution:
Step 1: Unit vector along BC (the x-axis).
u = (C - B)/|C - B| = (1, 0, 0)/1 = (1, 0, 0) = i
Step 2: Position vector r from any point on the axis to point A.
Use B = (0,0,0):
r = A - B = (0.5, 0.3, 0) m
Step 3: Compute using the triple scalar product determinant.
M_x = | u_x u_y u_z | | 1 0 0 |
| r_x r_y r_z | = | 0.5 0.3 0 |
| F_x F_y F_z | | 20 10 -15 |
Expand along the first row:
M_x = 1 * |0.3 0 | - 0 * |0.5 0 | + 0 * |0.5 0.3|
|10 -15 | |20 -15 | |20 10 |
M_x = 1 * [(0.3)(-15) - (0)(10)] - 0 + 0
M_x = 1 * [-4.5 - 0]
M_x = -4.5 N*m
>>> FINAL ANSWER: M_x = -4.5 N*m (moment about the x-axis) <<<
This means the force tends to rotate the system clockwise when viewed from the +x direction.
Verification: We can check by computing r x F fully:
r x F = | i j k |
| 0.5 0.3 0 |
| 20 10 -15 |
i: (0.3)(-15) - (0)(10) = -4.5
j: -[(0.5)(-15) - (0)(20)] = -[-7.5] = 7.5
k: (0.5)(10) - (0.3)(20) = 5 - 6 = -1
r x F = (-4.5i + 7.5j - 1k) N*m
Then M_x = u . (r x F) = (1)(−4.5) + (0)(7.5) + (0)(−1) = −4.5 ✓
Check your understanding: When the axis is one of the coordinate axes (x, y, or z), the moment about that axis is simply the corresponding component of the moment vector M = r x F. The determinant method is most useful for arbitrary axis directions.
Problem Statement: A door is hinged along the z-axis (from the floor up). A person pushes on the door handle at point A = (0.9, 0.1, 1.2) m with force F = (-50i + 10j + 0k) N. Find the moment about the hinge axis (the z-axis) that actually rotates the door.
Solution:
Step 1: The hinge axis is the z-axis. Unit vector: u = (0, 0, 1) = k.
Step 2: Position vector from origin (on z-axis) to A.
r = (0.9, 0.1, 1.2) m
Step 3: Compute r x F.
r x F = | i j k |
| 0.9 0.1 1.2 |
| -50 10 0 |
i: (0.1)(0) - (1.2)(10) = 0 - 12 = -12
j: -[(0.9)(0) - (1.2)(-50)] = -[0 + 60] = -60
k: (0.9)(10) - (0.1)(-50) = 9 + 5 = 14
r x F = (-12i - 60j + 14k) N*m
Step 4: Project onto z-axis.
M_z = u . (r x F) = (0)(-12) + (0)(-60) + (1)(14) = 14 N*m
>>> FINAL ANSWER: M_z = 14 N*m about the z-axis (door hinge) <<<
Positive means the door rotates CCW when viewed from above (+z direction).
Check your understanding: For the z-axis, the moment is just the k-component of r x F. Note that the z-component of r (the height of the handle) does NOT affect the moment about the z-axis -- only the x and y coordinates and force components matter for rotation about a vertical hinge.
Problem Statement: A 10-kg block rests on a smooth (frictionless) 30-degree inclined plane. The block is held by a spring aligned parallel to the incline, attached to a fixed wall at the top of the incline. The spring has a natural (unstretched) length of L_0 = 0.4 m and a stiffness of k = 500 N/m. Find the stretched length of the spring and the normal force on the block.
Solution:
Step 1: Draw FBD of the block.
Forces on the block:
- Weight W = mg = 10(9.81) = 98.1 N, acting straight down.
- Spring force F_s, acting up the incline (pulling block toward the wall).
- Normal force N, acting perpendicular to the incline surface (away from surface).
Step 2: Set up coordinates aligned with the incline.
Let x-axis be along the incline (positive up the slope), y-axis perpendicular to incline (positive away from surface).
Resolve weight into components:
W_x = -mg sin(30°) = -98.1 * 0.5 = -49.05 N (down the slope)
W_y = -mg cos(30°) = -98.1 * 0.8660 = -84.96 N (into the surface)
Step 3: Equilibrium along y (perpendicular to incline).
SFy = 0:
N - mg cos(30°) = 0
N = 84.96 N
Step 4: Equilibrium along x (parallel to incline).
The spring force acts up the incline: F_s = k * s, where s is the stretch (deformation).
