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More static problems

  1. Nov 22, 2006 #1
    I have no idea on how to start these two problems. I haven't done any work, but please give me some hints on how to get started.

    The two problems are here
    http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1461065539

    For problem 14, the anskwer are 1.2kg, .3kg, and .1kg
    for 15, the answers are 54.9N,95N,170N.

    For problem 14, I know that the right of the wire extends 3 times as far to suppor the penguin. It appears to me that we divide the mass by four each time with the exception of the last time. What is the reasoing behind it?

    For 15, I really have no idea. Maybe it's because I don't really understand the question.
     
  2. jcsd
  3. Nov 22, 2006 #2

    radou

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    Which part don't you understand?

    The force in a hinge has two components - it is convenient to set one of them to be horizontal, and the other vertical. You know the mass of the fence, and you know the tension in the cable. Think about the equations of equilibrium and try to solve the problem.
     
  4. Nov 23, 2006 #3

    Pyrrhus

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    Take notice that since the hinges are colineal (imagine a vertical line passing through them), you should take advantage that their vertical component won't have moment if you take moment about either of the hinges.
     
  5. Nov 23, 2006 #4

    andrevdh

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    I am not very happy with the answers for 14. For each of the crossbars to be in rotational equilibrium

    [tex]w_l \frac{l}{3} = w_r \frac{2l}{3}[/tex]

    which after cancellation comes to

    [tex]m_l = 2m_r[/tex]

    where the l and r subscripts refer to the masses hanging from the left and right end of the crossbar. Therefore for the topmost crossbar this comes to

    [tex]2.4 = m_2 + m_3 + m_4[/tex]

    or am I missing some basic principle?
     
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