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Torque problem - Direction and Magnitude of force exerted on socket

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A 23-kg sign is suspended from the end of a 7-kg pole that is 4m long. The pole is partially supported by a cable that is attached at a point 3m up along the pole from the socket. The angle between the pole and the vertical wall is 60°, and the angle between the cable and the wall is 70°. Determine the magnitude and direction of the force exerted by the wall socket on the bottom end of the pole.

    Forces in this problem would be:
    Fcable, Fsign, Fpole, Fsocket
    Fsign = 225.4N
    Fpole = 68.6N
    I don't have the force for the cable and we are trying to solve for the socket.


    2. Relevant equations
    [itex]\Sigma[/itex]Fx=0
    [itex]\Sigma[/itex]Fy=0
    [itex]\Sigma[/itex][itex]\Gamma[/itex]=0


    3. The attempt at a solution
    I know that this is a combination of a force and torque equation, but I'm not really sure how to start this problem. I have the picture from the textbook and my version of the free body diagram.
     

    Attached Files:

    Last edited: Nov 30, 2012
  2. jcsd
  3. Nov 30, 2012 #2

    haruspex

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    Please make some attempt at applying the equations. Find the horizontal and vertical components of each force acting on the pole. Put in algebraic unknowns where necessary (you will need one for the angle required as answer).
     
  4. Dec 1, 2012 #3
    So I've given the force equations a shot:

    [itex]\Sigma[/itex]Fy=0
    0=-Fcabley + Fsockety + Fpoley + Fsigny
    0=-Fcablesin50° + Fsocketsin30° + 68.6sin270° + 225.4sin270°
    0=-Fcablesin50° + Fsocketsin30° - 68.6 - 225.4
    294 = -Fcablesin50° + Fsocketsin30°

    [itex]\Sigma[/itex]Fx=0
    0=-Fcablex + Fsocketx + Fpolex + Fsignx
    0=-Fcablecos50° + Fsocketcos30° + 68.6cos270° + 255.4cos270°
    0=-Fcablecos50° + Fsocketcos30°

    Is this on the right path, I thought I was setting it up right, but I'm not sure I can get either component from the equations I made.
     
  5. Dec 1, 2012 #4

    haruspex

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    As a matter of technique, I urge you not to substitute in numbers until the very end. It's much easier to track the argument and spot mistakes if you keep it all in algebra as long as possible.
    Where does sin 50 come from? I understand that's the angle between the pole and cable, but what has that to do with the vertical component of anything?
    The signs don't seem consistent. Are you taking the positive y direction to be up or down?
    You cannot assume the force at the socket will be along the pole. Poles are rigid so can transmit torque. The question specifically asks for the direction of that force. Put in an unknown angle for it.
     
  6. Dec 3, 2012 #5
    I've been trying to work with this for a few days and I'm seriously lost as to why my force equations aren't coming out correctly.

    Fsocketx + Fsignx + Fpolex - Fcablex = 0
    Fsocket + 225.4Ncos250° + 68.6Ncos270 - Fcable = 0
    Fsocket = Fcablex + 77.09N

    Fsockety + Fsigny + Fpoley - Fcabley = 0
    Fsocket + 225.4Nsin250° + 68.6Nsin270° - Fcable = 0
    Fsocket - 280.4N = Fcable

    so if I were to substitute one equation into the other, then the Fsocket in both would cancel out so that doesn't really make any sense unless I'm setting up the entire force equation incorrectly.
     
  7. Dec 3, 2012 #6

    haruspex

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    I don't understand how you are calculating these angles.
    In which direction does the weight of the sign act? What then are its horizontal and vertical components?
    In which direction does the reaction from the socket act? (I answered this in a previous post.)
     
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