MCAT Physics Question — Swing
 by
 Nov 20, 2014
 MCAT Blog, MCAT Physics, MCAT Question of the Day
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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