MCAT Physics Question — Swing | Next Step Test Prep MCAT Physics Question — Swing | Next Step Test Prep

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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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