A child pushes, with F₁, his sister on a swing, generating torque, τ₁ which causes her to swing forward. If he then gets in a swing that hangs on a chain that is 20% shorter and his sister pushes with a force, F₂, that is directed at a 30º angle to the swing, what is the relationship between F₁ and F₂ if they generate the same torque?

A) F₁ = F₂

B) 2.5 F₁ = F₂

C) F₁ = 2.5 F₂

D) 2 F₁ = F₂

Explanation

The equation for torque is given by:

τ = rF sin θ

where r is the distance between the fulcrum and the force, F is the applied force, and θ is the angle between the applied force and the lever arm.

For the boy, we can write the following equation:

τ₁ = r₁F₁

The original question does not specify an angle for the boy. On the MCAT, if you’re not told otherwise then assume that θ is 90º in a torque problem and thus that sin θ is 1.

For the sister, we can write the following equation:

τ₂ = r₂F₂ sin θ₂

We are told that the length of the swing is 20% shorter and that she pushes with a force that is directed at 30º. Thus we may say that r₂ = 0.8r₁ and the equation becomes:

τ₂ = 0.8r₁F₂ sin 30

τ₂ = 0.8r₁F₂ (0.5)

τ₂ = 0.4r₁F₂

Finally, we’re told that both torques are the same, so we may set them equal:

τ₂ = 0.4r₁F₂ = τ₁ = r₁F₁

0.4r₁F₂ = r₁F₁

0.4F₂ = F₁

So, the answer is:

B) 2.5 F₁ = F₂

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