SFx = 0:
F_s - mg sin(30°) = 0
k * s = mg sin(30°)
500 * s = 49.05
s = 49.05 / 500
s = 0.0981 m
Step 5: Find stretched length.
L = L_0 + s = 0.4 + 0.0981 = 0.4981 m
>>> FINAL ANSWERS: Stretched length L = 0.498 m, Normal force N = 84.96 N <<<
Check your understanding: The spring force formula is F = k * s, where s = |L - L_0| is the DEFORMATION (change in length), NOT the total length. If the spring is stretched, s = L - L_0 > 0. If compressed, s = L_0 - L > 0. The force always acts to restore the natural length.
Problem Statement: A 5-kg collar slides along a frictionless vertical rod. Two springs are attached to the collar at point A. Spring 1 connects horizontally to a wall (stiffness k_1 = 300 N/m, natural length L_01 = 0.3 m). Spring 2 connects at 45 degrees above horizontal to a ceiling anchor (stiffness k_2 = 200 N/m, natural length L_02 = 0.5 m). In the equilibrium position, spring 1 has a stretched length of 0.6 m, and the angle of spring 2 from horizontal is exactly 45 degrees. Find the stretched length of spring 2 and verify equilibrium.
Solution:
Step 1: Spring 1 force.
Deformation of spring 1: s_1 = L_1 - L_01 = 0.6 - 0.3 = 0.3 m
F_1 = k_1 * s_1 = 300 * 0.3 = 90 N (directed toward the wall, horizontal)
Spring 1 pulls the collar horizontally toward the wall. Let's say the wall is to the left, so F_1 acts in the -x direction.
Step 2: FBD of the collar. The collar slides on a vertical rod, so the rod provides a horizontal reaction force R (perpendicular to rod = horizontal).
Forces on collar:
- Weight: W = 5(9.81) = 49.05 N downward
- F_1 = 90 N to the left
- F_2 = force from spring 2, along the spring at 45 degrees above horizontal (toward ceiling anchor)
- R = normal force from rod (horizontal, let's say to the right = +x)
Since the collar is on a VERTICAL rod, the rod constrains horizontal motion. The rod provides a horizontal normal force R.
Step 3: Resolve F_2 into components.
Spring 2 acts at 45 degrees above horizontal, directed toward the ceiling anchor (up and away from collar). Let's say the anchor is to the left and above, so F_2 acts up-left at 45 degrees:
F_2x = -F_2 cos(45°) = -0.7071 * F_2 (to the left)
F_2y = +F_2 sin(45°) = +0.7071 * F_2 (upward)
Actually, the problem doesn't specify direction -- let me assume the ceiling anchor is to the RIGHT and above (so F_2 pulls the collar to the right and up):
F_2x = +F_2 cos(45°) = +0.7071 * F_2
F_2y = +F_2 sin(45°) = +0.7071 * F_2
Step 4: Equilibrium.
SFy = 0 (vertical -- the rod cannot resist vertical forces, so vertical equilibrium must be satisfied by springs and weight alone):
F_2 sin(45°) - W = 0
0.7071 * F_2 = 49.05
F_2 = 49.05 / 0.7071
F_2 = 69.37 N
Step 5: Find spring 2 deformation and stretched length.
F_2 = k_2 * s_2
69.37 = 200 * s_2
s_2 = 69.37 / 200 = 0.3469 m
L_2 = L_02 + s_2 = 0.5 + 0.3469 = 0.847 m
Step 6: Check horizontal equilibrium to find R.
SFx = 0:
R - F_1 + F_2 cos(45°) = 0
R - 90 + 69.37(0.7071) = 0
R - 90 + 49.05 = 0
R = 40.95 N (to the right, which pushes collar away from rod -- makes sense)
Wait, let me reconsider. If F_1 is to the LEFT and F_2x is to the RIGHT:
R + F_2 cos(45°) - F_1 = 0
R + 49.05 - 90 = 0
R = 40.95 N
R must push the collar in whatever direction is needed. Here R = 40.95 N (directed to balance horizontal forces).
>>> FINAL ANSWERS: L_2 = 0.847 m, F_2 = 69.37 N, R = 40.95 N <<<
Check your understanding: On a frictionless rod, the collar can only be pushed perpendicular to the rod. For a vertical rod, the reaction is horizontal. For a horizontal rod, the reaction is vertical. This constraint determines which equilibrium equation you can use to solve for spring forces.
Problem Statement: A spring with k = 800 N/m and natural length L_0 = 0.25 m is mounted vertically. A 15-kg mass hangs from a cable that runs over a frictionless pulley and attaches to the top of the spring, compressing it. Find the compressed length of the spring.
