Source: ENGR/PHYS 216 Final Exam Practice Problems
Tags: particle equilibrium, cable tension, free body diagram, conservation of momentum, collision, moment of force, torque, cross product, statics, ENGR 216, Texas A&M
This section covers three major mechanics topics: solving for cable tensions using equilibrium of concurrent forces, applying conservation of linear momentum to collisions and multi-body problems, and computing moments (torques) about points using the cross product. These are among the most calculation-heavy problems on the exam.
Particle equilibrium
A particle is in equilibrium when the net force on it is zero: ΣFx = 0 and ΣFy = 0. This gives two equations to solve for unknowns like cable tensions or angles.
Free body diagram (FBD)
A sketch showing a body isolated from its surroundings with all external forces drawn as vectors. The essential first step for any statics problem.
Conservation of linear momentum
In the absence of external forces, the total momentum of a system is constant: Σ(m × v)_before = Σ(m × v)_after
Applies to collisions, explosions, and any interaction between objects in an isolated system.
Perfectly inelastic collision
A collision where the objects stick together after impact. Momentum is conserved, but kinetic energy is not. The baseball-cart problem is an example.
Moment of a force (torque)
The tendency of a force to cause rotation about a point. In 2D: M = r × F = rx·Fy − ry·Fx
where r is the position vector from the pivot to the point of force application. A positive result indicates counterclockwise rotation (by the standard convention).
Transmissibility
The principle that a force can be moved along its line of action without changing its moment about any point. This means you can pick any point on the line of action as the "application point" when computing moments.
The approach for any concurrent-force problem at a point:
Draw the FBD at the junction point
Resolve every force into x and y components
Set ΣFx = 0 and ΣFy = 0
Solve the system of equations
Worked example (Problem 17): Two cables AC and BC meet at C, each making 30° from vertical. At C, a horizontal force P = 100 lb acts to the left, and a 150 lb force acts at angle θ below horizontal.
For equal tensions (T_AC = T_BC = T), the horizontal cable components cancel:
ΣFx: −T sin30° + T sin30° − 100 + 150 cos θ = 0
Since the T terms cancel: 150 cos θ = 100, so cos θ = 2/3. θ = arccos(2/3) = 48.2°
From the vertical equilibrium: ΣFy: 2T cos30° − 150 sin θ = 0 T = 150 sin(48.2°) / (2 cos30°) = 64.5 lb
Worked example (Problem 18): Crane with cable AC at 5° from vertical, cable BC at α = 15° from horizontal, and 5 kN load at C.
ΣFx: −T_AC sin5° + T_BC cos15° = 0 ΣFy: T_AC cos5° + T_BC sin15° − 5 = 0
From the x-equation: T_AC = T_BC cos15° / sin5°
Substituting into the y-equation and solving:
T_BC = 0.443 kN T_AC = 4.90 kN
For collisions on a frictionless surface, define a positive direction (typically rightward), write the momentum equation, and solve.
Worked example (Problem 19): Block 1 (mass m₁) moves right at 4a. Block 2 (mass m₂) moves left at a. After collision: block 1 moves right at a, block 2 moves right at 5a.
Conservation of momentum: m₁(4a) + m₂(−a) = m₁(a) + m₂(5a)
Simplifying: m₁(4a − a) = m₂(5a + a) 3a·m₁ = 6a·m₂ m₁/m₂ = 2
Worked example (Problem 20): A 10 kg cart catches baseballs (0.145 kg each, 35 m/s). After 10 balls, find cart velocity.
This is a series of perfectly inelastic collisions. Total momentum is conserved:
p_total = 10 × 0.145 × 35 = 50.75 kg·m/s Total mass = 10 + 10 × 0.145 = 11.45 kg v_final = 50.75 / 11.45 = 4.43 m/s
The moment about a point O due to a force F applied at point P:
M_O = r_OP × F
In 2D, this reduces to a scalar: M = rx·Fy − ry·Fx.
The sign convention is typically counterclockwise positive. If the result is negative, the moment is clockwise.
Worked example (Problem 21): Points A(−4, −1) and B(2, 7). A 75 N force acts along line AB. Find the moment about the origin.
Direction AB = (6, 8), magnitude = 10. Unit vector = (0.6, 0.8). Force vector = 75 × (0.6, 0.8) = (45, 60) N.
Using point A as the application point: M_O = (−4)(60) − (−1)(45) = −240 + 45 = −195
|M_O| = 195.0 N·cm
Worked example (Problem 22): L-shaped bar OBA, with OB = 800 mm vertical and BA = 600 mm horizontal. A 90 N force at A must produce zero moment about O.
For zero moment, the force's line of action must pass through O. The direction from A to O is (−600, 800), and the angle from horizontal is:
θ = arctan(800/600) = 53.1°
Worked example (Problem 23): Triangular structure with B at bottom-left, force F = 220 N at angle α = 33° applied at the top. Dimensions L1 = 2.2 m (height), L2 = 1.0 m, L3 = 0.7 m.
With the force application point at position (L3 + L2, L1) = (1.7, 2.2) relative to B, and force directed at 33° above horizontal:
Fx = 220 cos33° = 184.5 N Fy = 220 sin33° = 119.8 N M_B = (1.7)(119.8) − (2.2)(184.5) = 203.7 − 405.9 = −202.22 N·m (clockwise)
⚠️ The exact answer to Problem 23 depends on the precise geometry from the figure. Verify the coordinates of the force application point relative to B.
Equilibrium conditions (2D): ΣFx = 0, ΣFy = 0
Conservation of momentum: Σ(mᵢvᵢ)_before = Σ(mᵢvᵢ)_after
Moment (2D cross product): M_O = rx·Fy − ry·Fx
where r = position vector from O to the point of force application.
Moment magnitude (alternative): |M| = |F| × d
where d is the perpendicular distance from the pivot to the line of action of the force.
⚠️ In cable problems, tension always pulls along the cable toward its anchor. The direction of each tension force in your FBD must point from the junction toward the support.
⚠️ Sign conventions in momentum problems: pick a direction as positive and stick with it. A velocity "to the left" is negative if rightward is positive.
⚠️ When computing moments, the position vector r goes FROM the pivot TO the application point (not the other way round). Reversing it flips the sign.
⚠️ A moment of zero does not mean the force is zero. It means the line of action passes through the pivot point.
⚠️ For the baseball-cart problem, each ball adds mass to the cart. You can solve it as a single collision (total momentum of all 10 balls = final momentum of cart + balls) because the result is the same.
Q: Two cables support a weight at a point. Cable A makes 45° with horizontal, cable B makes 30° with horizontal. If the weight is 200 N, set up the equilibrium equations.
A: ΣFx: T_A cos45° − T_B cos30° = 0. ΣFy: T_A sin45° + T_B sin30° − 200 = 0. Solve these two equations for T_A and T_B.
Q: Block A (3 kg, 5 m/s right) hits block B (2 kg, at rest). After collision, A moves at 1 m/s right. What is B's velocity?
A: 3(5) + 2(0) = 3(1) + 2(v_B). So 15 = 3 + 2v_B, giving v_B = 6 m/s to the right.
Q: A 50 N force acts at point (3, 4) m. What is the moment about the origin?
A: You need the force direction. If F = (50cos θ, 50sin θ), then M = 3·(50 sin θ) − 4·(50 cos θ).
Q: When does a force produce zero moment about a point?
A: When the line of action of the force passes through that point, making the perpendicular distance d = 0.
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