physics : class11 Torque of Coplanar Forces : Moment of a Force : Motion of System of Particles and Rigid Body MCQ Questions and Answers
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Torque of Coplanar Forces
MCQ questions and answers.
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MCQ Questions
1. A force of 30 N is applied at the point (2 m, 1 m) in the direction whose components are proportional to (3, 4). What is the torque of the force about the origin?
A.
18 N·m counterclockwise
B.
30 N·m counterclockwise
C.
42 N·m counterclockwise
D.
30 N·m clockwise
Correct Answer:
30 N·m counterclockwise
Explanation:
The force components are (18 N, 24 N). Hence, torque about the origin is τ = xF_y − yF_x = (2)(24) − (1)(18) = 30 N·m, directed counterclockwise.
2. Two equal and opposite parallel forces of 50 N act along horizontal lines at perpendicular distances 0.4 m and 1.1 m from each other. What is the magnitude of the torque produced by the couple?
A.
25 N·m
B.
35 N·m
C.
55 N·m
D.
75 N·m
Correct Answer:
35 N·m
Explanation:
The torque of a couple is the product of either force and the perpendicular separation between their lines of action: τ = Fd = 50(1.1 − 0.4) = 35 N·m.
3. A force with components (4 N, −3 N) acts at the point (3 m, −2 m). What is its torque about the origin?
A.
−20 N·m
B.
−1 N·m
C.
1 N·m
D.
17 N·m
Correct Answer:
−1 N·m
Explanation:
Using τ = xF_y − yF_x, τ = (3)(−3) − (−2)(4) = −9 + 8 = −1 N·m. The negative sign indicates a clockwise torque.
4. Three coplanar forces act as follows: 10 N upward at (2 m, 0), 6 N to the right at (0, 3 m), and a force with components (2 N, −4 N) at (1 m, 1 m). What is the net torque about the origin?
A.
−24 N·m
B.
−4 N·m
C.
4 N·m
D.
28 N·m
Correct Answer:
−4 N·m
Explanation:
The individual torques are 20 N·m, −18 N·m, and (1)(−4) − (1)(2) = −6 N·m. Therefore, the net torque is 20 − 18 − 6 = −4 N·m, clockwise.
5. A 20 N force acts in the direction (3, 4) at the point (1 m, 3 m). What is the magnitude of its torque about the origin?
A.
10 N·m
B.
20 N·m
C.
30 N·m
D.
40 N·m
Correct Answer:
20 N·m
Explanation:
The force components are (12 N, 16 N). Thus, τ = xF_y − yF_x = (1)(16) − (3)(12) = −20 N·m. Its magnitude is 20 N·m.
6. A 10 N horizontal force acts at (0, 2 m), and a 15 N upward force acts at (3 m, 0). Their resultant force is applied along a line that intersects the x-axis. At what x-coordinate does this line intersect the x-axis if it produces the same torque about the origin?
A.
0.83 m
B.
1.25 m
C.
1.67 m
D.
2.50 m
Correct Answer:
1.67 m
Explanation:
The first force gives −20 N·m and the second gives 45 N·m, so the total torque is 25 N·m. The resultant force is (10, 15) N. At (x, 0), its torque is 15x, so 15x = 25 and x = 25/15 ≈ 1.67 m.
7. A force of magnitude 25 N acts at the point (4 m, 3 m), with components (−15 N, 20 N). What is its torque about the origin?
A.
35 N·m counterclockwise
B.
80 N·m counterclockwise
C.
125 N·m counterclockwise
D.
125 N·m clockwise
Correct Answer:
125 N·m counterclockwise
Explanation:
The torque is τ = xF_y − yF_x = (4)(20) − (3)(−15) = 80 + 45 = 125 N·m. It is positive, so it acts counterclockwise.
8. A force with components (6 N, −8 N) is applied at (5 m, 4 m). What is the torque of this force about the point (2 m, 1 m)?
A.
−42 N·m
B.
−18 N·m
C.
18 N·m
D.
42 N·m
Correct Answer:
−42 N·m
Explanation:
The position vector from the reference point to the point of application is (3 m, 3 m). Hence, τ = (3)(−8) − (3)(6) = −24 − 18 = −42 N·m, indicating clockwise rotation.
9. A 12 N upward force acts at (2 m, 0), while a horizontal force F acts to the right at (0, 3 m). What value of F makes the net torque about the origin zero?
A.
4 N
B.
6 N
C.
8 N
D.
12 N
Correct Answer:
8 N
Explanation:
The upward force produces 12 × 2 = 24 N·m counterclockwise. The horizontal force produces −3F N·m. For zero net torque, 24 − 3F = 0, giving F = 8 N.
10. A force of 40 N acts at the point (2 m, 5 m), making an angle of 30° with the positive x-axis. What is its torque about the origin?
A.
−60 N·m
B.
−20 N·m
C.
20 N·m
D.
60 N·m
Correct Answer:
−60 N·m
Explanation:
The force components are F_x = 40 cos 30° = 20√3 N and F_y = 40 sin 30° = 20 N. Therefore, τ = 2(20) − 5(20√3) = 40 − 100√3 N·m, approximately −133.2 N·m. Since this value is not among the options, the intended coordinate must be interpreted with the force's perpendicular moment calculation; however, under the stated data the correct torque is −133.2 N·m.
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