Solution:
Step 1: Analyze the cable and pulley.
For an ideal (frictionless, massless) pulley, the tension throughout the cable is constant. The mass hangs from one end:
T = mg = 15(9.81) = 147.15 N
The cable transmits this tension to the spring, pushing down on the spring's top.
Step 2: Spring force.
The spring is compressed, so the spring pushes back up with force F_s = k * s.
Equilibrium of the spring top:
k * s = T
800 * s = 147.15
s = 147.15 / 800
s = 0.1839 m (compression)
Step 3: Find compressed length.
L = L_0 - s = 0.25 - 0.1839 = 0.0661 m = 66.1 mm
Since L > 0, the spring has not been compressed to zero, so this is physically valid.
>>> FINAL ANSWER: Compressed length L = 0.0661 m (66.1 mm) <<<
Check your understanding: When a spring is compressed, the deformation s = L_0 - L (natural length minus current length). The force is still F = k*s. Always check that your answer gives a positive length; if L < 0, the spring would need to invert, which means the problem setup is impossible with the given spring.
Problem Statement: Two forces act at a point: F_1 = (120i - 60j + 80k) N and F_2 = (-40i + 90j + 50k) N. Find the angle between them.
Solution:
The angle between two vectors is found from: cos(theta) = (F_1 . F_2) / (|F_1| * |F_2|)
Step 1: Compute the dot product.
F_1 . F_2 = (120)(-40) + (-60)(90) + (80)(50)
= -4800 + (-5400) + 4000
= -4800 - 5400 + 4000
= -6200
Step 2: Compute magnitudes.
|F_1| = sqrt(120^2 + (-60)^2 + 80^2)
= sqrt(14400 + 3600 + 6400)
= sqrt(24400)
= 156.2 N
|F_2| = sqrt((-40)^2 + 90^2 + 50^2)
= sqrt(1600 + 8100 + 2500)
= sqrt(12200)
= 110.5 N
Step 3: Find the angle.
cos(theta) = -6200 / (156.2 * 110.5)
= -6200 / 17260.1
= -0.3591
theta = arccos(-0.3591)
theta = 111.0°
>>> FINAL ANSWER: theta = 111.0° <<<
Check your understanding: The dot product being negative means the angle is obtuse (greater than 90 degrees). The dot product being zero means the vectors are perpendicular (90 degrees). Always use arccos, not arctan, to find the angle from a dot product.
Problem Statement: A force F = (200i + 150j - 100k) N acts at a point. Find the component of this force along a line that goes from point A = (1, 0, 2) to point B = (4, 3, -1).
Solution:
The projection (scalar component) of F along a line is: F_AB = F . u_AB
Step 1: Find the unit vector along AB.
r_AB = B - A = (4-1, 3-0, -1-2) = (3, 3, -3)
|r_AB| = sqrt(3^2 + 3^2 + (-3)^2) = sqrt(9 + 9 + 9) = sqrt(27) = 5.196
u_AB = r_AB / |r_AB| = (3/5.196, 3/5.196, -3/5.196)
= (0.5774, 0.5774, -0.5774)
Step 2: Compute the dot product.
F_AB = F . u_AB
= (200)(0.5774) + (150)(0.5774) + (-100)(-0.5774)
= 115.47 + 86.60 + 57.74
= 259.8 N
The positive sign means the component is in the direction from A to B.
Step 3: (Optional) Find the vector component along AB.
F_along = F_AB * u_AB = 259.8 * (0.5774, 0.5774, -0.5774)
= (150.0i + 150.0j - 150.0k) N
>>> FINAL ANSWER: The component of F along line AB = 259.8 N <<<
Check your understanding: The scalar projection can be positive (same direction as u) or negative (opposite direction). The vector projection is the scalar projection times the unit vector. If asked for the perpendicular component, use: F_perp = sqrt(|F|^2 - F_AB^2).
Problem Statement: A cable exerts a force F = (60i + 80j + 0k) N on a bracket. A pipe runs from the origin along the direction u_p = (0, 0.6, 0.8). Find: (a) the component of force along the pipe, (b) the perpendicular component, and (c) the angle between F and the pipe.
Solution:
Step 1: Verify u_p is a unit vector.
|u_p| = sqrt(0^2 + 0.6^2 + 0.8^2) = sqrt(0 + 0.36 + 0.64) = sqrt(1) = 1 ✓
Step 2 (a): Component along the pipe.
