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AP Physics Tension and Force Problem

  1. Nov 13, 2007 #1
    1. The problem statement:

    In the diagram (which I attached) a 50 kg square (2m on each side) is hung on a horizontal rod which is 3m long (of neglibible mass). A cable is attached to the end of rod and to a point on the wall 4m above the point where the hinged sign is.

    A. What is the Tension in the cable?

    B. What are the magnitudes and directions of the horizontal AND vertical components of force on the rod from the wall?

    2. Relevant equations

    -Confusion ??

    3. The attempt at a solution

    We know that the sign is in equilibrium. So to find the tension in the cable I thought that the y component of the tension would be equal to the weight of the sign. But this is wrong. Why? Do I need to use torque ? I have about 10 of these problems to do tonight so if I understood it, I'd be VERY thankful.
    Please help!
  2. jcsd
  3. Nov 13, 2007 #2
  4. Nov 13, 2007 #3

    Doc Al

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    Staff: Mentor

    If the only forces acting on the sign (and attached rod) where weight and cable tension, then your reasoning would be correct. But the rod is attached to the wall, so other forces come into play. The wall exerts a force on the rod (and the rod exerts a force on the sign).

    Absolutely! Pick the correct point as your pivot and you'll solve this in one step.
  5. Nov 13, 2007 #4
    Ok, I'm still very confused where to start at all. Do you have any suggestions at all? What's a pivot point? I'm very new to torque.

    How do I write the torque in the x-direction?

    Say I choose my point as the start of the block.. then Fblock(2m) + Fwall(-1) = 0 ?? is this right ??
  6. Nov 13, 2007 #5

    Doc Al

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    First thing to do is read up on torque in your textbook. Here are a couple of links that might help: Torque Concepts & Torque Equilibrium Examples

    No, that's not right. Choose the left end of the rod as your pivot. Treat the rod+sign as a single object. Where does the weight of the sign act?
  7. Nov 13, 2007 #6
    I read all about torque in my textbook, but I swear it doesn't help. My teacher never taught this but assigned us a lot of problems.

    ok so the net torque in the x-direction is

    0=W,sign(distance to its center of mass, or 2m) + F,wall(distance from end of rod, or 0)

    ?? This doesn't make sense - 0 = 490(2) + 0 ?? I'm not solving for anything !?

    I'm so SO SO SO SO confused
  8. Nov 13, 2007 #7

    Doc Al

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    In these kind of problems, torque can be clockwise or counterclockwise. (Each has a different sign.) Don't think of torque in the x-direction, just think of torque about some pivot.

    You left out the force due to the tension.
  9. Nov 13, 2007 #8
    well what would be the 'radius' of tension?

    0 = WEIGHTofsign(2m) + 0 + TENSION*(radius?)
  10. Nov 13, 2007 #9

    Doc Al

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    Staff: Mentor

    You want the distance from the pivot to the point where the tension acts on the rod (the right end). You also need the component of tension perpendicular to the rod, since tension acts at an angle.
  11. Nov 13, 2007 #10
    OK so i set it up like this !! I think I finally get it !! (at least for tension)

    0 = T*(sin53)*3 + -2(490)
    This gets me T = 409, which is the right answer!!

    Ok so now that I have tension..

    Ok, so I chose the middle of my block as my new point
    so I have

    0 = Fblock(which is zero because distance is zero) + -Fwall(2) + 409(Tension)sin53

    This gives me Fwall in the horizontal direction to be 163 which is right!! However I'm supposed to find the force of the wall in the vertical direction as well. Do you have any idea of how to do this?
  12. Nov 13, 2007 #11
    Re Tension

    I'm sorry I didn't do this right - I'm mixing something up - I'm getting 163 as my horizontal force component, but it should be vertical ! uh oh, do you know what I'm doing wrong
  13. Nov 13, 2007 #12

    Doc Al

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    Realize that what you are calling Fwall is the vertical component of the wall force, not the horizontal. (The horizontal component will not have a torque about the chosen pivot.)

    Rather than use a second torque equation like you did, I'd recommend using the equations for vertical and horizontal forces. The sum of the vertical forces must equal zero and the sum of horizontal forces must equal zero.
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