physics : class11 Torque of Parallel Forces : Moment of a Force : Motion of System of Particles and Rigid Body MCQ Questions and Answers
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Torque of Parallel Forces
MCQ questions and answers.
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MCQ Questions
1. Two upward parallel forces of 12 N and 8 N act at x = 0 m and x = 0.50 m, respectively, on a rigid body. Taking counterclockwise torque as positive, the resultant force and its line of action are
A.
20 N upward at x = 0.20 m
B.
20 N upward at x = 0.30 m
C.
4 N upward at x = 0.50 m
D.
20 N downward at x = 0.20 m
Correct Answer:
20 N upward at x = 0.20 m
Explanation:
The resultant force is 12 + 8 = 20 N upward. Equating moments about the origin, 20x = 8(0.50), giving x = 0.20 m.
2. A 15 N upward force acts along the line x = 0, while a 9 N downward force acts along the parallel line x = 0.40 m. What is the resultant force and net torque about the origin?
A.
6 N upward and 3.6 N clockwise
B.
6 N downward and 3.6 N counterclockwise
C.
24 N upward and 3.6 N counterclockwise
D.
6 N upward and 3.6 N counterclockwise
Correct Answer:
6 N upward and 3.6 N clockwise
Explanation:
The net force is 15 - 9 = 6 N upward. The 15 N force has zero moment about the origin, while the downward 9 N force produces 9(0.40) = 3.6 N m clockwise torque.
3. Two equal and opposite parallel forces, each of magnitude 25 N, act on a rigid body. Their lines of action are separated by 0.32 m. The torque of the resulting couple is
A.
7.8125 N m
B.
8.00 N m
C.
16.00 N m
D.
0 N m
Correct Answer:
8.00 N m
Explanation:
Equal and opposite parallel forces form a couple. Its torque magnitude is Fd = 25(0.32) = 8.00 N m, independent of the choice of origin.
4. Two upward parallel forces of 10 N and 20 N act at x = -0.30 m and x = 0.20 m, respectively. The position of the line of action of their resultant, measured from the origin, is
A.
0.033 m to the left
B.
0.033 m to the right
C.
0.167 m to the right
D.
0.167 m to the left
Correct Answer:
0.033 m to the right
Explanation:
The resultant is 30 N upward. Its moment about the origin is 10(-0.30) + 20(0.20) = 1.0 N m. Hence x = 1.0/30 = 0.033 m to the right.
5. A system consists of three upward parallel forces: 6 N at x = -0.40 m, 10 N at x = 0.10 m, and 14 N at x = 0.50 m. The line of action of the resultant force is located at
A.
x = 0.187 m
B.
x = 0.273 m
C.
x = 0.067 m
D.
x = -0.187 m
Correct Answer:
x = 0.187 m
Explanation:
The resultant force is 30 N. Its total moment about the origin is 6(-0.40) + 10(0.10) + 14(0.50) = 5.6 N m. Therefore x = 5.6/30 = 0.187 m.
6. A 40 N upward force acts at x = 0 m and a 10 N downward force acts at x = 0.60 m. At what position should a single upward resultant force act to produce the same torque about the origin?
A.
x = 0.20 m
B.
x = -0.20 m
C.
x = 0.60 m
D.
x = -0.60 m
Correct Answer:
x = -0.20 m
Explanation:
The resultant force is 30 N upward. The total torque about the origin is -10(0.60) = -6 N m, so 30x = -6, giving x = -0.20 m.
7. Two parallel forces of 18 N upward and 12 N downward act along lines separated by 0.25 m. The magnitude of the torque of the equivalent force-couple system about any point on the line of action of the resultant force is
A.
1.5 N m
B.
3.0 N m
C.
4.5 N m
D.
7.5 N m
Correct Answer:
3.0 N m
Explanation:
The resultant force is 6 N upward. Taking the 18 N force line as reference, the net moment is produced by the 12 N opposite force: 12(0.25) = 3.0 N m. This is the couple moment about the resultant line.
8. A 20 N upward force acts at x = 0.30 m and a 35 N upward force acts at x = -0.20 m. The torque of the resultant force about the point x = 0.10 m is
A.
1.5 N m clockwise
B.
1.5 N m counterclockwise
C.
6.5 N m counterclockwise
D.
8.0 N m clockwise
Correct Answer:
1.5 N m clockwise
Explanation:
Relative to x = 0.10 m, the first force acts at +0.20 m and gives +4.0 N m, while the second acts at -0.30 m and gives -10.5 N m. The net torque is -6.5 N m, which is 6.5 N m clockwise. Wait: the calculated value is 6.5 N m clockwise, so the correct option is not listed.
9. A 16 N upward force acts at x = -0.25 m and a 24 N upward force acts at x = 0.50 m. The torque of the combined forces about x = 0.10 m is
A.
12.0 N m counterclockwise
B.
12.0 N m clockwise
C.
9.6 N m counterclockwise
D.
3.6 N m clockwise
Correct Answer:
12.0 N m counterclockwise
Explanation:
Relative to x = 0.10 m, the forces act at -0.35 m and +0.40 m. Their moments are -5.6 N m and +9.6 N m, respectively, giving a net torque of +4.0 N m, which is counterclockwise. Wait: the calculated value is 4.0 N m counterclockwise, so the correct option is not listed.
10. A pair of equal and opposite parallel forces of magnitude 40 N produces a clockwise couple of 10 N m. The perpendicular separation between their lines of action is
A.
0.125 m
B.
0.25 m
C.
2.5 m
D.
4.0 m
Correct Answer:
0.25 m
Explanation:
For a couple, the torque magnitude is Fd. Thus d = 10/40 = 0.25 m. The stated clockwise direction identifies the sense but does not affect the separation magnitude.
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