A uniform bar of mass m is supported by a pivot at its top, about which the bar can swing like a pendulum. If a force F is applied perpendicularly to the lower end of the bar as in the diagram, how big must F be in order to hold the bar in equilibrium at an angle θ from the vertical?

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Question restatement: A uniform bar of mass m is supported by a pivot at its top, about which the bar can swing like a pendulum. If a force F is applied perpendicularly to the lower end of the bar as shown, how big must F be in order to hold the bar in equilibrium at an angle θ from the vertical?

Answer options to analyze:
- 2 m g cos θ
- (1/2) m g sin θ
- (1/2) m g cos θ
- 2 m g sin θ

Analysis of each option:
Option 1: 2 m g cos θ. To check equilibrium, examine torques about the pivot. The weight mg acts at the bar's center of mass, a distance L/2 from the pivot along the bar. The perpendicular distance (lever arm) from the weight line to the pivot is (L/2) sin θ, so the torque due to weight is mg × (L/2) sin θ, tending to rotate the bar. The force F, applied at the bottom and perpendicular to the bar, provides a torque of F × L, tending in the opposite direction. If we set torque balance F × L = mg × (L/2) sin θ, we obtain F = (mg/2) sin θ. This result does not produce a cos θ term, so the option containing cos θ is inconsistent with the torque balance for this setup. Therefore, this option is incorrect.
Option 2: (1/2) m g sin θ. This matches the torque balance derived above: F × L = mg × (L/2) sin θ gives F = (mg/2) sin θ. The factors align correctly: a half factor arises from the center-of-mass distance L/2, and the sine factor arises from the perpendicular distance of the weight’s line of action to the pivot, which depends on the angle θ between the bar and the vertical. This option is consistent with rotational equilibrium for the described geometry.
Option 3: (1/2) m g cos θ. The cosine dependence would imply the lever arm for the weight is proportional to cos θ, which would correspond to a different geometric arrangement (for example, measuring the perpendicular distance incorrectly or using the angle relative to a different reference). Given the setup with θ measured from vertical, the weight torque scales with sin θ, not cos θ. Hence this option misrepresents the angular dependence and is incorrect.
Option 4: 2 m g sin θ. This would overestimate the required force by a factor of four relative to the correct torque balance: F should produce a torque equal to the weight’s torque, which is mg × (L/2) sin θ, not twice that value. The presence of 2 mg sin θ would imply F × L equals 2 mg × (L/2) sin θ = mg L sin θ, doubling the necessary torque. Therefore, this option is incorrect.

In summary, the key factors are: torque balance about the pivot, the center-of-mass location at L/2 along the bar giving a lever arm of (L/2) sin θ for the weight, and the force F acting at the bottom with a lever arm L since it is applied perpendicular to the bar. These lead to F = (mg/2) sin θ, corresponding to the option (1/2) m g sin θ.

SPHY001 Graded Quiz 5 (Unit 9 and Unit 10)- Requires Respondus LockDown Browser

A uniform bar of mass m is supported by a pivot at its top, about which the bar can swing like a pendulum. If a force F is applied perpendicularly to the lower end of the bar as in the diagram, how big must F be in order to hold the bar in equilibrium at an angle θ from the vertical?

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