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Tension and Force

  1. Oct 28, 2004 #1

    tony873004

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    Figure 6-18: http://www.webassign.net/walker/06-18alt.gif

    The pulley system in Figure 6-18 is used to lift a crate of mass m = 52 kg. Note that a chain connects the upper pulley to the ceiling and a second chain connects the lower pulley to the crate. Assume that the masses of the chains, pulleys, and ropes are negligible.

    (a) Determine the force F required to lift the crate with constant speed.
    N
    (b) Determine the tension in each chain when the crate is being lifted with constant speed.
    N (top chain)
    N (bottom chain)

    The answers in the back of the book are 260 N for part A and 510 N for part b for both chains.

    The 510 part is easy. f=ma f=52*9.81 = 510. Each chain should have 510N of tension. No reason to divide it into half the tension for each chain.

    But the first part? Where'd they get 260 from? Gravity is pulling down with f=ma or 510N. So if you don't pull the rope with at least 510N, you're not going to lift the box. Shouldn't 260N of force on the rope just put 260N of tension on the chains while not budging the box one bit? 260 is approximately but not exactly half of 510. I don't know if that yields a clue?

    Anyone???
     
  2. jcsd
  3. Oct 28, 2004 #2

    Doc Al

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    pulleys multiply force

    Gravity is pulling down on the crate with 510 N. But how many rope segments are pulling up on the crate? (Think of the crate plus bottom chain & pulley as being one composite object.)
    If you exert a force on the rope, that force is transmitted via the tension in the rope. Pulling with 260 N creats a tension of 260 N. But, due to the pulley arrangement, that tension pulls twice on the crate. That's the entire point of using a pulley: it gives you a mechanical advantage that multiplies the force you can exert.
     
  4. Oct 28, 2004 #3

    tony873004

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    Thanks Doc Al. That 2nd rope pulling up makes sense.
    I'm still confused as to why the force needed isn't exactly half the weight. Shouldn't the answer for part A be exactly half the answer for part B? We're supposed to assume frictionless and massless pullies, ropes, and chains. Maybe the back of the book got it wrong. It wouldn't be thier first mistake :rofl:
     
  5. Oct 28, 2004 #4

    Doc Al

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    Sure the tension should equal exactly half the weight, but your numbers only have 2 significant figures: so 510/2 = 255 becomes 260. :wink:
     
  6. Oct 28, 2004 #5

    tony873004

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    Thank you very much. That clears up everything :smile:
     
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