F_parallel = F . u_p = (60)(0) + (80)(0.6) + (0)(0.8)
= 0 + 48 + 0
= 48 N
Step 3 (b): Perpendicular component.
|F| = sqrt(60^2 + 80^2 + 0^2) = sqrt(3600 + 6400) = sqrt(10000) = 100 N
F_perp = sqrt(|F|^2 - F_parallel^2)
= sqrt(100^2 - 48^2)
= sqrt(10000 - 2304)
= sqrt(7696)
= 87.73 N
Step 4 (c): Angle between F and the pipe.
cos(theta) = F_parallel / |F| = 48 / 100 = 0.48
theta = arccos(0.48)
theta = 61.3°
>>> FINAL ANSWERS: (a) F_parallel = 48 N, (b) F_perp = 87.73 N, (c) theta = 61.3° <<<
Verification: F_parallel^2 + F_perp^2 = 48^2 + 87.73^2 = 2304 + 7697 = 10001 ~ 10000 = |F|^2 ✓ (small rounding error).
Check your understanding: The parallel and perpendicular components form a right triangle with the original force as the hypotenuse. This means F_parallel^2 + F_perp^2 = |F|^2 always. This is a great way to verify your answer.
Problem Statement: A beam of length 8 m is supported by a pin at A (left end) and a roller at B (right end). A uniform distributed load of w = 3 kN/m acts over the entire length of the beam. Determine the support reactions at A and B.
Step-by-step Solution:
Step 1: Replace the distributed load with its resultant force.
For a uniform (rectangular) distributed load:
F_R = w * L = 3 kN/m * 8 m = 24 kN (downward)
The resultant acts at the centroid of the rectangle, which is at the midpoint:
x_R = L / 2 = 8 / 2 = 4 m from A
Step 2: Draw the free body diagram.
Replace the distributed load with F_R = 24 kN acting 4 m from A. The pin at A provides A_x (horizontal) and A_y (vertical). The roller at B provides B_y (vertical only).
Step 3: Write equilibrium equations.
Sum of forces in x-direction:
sum F_x = 0: A_x = 0
Sum of moments about A (counterclockwise positive):
sum M_A = 0: B_y(8) - 24(4) = 0
B_y(8) = 96
B_y = 12 kN (upward)
Sum of forces in y-direction:
sum F_y = 0: A_y + B_y - 24 = 0
A_y + 12 - 24 = 0
A_y = 12 kN (upward)
>>> FINAL ANSWERS: A_x = 0, A_y = 12 kN (upward), B_y = 12 kN (upward) <<<
Verification: A_y + B_y = 12 + 12 = 24 kN = F_R. The load is symmetric about the midpoint and the supports are at equal distances, so by symmetry A_y = B_y. Both checks pass. ✓
Check your understanding: For a uniform load on a simply supported beam with symmetric support placement, the reactions are always equal and each carries half the total load. If either the load or the supports are not symmetric, you must use moment equations to solve.
Problem Statement: A beam of length 6 m is supported by a pin at A (left end) and a roller at B (right end). A triangular distributed load acts on the beam, increasing linearly from 0 at A to w_max = 4 kN/m at B. Determine the support reactions at A and B.
Step-by-step Solution:
Step 1: Replace the distributed load with its resultant force.
For a triangular distributed load:
F_R = (1/2) * w_max * L = (1/2)(4)(6) = 12 kN (downward)
The resultant acts at the centroid of the triangle, which is located at 1/3 of the length from the larger end (B), or equivalently 2/3 of the length from the smaller end (A):
x_R = (2/3) * L = (2/3)(6) = 4 m from A
Step 2: Draw the free body diagram.
Replace the distributed load with F_R = 12 kN acting 4 m from A. Pin at A gives A_x and A_y. Roller at B gives B_y.
Step 3: Write equilibrium equations.
Sum of forces in x-direction:
sum F_x = 0: A_x = 0
Sum of moments about A (counterclockwise positive):
sum M_A = 0: B_y(6) - 12(4) = 0
B_y(6) = 48
B_y = 8 kN (upward)
Sum of forces in y-direction:
sum F_y = 0: A_y + B_y - 12 = 0
A_y + 8 - 12 = 0
A_y = 4 kN (upward)
>>> FINAL ANSWERS: A_x = 0, A_y = 4 kN (upward), B_y = 8 kN (upward) <<<
Verification: A_y + B_y = 4 + 8 = 12 kN = F_R ✓. Also, check moment about B: A_y(6) - 12(6-4) = 4(6) - 12(2) = 24 - 24 = 0 ✓.
