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Homework Help: Chain consisting of five links

  1. Jan 27, 2008 #1
    So the first part of this question asks: A chain consisting of 5 links, each of mass .19kg, is lifted vertically with a constant acceleration of a=2.8 m/s^2. Find the magnitude of the force that link 3 exerts on link 2.

    I found this answer to be 4.7 N with the following formula:
    F(link 3 exerts on link 2) = ma + mg + F (link 2 exerts on link 1)

    I can't seem to get the second part of this question: What is the magnitude of the force F that must be exerted on the top link to achieve this acceleration?

    I used the same logic as above, namely:
    F = ma + mg + F(link 5 exerts on link 4)

    or, can you simply just use the formula F=ma, where m=(5)(.19kg) and a = 2.8m/s^2 ?

  2. jcsd
  3. Jan 27, 2008 #2
    You can use a singe F=ma to describe the net force applied to the entire chain, and if the only forces acting on the chain are the force pulling the top link pulling up and graving pulling the entire chain down, the net force acting on the chain = the force pulling up on the top link of the chain + the force the force of gravity pulling down on the chain. Of course, the rules of vector addition apply, and the force of gravity is opposite the force pulling the chain, so be careful when you do the math with that addition sign.
  4. Jan 27, 2008 #3
    Ok, so you mean something like

    F - mg = ma ---> F = m(a+g) ?
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