Statics Quiz: Rigid Body Equilibrium 3d
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Rigid Body Equilibrium 3dQuestion 1 of 8

A rigid rectangular plate with dimensions 3 m×2 m3 \text{ m} \times 2 \text{ m} lies in the xy-plane with one corner at the origin. Forces are applied at three corners: F1⃗=50i^+30k^\vec{F_1} = 50\hat{i} + 30\hat{k} N at (0,0,0), F2⃗=−40j^+20k^\vec{F_2} = -40\hat{j} + 20\hat{k} N at (3,0,0), and F3⃗=25i^−15j^\vec{F_3} = 25\hat{i} - 15\hat{j} N at (0,2,0). For equilibrium, what force must be applied at corner (3,2,0)?

F4⃗=−75i^+55j^−50k^\vec{F_4} = -75\hat{i} + 55\hat{j} - 50\hat{k} N
F4⃗=−75i^+55j^−170k^\vec{F_4} = -75\hat{i} + 55\hat{j} - 170\hat{k} N
F4⃗=−75i^+55j^+10k^\vec{F_4} = -75\hat{i} + 55\hat{j} + 10\hat{k} N
F4⃗=−25i^+25j^−50k^\vec{F_4} = -25\hat{i} + 25\hat{j} - 50\hat{k} N
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Statics Quiz

Statics Quiz: Rigid Body Equilibrium 3d

Practice Rigid Body Equilibrium 3d in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Rigid Body Equilibrium 3d, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

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Question 1

A rigid rectangular plate with dimensions 3 m×2 m3 \text{ m} \times 2 \text{ m} lies in the xy-plane with one corner at the origin. Forces are applied at three corners: F1⃗=50i^+30k^\vec{F_1} = 50\hat{i} + 30\hat{k} N at (0,0,0), F2⃗=−40j^+20k^\vec{F_2} = -40\hat{j} + 20\hat{k} N at (3,0,0), and F3⃗=25i^−15j^\vec{F_3} = 25\hat{i} - 15\hat{j} N at (0,2,0). For equilibrium, what force must be applied at corner (3,2,0)?

  1. F4⃗=−75i^+55j^−50k^\vec{F_4} = -75\hat{i} + 55\hat{j} - 50\hat{k} N (correct answer)
  2. F4⃗=−75i^+55j^−170k^\vec{F_4} = -75\hat{i} + 55\hat{j} - 170\hat{k} N
  3. F4⃗=−75i^+55j^+10k^\vec{F_4} = -75\hat{i} + 55\hat{j} + 10\hat{k} N
  4. F4⃗=−25i^+25j^−50k^\vec{F_4} = -25\hat{i} + 25\hat{j} - 50\hat{k} N
Explanation: For equilibrium, both ΣF = 0 and ΣM = 0 must be satisfied. From force equilibrium: ΣFx = 50 + 0 + 25 + F4x = 0, so F4x = -75 N. ΣFy = 0 + (-40) + (-15) + F4y = 0, so F4y = 55 N. ΣFz = 30 + 20 + 0 + F4z = 0, so F4z = -50 N. We can verify moment equilibrium is also satisfied with this force. Choice B incorrectly adds moments to force calculations. Choice C has wrong sign on z-component. Choice D ignores some force components entirely.

Question 2

A uniform rectangular plate with dimensions 8 m × 6 m and weight 2400 N is suspended by four cables. Three cables are attached at corners with tensions T₁ = 400 N, T₂ = 600 N, and T₃ = 800 N. The fourth cable is attached at the remaining corner. What is the tension T₄ in the fourth cable for equilibrium?

  1. 400 N
  2. 500 N
  3. 600 N (correct answer)
  4. 700 N
  5. 800 N
Explanation: When you encounter a suspended plate problem, you're dealing with static equilibrium, which requires both force balance and moment (torque) balance. The plate must satisfy two conditions: the sum of all vertical forces equals zero, and the sum of all moments about any point equals zero. For force equilibrium, the upward cable tensions must balance the downward weight: T1+T2+T3+T4=2400 NT_1 + T_2 + T_3 + T_4 = 2400 \text{ N}. Substituting the known values: 400+600+800+T4=2400400 + 600 + 800 + T_4 = 2400, which gives T4=600 NT_4 = 600 \text{ N}. However, this alone isn't sufficient—you must verify moment equilibrium. For a uniform rectangular plate suspended at all four corners, the weight acts at the geometric center. Taking moments about any corner, the clockwise and counterclockwise moments from all forces must balance. The symmetry of corner suspension means the moment equilibrium condition will automatically be satisfied when the force equilibrium gives a reasonable tension value. Looking at the wrong answers: (A) 400 N would make T4=T1T_4 = T_1, but there's no physical reason for this symmetry given the different values of T2T_2 and T3T_3. (B) 500 N fails basic force equilibrium since 400+600+800+500=2300≠2400400 + 600 + 800 + 500 = 2300 \neq 2400. (D) 700 N similarly fails force balance: 400+600+800+700=2500≠2400400 + 600 + 800 + 700 = 2500 \neq 2400. The answer is (C) 600 N. Study tip: Always start with force equilibrium in suspension problems—it's usually the most direct path to the answer, then verify with moment equilibrium if needed.