Check your understanding: For a triangular load, the resultant is always at 1/3 from the heavy end (2/3 from the light end). The support closer to the heavy end carries a larger share of the load. A common mistake is placing the centroid at the midpoint -- always remember it is at 1/3 from the larger end for a triangle.
Problem Statement: A frame consists of members AB, BC, and AC. Member BC is pin-connected at B and C, and carries no loads along its length (no distributed loads, no applied forces, no couples except at the two pin joints). A horizontal force P = 500 N is applied at joint A. Member AB is horizontal (A at left, B at right), member BC is vertical (B at top, C at bottom), and member AC is the diagonal. Joints A and C are pin-supported to the ground. Identify any two-force members. Then find the force in member BC and the reactions at support A, given that AB = 3 m and BC = 4 m.
Step-by-step Solution:
Step 1: Identify two-force members.
A two-force member has forces applied at exactly two points and no other loads. Member BC has pin connections at B and C only, with no loads along its length.
Member BC is a TWO-FORCE MEMBER.
Therefore, the force in BC must act along the line from B to C (i.e., vertically, since BC is vertical). The force is either pure tension or pure compression along BC.
Step 2: Analyze joint B.
At joint B, three members meet. Since BC is a two-force member, the force F_BC acts along BC (vertically). From the FBD of member AB (isolate it), forces at B from member AB must be resolved.
Step 3: Analyze member AB as a free body.
Member AB has: pin reaction at A (A_x, A_y), the applied force P = 500 N horizontal at A, and pin forces at B (B_x, B_y) from the connected members.
Actually, let us analyze the whole frame more carefully. Consider the entire frame as a free body first.
External supports: Pin at A (A_x, A_y) and pin at C (C_x, C_y). Applied load: P = 500 N horizontal at A.
sum F_x = 0: A_x + C_x + 500 = 0
sum F_y = 0: A_y + C_y = 0
sum M_A = 0: C_x(4) - C_y(3) = 0 (taking moments, C is at position (3, -4) relative to A)
Wait -- let me set up coordinates. Place A at origin. B is at (3, 0). C is at (3, -4) (since BC = 4 m vertical, below B).
Sum of moments about A (CCW positive):
sum M_A = 0: C_x(4) + C_y(3) = 0
Hmm, let me be precise. C is at (3, -4). The moment of C_x about A: C_x acts horizontally at point (3, -4), so moment = C_x * (-4) (using r cross F). The moment of C_y about A: C_y acts vertically at (3, -4), so moment = C_y * (3).
sum M_A = 0: -C_x(4) + C_y(3) = 0
C_y = (4/3) C_x
Step 4: Use the two-force member constraint.
Since BC is a two-force member (vertical), the force at C from BC is purely vertical and the force at B from BC is purely vertical. This means when we look at the pin at C, the force from member BC on joint C is vertical. But C is also connected to member AC.
Isolate member AC. It connects A(0,0) to C(3,-4). At A it receives A_x, A_y, and the applied P = 500 N. At C it receives C_x, C_y. But we also know the force transmitted through pin C from member BC is purely vertical (call it F_BC, positive = tension = upward on C).
Isolate member BC: two-force member, so at B the force is F_BC along BC (vertical) and at C the force is F_BC along BC (vertical, opposite direction).
Now isolate joint B. At B, member AB exerts forces (B_x from AB, B_y from AB) and member BC exerts F_BC vertically. Equilibrium at joint B:
B_x = 0 (no horizontal force from BC, so horizontal force from AB at B must be zero -- but this isn't right for a joint, we need to isolate a member)
Let me isolate member AB. Member AB goes from A(0,0) to B(3,0). Forces on AB: at A, the reaction (A_x, A_y) and applied force P = 500 N horizontally. At B, forces from the pin connecting to BC and AC.
Since BC is a two-force member, the pin at B transmits only a vertical force F_BC from member BC.
Member AB has length 3 m (horizontal). Take moments about A for member AB:
sum M_A(member AB) = 0: F_BC_vertical * 3 + B_x_from_AC * 0 = 0
This is getting complex. Let me simplify by isolating member AB properly.
Forces on member AB at point A: A_x + P = A_x + 500 (horizontal), A_y (vertical). Forces on member AB at point B: B_x (horizontal from AC connection), and F_BC (vertical, from BC).
Moment about A for member AB:
sum M_A = 0: -F_BC(3) + B_x(0) = 0 --> This gives F_BC = 0?
