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Midterm problem!

  1. Jan 25, 2005 #1

    My midterm is tomorrow and i dont know how to do this! See if one of you does.

    A 850 N person sits 1.2 m from one end of a swing (5 m long) supported at each end by a chain. The swing weighs 250 N and there is a 40 N can of soda .5 m from the end opposite the person. How much force does each chain exert?

    Any and all help is greatly appreciated!!

  2. jcsd
  3. Jan 25, 2005 #2


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    Wo- that's a toughy :tongue2:. Here's what I'd do.

    There are two things to help us understand the problem here.

    First, the center of mass (CM) of the swing is located at its geometrical center, (i.e. 2.5m from each end) and it serves as its axis of rotation. Now, clearly the swing is not more rotating than it is accelerating linearly, so the sum of the torques about the CM must be 0. The person exerts a positive torque about the CM, and the can exerts a negative one. So the tension will have to exert a torque that cancel the sum of those torques. You can surely figure out which one of the two chains will apply that torque, and the value of the force (tension) producing it.

    Second thing is the weight of the swing. This force, in so far as we consider that it is a force acting solely at the CM, exerts no torque about the CM. This is where I don't really know how to justify it, but I'd say the chains equally split the burden of supporting the bench, meaning you must add to each of the tensions found previously, 125 N.

    So, basically, the chain near the soda has tension 125 N and the chain near the person has tension higher than that.
  4. Jan 25, 2005 #3
    try this....

    i'll assume the whole system is at rest, so i don't blow a brain cell on this one!

    each chain will support "its part" of the masses hanging from it!

    assuming the swing is of uniform density (if not, go out for pizza...), each chain supports what fraction of the swing alone? (half!)

    if the soda can or the person were (or could) be supported by either chain alone (i.e., they could sit right at the end of the swing, or straddle the chain), that chain would support 100% of the weight and the other chain would be supporting none of their weight. if the can or the person sat in the middle, each chain would support half of the weight.

    can you see what's coming? the one chain or the other chain has to support between zero and 100% of the weight of anything sitting on the swing, (plus the swing), and the fraction of the weight of the soda can OR the person is directly (and linearly) related to their position on the swing.

    i'd prefer you use this logic to figure out how to solve the problem from here on, ok?

    go for it!
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