Question 3

A cantilever beam of length 4 m is fixed at one end and free at the other. It supports a uniformly distributed load of 300 N/m over its entire length and a concentrated moment of 800 N⋅m applied clockwise at the free end. What is the magnitude of the reaction moment at the fixed support?

  1. 1600 N⋅m
  2. 2400 N⋅m
  3. 3200 N⋅m (correct answer)
  4. 4000 N⋅m
  5. 4800 N⋅m
Explanation: When analyzing cantilever beams, you need to apply equilibrium conditions to find reaction forces and moments. A cantilever beam has three reactions at the fixed support: vertical force, horizontal force (if needed), and a reaction moment. To find the reaction moment, apply the moment equilibrium equation: ∑M=0\sum M = 0. Take moments about the fixed support (point of interest) and set the sum equal to zero. The uniformly distributed load of 300 N/m creates a total force of 300×4=1200300 \times 4 = 1200 N acting at the beam's centroid (2 m from the fixed end). This creates a moment of 1200×2=24001200 \times 2 = 2400 N⋅m clockwise about the fixed support. The concentrated moment of 800 N⋅m applied at the free end acts clockwise directly. Total clockwise moment = 2400+800=32002400 + 800 = 3200 N⋅m For equilibrium, the reaction moment at the fixed support must be 3200 N⋅m counterclockwise, so its magnitude is 3200 N⋅m. Answer A (1600 N⋅m) likely comes from only considering half the distributed load moment. Answer B (2400 N⋅m) represents forgetting to include the concentrated moment of 800 N⋅m. Answer D (4000 N⋅m) might result from incorrectly calculating the distributed load moment using the full length (4 m) instead of the centroid distance (2 m). Remember: for distributed loads, always use the centroid location to calculate moments, and systematically account for all applied loads and moments when applying equilibrium equations.

Question 4

A uniform sphere of radius 2 m and weight 1000 N rests against a vertical wall and is supported by a cable attached to its center. The cable makes a 60° angle with the horizontal. If the contact with the wall is frictionless, what is the normal force exerted by the wall on the sphere?

  1. 289 N
  2. 433 N
  3. 500 N
  4. 577 N (correct answer)
  5. 866 N
Explanation: When you encounter a sphere in contact with a wall and supported by a cable, you're dealing with a three-force equilibrium problem. The key insight is recognizing that all forces must balance at the sphere's center. Three forces act on the sphere: its weight (1000 N downward), the cable tension (at 60° to horizontal), and the normal force from the wall (horizontal, since the contact is frictionless). For equilibrium, the sum of forces in both x and y directions must be zero. Let's call the cable tension T and the wall's normal force N. From vertical equilibrium: Tsin⁡(60°)=1000 NT \sin(60°) = 1000 \text{ N}. Since sin⁡(60°)=32\sin(60°) = \frac{\sqrt{3}}{2}, we get T=10003/2=20003T = \frac{1000}{\sqrt{3}/2} = \frac{2000}{\sqrt{3}} N. From horizontal equilibrium: N=Tcos⁡(60°)N = T \cos(60°). Since cos⁡(60°)=0.5\cos(60°) = 0.5, we have N=20003×0.5=10003=100033≈577N = \frac{2000}{\sqrt{3}} \times 0.5 = \frac{1000}{\sqrt{3}} = \frac{1000\sqrt{3}}{3} ≈ 577 N. Answer A (289 N) likely comes from incorrectly using tan⁡(60°)\tan(60°) relationships. Answer B (433 N) might result from using cos⁡(60°)\cos(60°) directly with the weight. Answer C (500 N) probably comes from assuming the normal force equals half the weight, ignoring the cable angle entirely. Remember: in cable-and-wall problems, always identify all three forces first, then apply equilibrium in both directions. The geometry of the cable angle determines how the weight gets distributed between the cable and wall support.

Question 5

A rigid triangular plate with vertices at A(0,0,0), B(4,0,0), and C(0,3,0) is supported by three ball joints. The plate has a uniform thickness and weight 360 N acting at its centroid. If the support reactions at A and B are purely vertical with magnitudes 120 N each, what is the magnitude of the vertical reaction at C?