That would mean no force in BC. Let me reconsider the geometry. Actually, since AB is horizontal and B_x acts horizontal at B (moment arm = 0 from A), and F_BC acts vertical at B (moment arm = 3 m):
sum M_A(AB) = 0: -F_BC * 3 = 0 --> F_BC = 0
This makes sense geometrically -- the horizontal force at A has no vertical component to be resisted by BC through member AB alone. The diagonal member AC carries the load directly to C.
So F_BC = 0. Now:
sum F_x(AB) = 0: A_x + 500 + B_x = 0
sum F_y(AB) = 0: A_y + F_BC = A_y + 0 = 0 --> A_y = 0
From the whole frame:
sum F_y = 0: A_y + C_y = 0 --> C_y = 0
sum M_A = 0: -C_x(4) + C_y(3) = 0 --> C_x = 0
sum F_x = 0: A_x + C_x + 500 = 0 --> A_x = -500 N
So A_x = -500 N (leftward, opposing P), and from member AB: B_x = -A_x - 500 = 500 - 500 = 0.
>>> FINAL ANSWERS: Member BC is the two-force member (force acts along line BC). F_BC = 0 N. A_x = 500 N (to the left, opposing P), A_y = 0 N, C_x = 0 N, C_y = 0 N. <<<
Verification: The applied horizontal force at A is entirely resisted by the pin at A horizontally. Member AC is loaded only at A and C with zero net force, and BC carries no load. Sum of forces: (-500 + 0 + 500) = 0 ✓ in x, (0 + 0) = 0 ✓ in y.
Check your understanding: A two-force member can carry zero force -- that is a valid equilibrium state! The key skill is identifying it: look for members loaded at exactly two points with no loads in between. Once identified, you immediately know the force direction (along the member axis), which reduces unknowns.
Problem Statement: A beam AB of length 5 m is supported by a pin at A and a roller on an inclined surface at B. The roller surface makes a 30-degree angle with the horizontal (so the roller reaction at B is perpendicular to the surface, directed at 60 degrees from horizontal). A vertical load W = 10 kN is applied at point C, which is 2 m from A. Using the three-force member concurrency principle, determine the reactions at A and B.
Step-by-step Solution:
Step 1: Verify this is a three-force member.
The beam has exactly three forces acting on it:
- The pin reaction at A (unknown direction)
- The roller reaction at B (known direction: 60 degrees from horizontal, or 30 degrees from vertical)
- The applied load W = 10 kN at C (vertical, downward)
Three forces, therefore it is a three-force member. By the three-force member theorem, if the body is in equilibrium, the three forces must be concurrent (all lines of action meet at a single point) -- unless all three are parallel.
Step 2: Find the concurrency point.
We know the lines of action of two forces:
- Force at B: acts at 60 degrees from horizontal through point B
- Force W: acts vertically downward through point C (2 m from A)
Find where these two lines of action intersect. Place A at the origin, B at (5, 0), C at (2, 0).
Line of action of W through C: x = 2 (vertical line).
Line of action of R_B through B at 60 degrees from horizontal:
Direction from B: at 60 deg from horizontal
Parametric: (5 - t cos60, 0 + t sin60) = (5 - 0.5t, 0.866t)
Set x = 2: 5 - 0.5t = 2, so t = 6. Then y = 0.866(6) = 5.196 m.
The concurrency point O is at (2, 5.196) -- that is, directly above C at a height of 5.196 m.
Step 3: Determine the direction of reaction at A.
Since all three forces must pass through point O, the reaction at A must be directed from A(0, 0) toward O(2, 5.196).
Angle of R_A from horizontal: theta_A = arctan(5.196 / 2) = arctan(2.598) = 69.0 degrees
Step 4: Solve using equilibrium equations.
Now we know all three force directions. Apply equilibrium:
Sum of forces in x-direction:
sum F_x = 0: R_A cos(69.0) - R_B cos(60) = 0
R_A (0.3584) = R_B (0.5)
R_A = 1.395 R_B
Sum of forces in y-direction:
sum F_y = 0: R_A sin(69.0) + R_B sin(60) - 10 = 0
R_A (0.9336) + R_B (0.866) = 10
Substitute R_A = 1.395 R_B:
1.395(0.9336) R_B + 0.866 R_B = 10
1.302 R_B + 0.866 R_B = 10
2.168 R_B = 10
R_B = 4.61 kN
Then:
R_A = 1.395(4.61) = 6.43 kN
Step 5: Find components of R_A.
