Statics Quiz: Particle Equilibrium 3d
4 questions · exam conditions
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Particle Equilibrium 3dQuestion 1 of 4

A particle is subjected to four forces in 3D space. Three of the forces are: F1⃗=20i^+15j^−10k^\vec{F_1} = 20\hat{i} + 15\hat{j} - 10\hat{k} N, F2⃗=−12i^+8j^+25k^\vec{F_2} = -12\hat{i} + 8\hat{j} + 25\hat{k} N, and F3⃗=5i^−20j^+12k^\vec{F_3} = 5\hat{i} - 20\hat{j} + 12\hat{k} N. For equilibrium, what must be the fourth force F4⃗\vec{F_4}?

−13i^−3j^−27k^-13\hat{i} - 3\hat{j} - 27\hat{k} N
13i^+3j^+27k^13\hat{i} + 3\hat{j} + 27\hat{k} N
−13i^+3j^−27k^-13\hat{i} + 3\hat{j} - 27\hat{k} N
13i^−3j^+27k^13\hat{i} - 3\hat{j} + 27\hat{k} N
−13i^−3j^+27k^-13\hat{i} - 3\hat{j} + 27\hat{k} N
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Statics Quiz: Particle Equilibrium 3d

Practice Particle 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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Question 1

A particle is subjected to four forces in 3D space. Three of the forces are: F1⃗=20i^+15j^−10k^\vec{F_1} = 20\hat{i} + 15\hat{j} - 10\hat{k} N, F2⃗=−12i^+8j^+25k^\vec{F_2} = -12\hat{i} + 8\hat{j} + 25\hat{k} N, and F3⃗=5i^−20j^+12k^\vec{F_3} = 5\hat{i} - 20\hat{j} + 12\hat{k} N. For equilibrium, what must be the fourth force F4⃗\vec{F_4}?

  1. −13i^−3j^−27k^-13\hat{i} - 3\hat{j} - 27\hat{k} N (correct answer)
  2. 13i^+3j^+27k^13\hat{i} + 3\hat{j} + 27\hat{k} N
  3. −13i^+3j^−27k^-13\hat{i} + 3\hat{j} - 27\hat{k} N
  4. 13i^−3j^+27k^13\hat{i} - 3\hat{j} + 27\hat{k} N
  5. −13i^−3j^+27k^-13\hat{i} - 3\hat{j} + 27\hat{k} N
Explanation: When you encounter equilibrium problems in 3D space, remember that the fundamental principle is that the sum of all forces must equal zero. This means ∑F⃗=0⃗\sum \vec{F} = \vec{0}, or equivalently, F1⃗+F2⃗+F3⃗+F4⃗=0⃗\vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} = \vec{0}. To find F4⃗\vec{F_4}, you need to rearrange this equation: F4⃗=−(F1⃗+F2⃗+F3⃗)\vec{F_4} = -(\vec{F_1} + \vec{F_2} + \vec{F_3}). First, add the three given forces component by component: F1⃗+F2⃗+F3⃗=(20−12+5)i^+(15+8−20)j^+(−10+25+12)k^\vec{F_1} + \vec{F_2} + \vec{F_3} = (20-12+5)\hat{i} + (15+8-20)\hat{j} + (-10+25+12)\hat{k} =13i^+3j^+27k^= 13\hat{i} + 3\hat{j} + 27\hat{k} Therefore: F4⃗=−(13i^+3j^+27k^)=−13i^−3j^−27k^\vec{F_4} = -(13\hat{i} + 3\hat{j} + 27\hat{k}) = -13\hat{i} - 3\hat{j} - 27\hat{k} N Answer A gives exactly this result. Answer B represents the sum of the first three forces without the negative sign—this would actually double the net force instead of achieving equilibrium. Answer C incorrectly negates only the i and k components while keeping the j component positive, suggesting a sign error in calculations. Answer D has the wrong signs on all components, representing the opposite of what's needed for equilibrium. For equilibrium problems, always remember that the unknown force must be equal and opposite to the resultant of all other forces. Double-check your arithmetic by verifying that when you add all four forces together, each component (i, j, k) sums to exactly zero.

Question 2

A particle at the origin is held in equilibrium by three forces. Force A has magnitude 150 N and acts along the line from origin to point (3, 4, 0). Force B has magnitude 200 N and acts along the line from origin to point (0, 3, 4). What must be the components of force C to maintain equilibrium?

  1. Cx=−90C_x = -90 N, Cy=−240C_y = -240 N, Cz=−160C_z = -160 N (correct answer)
  2. Cx=−90C_x = -90 N, Cy=−120C_y = -120 N, Cz=−160C_z = -160 N
  3. Cx=90C_x = 90 N, Cy=240C_y = 240 N, Cz=160C_z = 160 N
  4. Cx=−45C_x = -45 N, Cy=−180C_y = -180 N, Cz=−120C_z = -120 N
  5. Cx=−135C_x = -135 N, Cy=−180C_y = -180 N, Cz=−80C_z = -80 N
Explanation: When you encounter equilibrium problems with multiple forces, remember that the fundamental principle is that all forces must sum to zero in each direction. This means ∑Fx=0\sum F_x = 0, ∑Fy=0\sum F_y = 0, and ∑Fz=0\sum F_z = 0. To solve this problem, you first need to find the unit vectors for forces A and B, then convert them to component form. For Force A acting toward point (3, 4, 0), the distance is 32+42=5\sqrt{3^2 + 4^2} = 5. So the unit vector is (35,45,0)(\frac{3}{5}, \frac{4}{5}, 0), making Force A = (90,120,0)(90, 120, 0) N. For Force B acting toward point (0, 3, 4), the distance is 32+42=5\sqrt{3^2 + 4^2} = 5. The unit vector is (0,35,45)(0, \frac{3}{5}, \frac{4}{5}), making Force B = (0,120,160)(0, 120, 160) N. For equilibrium, Force C must balance the sum of A and B: Cx=−(90+0)=−90C_x = -(90 + 0) = -90 N, Cy=−(120+120)=−240C_y = -(120 + 120) = -240 N, and Cz=−(0+160)=−160C_z = -(0 + 160) = -160 N. This confirms answer A is correct. Answer B incorrectly calculates Cy=−120C_y = -120 N, likely from forgetting to include both y-components. Answer C has the right magnitudes but wrong signs—this would create a net force of twice the original instead of equilibrium. Answer D appears to use incorrect force magnitudes, possibly from calculation errors in the unit vectors. Remember: always double-check your unit vector calculations and verify that your final answer actually produces zero net force in all three directions.

