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Springs question

  1. Feb 22, 2006 #1
    ok guys I'm working on this problem:

    [​IMG]

    now, I think the answer is 7.0149 because the force of the block is 47, and when u devide by the bigger spring constant, 6.7, u get 7.0149
    The other one expands more. I think the question is worded weird... is this right?

    also, how exactly would I go about solving for the combined spring constant?

    thanks
     
  2. jcsd
  3. Feb 22, 2006 #2
    Both springs will exert a force on the block, you need to find the value of x such that the sum of their forces counteracts the gravitational force.
     
  4. Feb 22, 2006 #3
    sorry, but I still dont get it. I understand what you're saying, but my main problem is, in my mind, I see the block being crooked
     
  5. Feb 22, 2006 #4
    I think you're right, that could happen. I don't think the problem intends for you to take that tact though, there's no information given on the relative distances of the springs from the center of the rod. You'd have to take into account the torques exerted by the springs about the rod's CM, and for that you'd need to know how far they are. I say comfort yourself in the fact that the spring with the smaller spring constant seems to be further out, so it'll be exerting a smaller force with a larger distance from the CM which could make up for it in terms of torque.

    In short, ignore the rotation problem :wink:
     
  6. Feb 22, 2006 #5
    still lost lol. So do I just use the smaller one? What is wrong with my math in my first post?
     
  7. Feb 22, 2006 #6
    Draw a free body diagram for the rod. You have three forces acting on it, one for each spring and one for the force of gravity. If the rod isn't accelerating the sum of all the forces acting on it is equal to zero.

    By the way, while re-reading the problem I noticed that it explicitly states that "the 47N weight stretches each spring equally" so don't worry about any possibility of it being at an angle.
     
  8. Feb 23, 2006 #7
    For the record:
    [tex]F_1=k_1x[/tex]
    [tex]F_2=k_2x[/tex]
    [tex]F_1+F_2=47N[/tex]
    Three equations, three unknowns.

    -Dan
     
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