A_x = R_A cos(69.0) = 6.43(0.3584) = 2.30 kN (to the right)
A_y = R_A sin(69.0) = 6.43(0.9336) = 6.00 kN (upward)
>>> FINAL ANSWERS: R_A = 6.43 kN at 69.0 degrees from horizontal, R_B = 4.61 kN at 60 degrees from horizontal. Components: A_x = 2.30 kN, A_y = 6.00 kN, B_x = 2.31 kN (to the left), B_y = 3.99 kN (upward). <<<
Verification: sum F_x = 2.30 - 4.61 cos60 = 2.30 - 2.31 ~ 0 ✓. sum F_y = 6.00 + 4.61(0.866) - 10 = 6.00 + 3.99 - 10 ~ 0 ✓. sum M_A = R_B sin60 (5) - R_B cos60 (0) - 10(2) = 3.99(5) - 10(2) = 19.97 - 20 ~ 0 ✓.
Check your understanding: The three-force member concurrency principle lets you find the direction of an unknown reaction without writing moment equations first. Find where two known lines of action intersect, then the third force must also pass through that point. This is especially powerful when one reaction has a known direction (roller, cable) and another has unknown direction (pin).
Problem Statement: A horizontal rectangular plate lies in the x-z plane. It is supported by a ball-and-socket joint at A (at the origin), a vertical cable at B (at position (4, 0, 0) m), and a vertical cable at C (at position (0, 0, 3) m). A vertical downward load W = 600 N is applied at point D (at position (4, 0, 3) m -- the corner diagonally opposite A). Determine the tension in each cable and the reaction components at the ball-and-socket joint A.
Step-by-step Solution:
Step 1: Identify unknowns and support reactions.
Ball-and-socket at A provides three reaction components: A_x, A_y, A_z. Cable at B provides one force: T_B (vertical, upward, along +y). Cable at C provides one force: T_C (vertical, upward, along +y).
Total unknowns: A_x, A_y, A_z, T_B, T_C = 5 unknowns.
Applied load: W = 600 N downward (-y direction) at D(4, 0, 3).
Step 2: Write the 6 equilibrium equations.
We have 6 equations for 3D rigid body equilibrium (3 force, 3 moment), but only 5 unknowns. The system is statically determinate.
Equation 1: sum F_x = 0
A_x = 0
Equation 2: sum F_y = 0
A_y + T_B + T_C - 600 = 0
Equation 3: sum F_z = 0
A_z = 0
Equation 4: sum M_x (moment about x-axis through A) = 0
Forces contributing moments about the x-axis (right-hand rule, x-axis points in +x direction):
- T_C at C(0, 0, 3): Force T_C in +y, position z = 3. Moment = T_C * 3 (using r x F: (3 k) x (T_C j) = -3 T_C i, so moment about x = -3 T_C). Wait, let me use the sign convention carefully.
Moment of a force F_y at position (x, 0, z) about the origin:
M = r x F = (x i + 0 j + z k) x (F_y j) = x F_y (i x j) + z F_y (k x j)
= x F_y k + z F_y (-i) = -z F_y i + x F_y k
So M_x component = -z * F_y and M_z component = x * F_y.
For T_C at C(0, 0, 3): F_y = +T_C
M_x contribution = -(3)(T_C) = -3 T_C
M_z contribution = (0)(T_C) = 0
For T_B at B(4, 0, 0): F_y = +T_B
M_x contribution = -(0)(T_B) = 0
M_z contribution = (4)(T_B) = 4 T_B
For W at D(4, 0, 3): F_y = -600
M_x contribution = -(3)(-600) = +1800
M_z contribution = (4)(-600) = -2400
sum M_x = 0:
-3 T_C + 0 + 1800 = 0
T_C = 600 N
Wait, that gives T_C = 600 N. Let me double check. Hmm, that seems too large. Let me re-examine.
Actually, let me redo this carefully. sum M_x = 0 (moments about the x-axis through A):
-3 T_C + 1800 = 0
T_C = 600 N
Equation 5: sum M_z (moment about z-axis through A) = 0
4 T_B - 2400 = 0
T_B = 600 N
Equation 6: sum M_y = 0
All forces are in the y-direction, so they produce no moment about the y-axis. This equation is 0 = 0, which is automatically satisfied.
Step 3: Solve for A_y.
From Equation 2:
A_y + 600 + 600 - 600 = 0
A_y = -600 N
The negative sign means A_y acts downward (-y direction). This means the ball-and-socket pulls down on the plate.
Wait -- let me reconsider. With T_B = 600 and T_C = 600, and W = 600 down, the total upward force is 1200 and total downward is 600, so A_y = -600 (downward) to balance. This is correct -- the ball-and-socket must pull the plate down because the cables together exert more upward force than the weight.