Question 3

A mass hangs from three cables meeting at point O. The cables make the following angles with the vertical: Cable 1: 30° in a plane making 0° with the x-axis, Cable 2: 45° in a plane making 120° with the x-axis, Cable 3: 60° in a plane making 240° with the x-axis. If the mass is 1000 N, what is the tension in Cable 1?

  1. 577577 N
  2. 667667 N (correct answer)
  3. 756756 N
  4. 845845 N
  5. 933933 N
Explanation: When you encounter a 3D static equilibrium problem with cables, you need to resolve forces in all three coordinate directions and apply the equilibrium condition that the sum of forces equals zero in each direction. First, convert each cable's orientation to 3D force components. For Cable 1 (30° from vertical, 0° plane): the components are T1(sin⁡30°cos⁡0°,sin⁡30°sin⁡0°,cos⁡30°)=T1(0.5,0,0.866)T_1(\sin 30°\cos 0°, \sin 30°\sin 0°, \cos 30°) = T_1(0.5, 0, 0.866). For Cable 2 (45° from vertical, 120° plane): T2(−0.354,0.612,0.707)T_2(-0.354, 0.612, 0.707). For Cable 3 (60° from vertical, 240° plane): T3(−0.25,−0.433,0.5)T_3(-0.25, -0.433, 0.5). Setting up equilibrium equations:
  • x-direction: 0.5T1−0.354T2−0.25T3=00.5T_1 - 0.354T_2 - 0.25T_3 = 0
  • y-direction: 0.612T2−0.433T3=00.612T_2 - 0.433T_3 = 0
  • z-direction: 0.866T1+0.707T2+0.5T3=10000.866T_1 + 0.707T_2 + 0.5T_3 = 1000
Solving the y-equation gives T3=1.414T2T_3 = 1.414T_2. Substituting into the x-equation yields T1=0.354T2T_1 = 0.354T_2. Using the z-equation: 0.866(0.354T2)+0.707T2+0.5(1.414T2)=10000.866(0.354T_2) + 0.707T_2 + 0.5(1.414T_2) = 1000, which gives T2=667T_2 = 667 N, so T1=667T_1 = 667 N. The correct answer is B) 667 N. A) 577 N represents using incorrect trigonometric relationships. C) 756 N likely comes from neglecting one force direction. D) 845 N suggests errors in the 3D geometry setup. For 3D cable problems, always establish your coordinate system clearly, convert angles to components systematically, and solve the three equilibrium equations simultaneously—never assume symmetry without verification.

Question 4

A particle is in equilibrium under four coplanar forces acting in the xy-plane and one force acting in the z-direction. The xy-plane forces are: F1⃗=60i^+80j^\vec{F_1} = 60\hat{i} + 80\hat{j} N, F2⃗=−40i^+30j^\vec{F_2} = -40\hat{i} + 30\hat{j} N, F3⃗=−25i^−45j^\vec{F_3} = -25\hat{i} - 45\hat{j} N, F4⃗=5i^−65j^\vec{F_4} = 5\hat{i} - 65\hat{j} N. What must be the z-component of the fifth force?

  1. 00 N (correct answer)
  2. 1515 N
  3. −15-15 N
  4. 2525 N
  5. −25-25 N
Explanation: When you encounter equilibrium problems with multiple forces, remember that equilibrium means the net force in every direction must be zero. This applies separately to each coordinate direction. Since the particle is in equilibrium, the sum of all force components in each direction must equal zero. Let's first examine what happens in the xy-plane by adding up all the given forces: Fxy⃗=(60−40−25+5)i^+(80+30−45−65)j^=0i^+0j^\vec{F_{xy}} = (60-40-25+5)\hat{i} + (80+30-45-65)\hat{j} = 0\hat{i} + 0\hat{j} The x-components sum to zero: 60−40−25+5=060-40-25+5 = 0 The y-components sum to zero: 80+30−45−65=080+30-45-65 = 0 Since the four coplanar forces already balance perfectly in the xy-plane, the fifth force cannot have any x or y components—it can only act in the z-direction. More importantly, since there are no other z-direction forces acting on the particle, the z-component of the fifth force must also be zero for equilibrium in the z-direction. Answer A (00 N) is correct because equilibrium requires zero net force in all directions, and no other z-forces exist to balance against. Answer B (1515 N) would create an unbalanced upward force, violating equilibrium. Answer C (−15-15 N) would create an unbalanced downward force, also violating equilibrium. Answer D (2525 N) would similarly create an unbalanced upward force. Study tip: In equilibrium problems, always check each coordinate direction separately. If forces in one direction already balance without involving a particular force component, that component must be zero.