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Problem with Torque

  1. Oct 10, 2015 #1
    I used symbol M for torque by my mistake in equations,(because M is used in my language),If I am right americans use symbol T for torque.

    At equation (1) I get two unknowns : force T and force Rbx
    tried more equations but it gets more complicated,what am I doing wrong...

    note:I've choosen point A for calculating torque!
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2015 #2
     
  4. Oct 10, 2015 #3

    SteamKing

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    Speaking as an American, I use M for 'moment', rather than torque. Using T for 'torque' gets confusing, since T is often used for 'Tension'.

    In your equation (1), you have written (√3 / 3) * RBx as one of the components of the moment about A. Since you have already included the forces in the wire in calculating the moment about A, this is an unnecessary and incorrect inclusion in your moment summation.

    The correct moment summation about A allows you to calculate the magnitude of the tension T in the wire, which can then be used to determine the reaction at A. You are also including the vertical component of the tension in the wire twice in equation (2), once as RBy and again as Ty.

    A proper free body drawing of the beam ADC would include the beam and the components of the tension in the wire at C. Since the beam is in equilibrium and there is only the wire between C and B, once you determine the tension in the wire, the reactions at point B are automatically known. There is no need to include the components of RB in your force summation.
     
  5. Oct 10, 2015 #4
    So RBy and Ty are basically the same forces? I thought RBy is the force of the reaction that wall produces on the rod,but when I think of it must be the same thing because Tension is actually produced by gravitational force that pulls down the rod and they are obviously both in the same direction.(hope you understand my thought)

    So I get from equation (1) that T= (2.58)/(3*10^4) =7.75 kN ?
     
  6. Oct 10, 2015 #5

    SteamKing

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    That's why it's important to learn how to draw a proper free body diagram: it allows you to isolate only the forces acting directly on the member of interest.
    The reactions at point A directly influence what happens to the beam, from the standpoint of its equilibrium. The same cannot be said of the reactions at B, because their influence has already been taken into account by including the tension in the wire, which acts to keep the beam in equilibrium under the action of the load G.

    You've made a slight mistake in writing your moments from one line to the next.

    The moment due to load G = 2 m × 10 kN = 20 kN-m = 2 × 104 N-m

    T = 2 × 104 N-m / 2.58 m = 7.75 × 103 N = 7.75 kN
     
  7. Oct 10, 2015 #6
    Yes ,I've noticed that mistake in equation (1),that force RB only confused me,
    and thank you for your help.
    Ivan.
     
  8. Oct 10, 2015 #7
     
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