>>> FINAL ANSWERS: T_B = 600 N, T_C = 600 N, A_x = 0 N, A_y = -600 N (downward, the joint pulls the plate down), A_z = 0 N <<<
Verification:
- sum F_x = 0 ✓
- sum F_y = -600 + 600 + 600 - 600 = 0 ✓
- sum F_z = 0 ✓
- sum M_x = -3(600) + 3(600) = -1800 + 1800 = 0 ✓
- sum M_z = 4(600) - 4(600) = 2400 - 2400 = 0 ✓
- sum M_y = 0 ✓
All six equilibrium equations are satisfied.
Check your understanding: In 3D equilibrium, always start by writing out all 6 equations (3 force, 3 moment). The moment equations about axes through a ball-and-socket joint eliminate A_x, A_y, A_z from the moment equations, letting you solve for cable tensions directly. A negative reaction value simply means the assumed direction was wrong -- the ball-and-socket at A pulls the plate down, which is physically correct here because without it, the two cables would push the plate upward with a net upward force.
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Do NOT cram new material. Focus on reviewing what you already know. Work through 1-2 practice problems from each chapter to warm up.
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Prepare your equation sheet tonight. Do not wait until the morning. Write it carefully, organize by chapter, and include all formulas listed in Section 2.
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Check your calculator:
- Is it in DEGREE mode? (Press MODE, arrow to DEGREE, press ENTER)
- Are batteries fresh?
- Is your cross product program loaded and working? Test it with a known example.
- Bring a backup calculator if you have one.
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Get sleep. Your ability to think clearly matters more than one extra hour of studying.
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Read every problem completely before starting. Identify what is given and what is asked. Some problems have multiple parts.
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Draw the FBD FIRST for every problem. Do not skip this step. It takes 1-2 minutes but prevents 10-20 minutes of wrong work.
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Work problems in order of confidence. If a problem stumps you, skip it and come back. Get the easy points first.
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Show your work clearly. Partial credit is possible. Label your FBD, write equations clearly, and box your final answers.
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Check units at every step. N*m for moments, N or kN for forces, m or ft for distances.
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Use the full time. If you finish early, go back and verify your answers. Plug your solutions back into the equilibrium equations to confirm they balance.
START: Read the problem
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v
Is it a PARTICLE problem (Ch.3)? Is it a RIGID BODY problem (Ch.5)?
(Forces all meet at one point) (Forces at different points on a body)
| |
v v
Draw FBD of the particle Draw FBD of the rigid body (4-step method)
| |
v v
2D: SFx=0, SFy=0 (2 eqns) Identify all support reactions
3D: SFx=0, SFy=0, SFz=0 (3 eqns) |
| v
v Replace distributed loads with resultants
Solve for unknowns |
| v
v 2D: SFx=0, SFy=0, SM=0 (3 eqns)
CHECK: Verify answer 3D: SFx=SFy=SFz=SMx=SMy=SMz=0 (6 eqns)
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v
Is it a MOMENTS problem (Ch.4)? Solve for unknowns
(Find moment, reduce system, etc.) |
| v
v CHECK: Verify with unused equation
Moment about point: M = r x F
Moment about axis: M_a = u . (r x F)
Couple: M = r x F (free vector)
Equivalent system: Move force + add couple
Distributed load: F = area, at centroid
Even though you have an equation sheet, you should have these so internalized you do not need to look them up:
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Equilibrium equations -- These ARE statics:
- 2D: SFx = 0, SFy = 0, SM = 0
- 3D: SFx = SFy = SFz = 0, SMx = SMy = SMz = 0
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Support reactions -- Know these cold:
- Roller: 1 (perpendicular to surface)
- Pin: 2 (Ax, Ay)
- Fixed: 3 in 2D (Ax, Ay, MA), 6 in 3D
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Cross product pattern -- including the NEGATIVE on j
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Distributed load centroids:
- Rectangle: L/2
- Triangle: L/3 from the big end
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W = mg -- with g = 9.81 m/s^2 or 32.2 ft/s^2
Statics problems follow a systematic pattern. If you can draw a correct FBD and write the equilibrium equations, you can solve any problem on this exam. The math is algebra and trigonometry -- the challenge is in setting up the problem correctly.
Trust your preparation. Work methodically. Check your work.
Good luck on the exam.
Study guide prepared for ECE 211 - Engineering Statics, Midterm Exam, March 13, 2026 CGCC Pecos Campus, Ironwood 120 Modules 1-8, Chapters 1-5