Why is the doorknob placed far from the hinge?

Answer: It increases the torque for a given force (longer lever arm). A larger lever arm means a smaller force can produce the required torque to rotate the door open.

Two equal and opposite parallel forces act on a rigid body but their lines of action <strongdon't</strong coincide. The result is

Answer: Pure rotation (a couple). Equal and opposite forces with different lines of action form a "couple" — net force zero, but nonzero torque → pure rotational acceleration.

A 30\text{N} force is applied perpendicular to a 0.40\text{m} wrench. What torque results?

Answer: 12\text{N·m} \tau = r F = (0.40)(30) = 12\text{N·m}.

AP Physics 1 · Topic 5.3

Torque Practice

Q: Two equal and opposite parallel forces act on a rigid body but their lines of action <strongdon't</strong coincide. The result is

Answer: Pure rotation (a couple). Equal and opposite forces with different lines of action form a "couple" — net force zero, but nonzero torque → pure rotational acceleration.

Q: A 30\text{N} force is applied perpendicular to a 0.40\text{m} wrench. What torque results?

Answer: 12\text{N·m} \tau = r F = (0.40)(30) = 12\text{N·m}.

Part of Torque and Rotational Dynamics.(TOP-5.C)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$Why is the doorknob placed far from the hinge?$
- A
  $It looks better.$
- B
  $It reduces door weight.$
- C
  $It changes the moment of inertia.$
- D
  $It increases the torque for a given force (longer lever arm).$
  check_circle
Why
$A larger lever arm means a smaller force can produce the required torque to rotate the door open.$
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Sample 2difficulty 1/5
$Two equal and opposite parallel forces act on a rigid body but their lines of action <strong> don't </strong> coincide. The result is$
- A
  $No motion.$
- B
  $Pure rotation (a couple).$
  check_circle
- C
  $Both translation and rotation.$
- D
  $Pure translation.$
Why
$Equal and opposite forces with different lines of action form a "couple" — net force zero, but nonzero torque \to pure rotational acceleration.$
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Sample 3difficulty 1/5
$30~\text{N}$
- A
  $120~\text{N·m}$
- B
  $30~\text{N·m}$
- C
  $7.5~\text{N·m}$
- D
  $12~\text{N·m}$
  check_circle
Why
$\tau = r F = (0.40)(30) = 12~\text{N·m}$
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Sample 4difficulty 1/5
$40~\text{N}$
- A
  $40~\text{N·m}$
- B
  $10~\text{N·m}$
  check_circle
- C
  $160~\text{N·m}$
- D
  $1.6~\text{N·m}$
Why
$\tau = r F \sin\theta = (0.25)(40)(\sin 90^\circ) = 10~\text{N·m}$
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Sample 5difficulty 1/5
$20~\text{N}$
- A
  $4~\text{N·m}$
  check_circle
- B
  $8~\text{N·m}$
- C
  $0$
- D
  $2~\text{N·m}$
Why
$20 \cdot 0.30 = 6$
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