A child pushes, with F_{1}, his sister on a swing, generating torque, τ_{1} 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, F_{2}, that is directed at a 30º angle to the swing, what is the relationship between F_{1} and F_{2} if they generate the same torque?

A) F_{1} = F_{2}

B) 2.5 F_{1} = F_{2}

C) F_{1} = 2.5 F_{2}

D) 2 F_{1} = F_{2}

**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:

τ_{1} = r_{1}F_{1}

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:

τ_{2} = r_{2}F_{2} sin θ_{2}

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 r_{2} = 0.8r_{1} and the equation becomes:

τ_{2} = 0.8r_{1}F_{2} sin 30

τ_{2} = 0.8r_{1}F_{2} (0.5)

τ_{2} = 0.4r_{1}F_{2}

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

τ_{2} = 0.4r_{1}F_{2 }= τ_{1} = r_{1}F_{1}

0.4r_{1}F_{2 }= r_{1}F_{1}

0.4F_{2 }= F_{1}

So, the answer is:

**B) 2.5 F _{1} = F_{2}**

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