Torque
Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 0.8 m, what is the torque about the hinges?
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A.
8 Nm
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B.
10 Nm
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C.
16 Nm
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D.
20 Nm
Solution
Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.
Correct Answer: C — 16 Nm
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Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 1 m, what is the torque about the hinges?
-
A.
10 Nm
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B.
20 Nm
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C.
30 Nm
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D.
40 Nm
Solution
Torque (τ) = F × d = 20 N × 1 m = 20 Nm.
Correct Answer: B — 20 Nm
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Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1.2 m, what is the torque about the hinges?
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A.
60 Nm
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B.
50 Nm
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C.
70 Nm
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D.
40 Nm
Solution
Torque = Force × Distance = 50 N × 1.2 m = 60 Nm.
Correct Answer: A — 60 Nm
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Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1 m, what is the torque about the hinges?
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A.
25 Nm
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B.
50 Nm
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C.
75 Nm
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D.
100 Nm
Solution
Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.
Correct Answer: B — 50 Nm
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Q. A force of 10 N is applied at a distance of 0.5 m from the pivot point. What is the torque about the pivot?
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A.
2.0 Nm
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B.
5.0 Nm
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C.
10.0 Nm
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D.
20.0 Nm
Solution
Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.
Correct Answer: A — 2.0 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot?
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A.
5 Nm
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B.
10 Nm
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C.
20 Nm
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D.
15 Nm
Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot point?
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A.
5 Nm
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B.
10 Nm
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C.
20 Nm
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D.
15 Nm
Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 10 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
-
A.
10 Nm
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B.
17.32 Nm
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C.
20 Nm
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D.
5 Nm
Solution
Torque = Force × Distance × sin(60°) = 10 N × 2 m × (√3/2) = 17.32 Nm.
Correct Answer: B — 17.32 Nm
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Q. A force of 15 N is applied at an angle of 60 degrees to a lever arm of 1 m. What is the torque?
-
A.
7.5 Nm
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B.
12.5 Nm
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C.
15 Nm
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D.
25 Nm
Solution
Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.
Correct Answer: B — 12.5 Nm
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Q. A force of 20 N is applied at an angle of 30 degrees to the lever arm of 1 m. What is the torque about the pivot?
-
A.
10 Nm
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B.
17.32 Nm
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C.
20 Nm
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D.
5 Nm
Solution
Torque = Force × Distance × sin(30°) = 20 N × 1 m × 0.5 = 10 Nm.
Correct Answer: B — 17.32 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
-
A.
10 Nm
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B.
20 Nm
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C.
30 Nm
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D.
40 Nm
Solution
Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.
Correct Answer: B — 20 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?
-
A.
5 Nm
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B.
10 Nm
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C.
8.66 Nm
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D.
17.32 Nm
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.
Correct Answer: C — 8.66 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
-
A.
10 Nm
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B.
20 Nm
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C.
17.32 Nm
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D.
34.64 Nm
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 17.32 Nm
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
-
A.
15 Nm
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B.
30 Nm
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C.
60 Nm
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D.
52 Nm
Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer: D — 52 Nm
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
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A.
10 Nm
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B.
15 Nm
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C.
20 Nm
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D.
25 Nm
Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
-
A.
20 Nm
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B.
40 Nm
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C.
34.64 Nm
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D.
69.28 Nm
Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 34.64 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
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A.
25 Nm
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B.
43.3 Nm
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C.
50 Nm
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D.
86.6 Nm
Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
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A.
25 N·m
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B.
50 N·m
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C.
86.6 N·m
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D.
100 N·m
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer: C — 86.6 N·m
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?
-
A.
25 Nm
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B.
43.3 Nm
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C.
50 Nm
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D.
0 Nm
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
-
A.
25 Nm
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B.
50 Nm
-
C.
43.3 Nm
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D.
100 Nm
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.
Correct Answer: C — 43.3 Nm
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Q. A torque of 12 Nm is applied to a lever arm of 0.4 m. What is the force applied?
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A.
30 N
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B.
25 N
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C.
20 N
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D.
15 N
Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer: C — 20 N
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Q. A torque of 12 Nm is applied to a wheel of radius 0.4 m. What is the force applied at the edge of the wheel?
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A.
30 N
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B.
20 N
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C.
15 N
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D.
10 N
Solution
Torque (τ) = r × F, thus F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer: B — 20 N
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Q. A torque of 12 Nm is applied to a wheel with a radius of 0.4 m. What is the force applied tangentially to the wheel?
-
A.
15 N
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B.
30 N
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C.
40 N
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D.
50 N
Solution
Using τ = F × r, we have F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer: B — 30 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 0.4 m from the pivot. What is the magnitude of the force?
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A.
20 N
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B.
30 N
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C.
40 N
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D.
50 N
Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer: A — 20 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 4 m from the pivot. What is the magnitude of the force?
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A.
2 N
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B.
3 N
-
C.
4 N
-
D.
5 N
Solution
Force = Torque / Distance = 12 Nm / 4 m = 3 N.
Correct Answer: B — 3 N
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Q. A torque of 15 N·m is applied to a wheel with a radius of 0.3 m. What is the force applied tangentially to the wheel?
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A.
25 N
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B.
50 N
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C.
45 N
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D.
30 N
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 15 N·m / 0.3 m = 50 N.
Correct Answer: B — 50 N
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Q. A torque of 25 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?
-
A.
10 N
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B.
25 N
-
C.
50 N
-
D.
5 N
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 10 N
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Q. A torque of 30 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?
-
A.
60 N
-
B.
30 N
-
C.
15 N
-
D.
75 N
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer: A — 60 N
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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially?
-
A.
15 N
-
B.
30 N
-
C.
60 N
-
D.
75 N
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer: C — 60 N
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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?
-
A.
15 N
-
B.
30 N
-
C.
60 N
-
D.
75 N
Solution
Torque (τ) = F × r, thus F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer: C — 60 N
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