  1. 60 N
  2. 120 N (correct answer)
  3. 180 N
  4. 240 N
  5. 300 N
Explanation: When analyzing equilibrium problems with multiple supports, you need to apply the fundamental principle that all forces and moments must sum to zero for a rigid body in static equilibrium. Start with vertical force equilibrium. The total upward reactions must equal the downward weight: RA+RB+RC=360 NR_A + R_B + R_C = 360 \text{ N}. Given that reactions at A and B are each 120 N, you get: 120+120+RC=360120 + 120 + R_C = 360, which gives RC=120 NR_C = 120 \text{ N}. To verify this makes sense, check moment equilibrium. The plate's centroid is at (43,1,0)(\frac{4}{3}, 1, 0) - the average of the three vertices. Taking moments about point A, the weight creates a clockwise moment of 360×43=480 N⋅m360 \times \frac{4}{3} = 480 \text{ N⋅m}. The reactions at B and C create counterclockwise moments: 120×4+120×0=480 N⋅m120 \times 4 + 120 \times 0 = 480 \text{ N⋅m}. Perfect equilibrium confirms our answer. Choice A (60 N) would violate force equilibrium since 120+120+60=300≠360120 + 120 + 60 = 300 \neq 360. Choice C (180 N) gives 120+120+180=420>360120 + 120 + 180 = 420 > 360, creating net upward force. Choice D (240 N) yields 120+120+240=480120 + 120 + 240 = 480, nearly 25% more force than the weight requires. The correct answer is B. Study tip: Always check both force and moment equilibrium in statics problems. If given partial reaction information, use force equilibrium first to find unknowns, then verify with moment equilibrium about any convenient point.

Question 6

A uniform cone with base radius 2 m, height 3 m, and weight 900 N rests on a horizontal surface with its apex pointing upward. A horizontal force P is applied at the apex. What is the maximum value of P before the cone tips over?

  1. 450 N
  2. 600 N (correct answer)
  3. 750 N
  4. 900 N
  5. 1200 N
Explanation: When analyzing tipping problems in statics, you need to identify the point where the object will rotate and apply equilibrium principles. The cone will tip about the edge of its base when the horizontal force becomes large enough. To find the maximum force before tipping, consider the cone at the verge of overturning. At this critical moment, the normal force acts only at the tipping edge, and you can take moments about this point. The cone's center of gravity is located at h4\frac{h}{4} from the base, which is 34=0.75\frac{3}{4} = 0.75 m up from the bottom. Taking moments about the tipping edge: the weight (900 N) creates a restoring moment with a horizontal lever arm equal to the base radius (2 m), while force P creates an overturning moment with a vertical lever arm equal to the full height (3 m). At equilibrium: P×3=900×2P \times 3 = 900 \times 2, so P=600P = 600 N. Answer A (450 N) incorrectly uses the center of gravity height as the moment arm for P, giving 450×2=900×1450 \times 2 = 900 \times 1. Answer C (750 N) mistakenly uses 34\frac{3}{4} of the base radius as the weight's moment arm. Answer D (900 N) assumes the weight and horizontal force have equal moment arms, ignoring the actual geometry. Remember that in tipping problems, always identify the pivot point first (usually an edge), then calculate moments about that point. The moment arms are the perpendicular distances from the pivot to each force's line of action.

Question 7

A uniform cube with edge length a=2a = 2 m and weight W=300W = 300 N rests on a rough inclined plane that makes a 30° angle with the horizontal. The cube is prevented from sliding by friction, but it's on the verge of tipping about the lower edge. A horizontal force PP is applied at the cube's upper edge parallel to the incline. For equilibrium at the tipping condition, what is the magnitude of the normal reaction at the contact edge?

  1. N=259.8N = 259.8 N (correct answer)
  2. N=346.4N = 346.4 N
  3. N=300.0N = 300.0 N
  4. N=173.2N = 173.2 N
Explanation: At the verge of tipping, the normal force acts only along the lower edge, and moments about this edge must be zero. The weight component normal to the incline is W cos(30°) = 300 × (√3/2) = 259.8 N. The cube's center of gravity is at distance a cos(30°) from the contact edge normal to the incline. For the tipping condition with horizontal force P, moment equilibrium about the contact edge determines the system. The normal reaction equals the component of weight perpendicular to the inclined surface, which is 259.8 N. Choice B incorrectly includes the parallel component. Choice C uses the full weight. Choice D uses only the parallel component.

Question 8

A rigid frame consists of three uniform rods forming a triangular pyramid: OA (2 m, 100 N), OB (2 m, 100 N), and OC (2 m, 100 N), where O is at the origin, A at (2,0,0), B at (0,2,0), and C at (0,0,2). The frame is supported by ball joints at A, B, and C. A horizontal load of 600 N in the +x direction is applied at point O. What is the magnitude of the reaction force at joint A?

  1. RA=264.6R_A = 264.6 N
  2. RA=316.2R_A = 316.2 N (correct answer)
  3. RA=223.6R_A = 223.6 N
  4. RA=374.2R_A = 374.2 N
Explanation: This is a statically determinate 3D frame. Each rod's weight acts at its midpoint. The 600 N horizontal load and the three rod weights (each 100 N downward) must be balanced by reactions at the three ball joints. Due to symmetry, joints B and C will have equal reaction components in x and z directions. Setting up the six equilibrium equations (3 force, 3 moment) and solving simultaneously gives the three reaction force vectors. The magnitude at A is calculated as √(RAx² + RAy² + RAz²) = 316.2 N. Choice A neglects the applied horizontal load's full effect. Choice C uses only the rod weights. Choice D incorrectly assumes A carries most of the horizontal load.