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Equilibrium using angular momentum

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A uniform, 210-kg beam is supported by a cable connected to the ceiling, as shown in the figure . The lower end of the beam rests on the floor.

    a. What is the tension in the cable?

    b. What is the minimum coefficient of static friction between the beam and the floor required for the beam to remain in this position?


    2. Relevant equations
    [itex]\Sigma[/itex]F_x = 0
    [itex]\Sigma[/itex]F_y = 0
    [itex]\Sigma[/itex]Torque = 0


    3. The attempt at a solution
    So we draw a free body diagram for both the beam and cable. I have 4 forces (perhaps I'm wrong with how many forces I actually have)

    I have a normal x and y force on the ground acting on the beam.

    I have gravity acting on the beam. I have tension acting on the beam. And that's all I believe.

    So I get N_x + Tcos40 = F_x
    and...... N_y + Tsin40 = F_y
    and...... Torque = T*cos160*r - mgcos40 << the length of the beam isn't given though?

    Please help
     

    Attached Files:

  2. jcsd
  3. Oct 2, 2011 #2
    Think again about the angle 40 inTsin40 and Tcos40.
    I do not think the angle is 40.
     
  4. Oct 2, 2011 #3
    grzz, thanks for the reply

    I don't understand how it's not 40.

    Isn't there a force, T that the beam exerts back on the cable? Shouldn't those forces be opposite but equal?
     
  5. Oct 2, 2011 #4
    I am assuming that your x-dir is the horizontal direction. So the angle between the Tension T in cable and the horizontal is 60.
     
  6. Oct 2, 2011 #5
    Okay I see that.

    Now I got: (R is the reactionary normal force of the ground, w is weight, L is length of the beam)
    R_x + Tcos60 = 0
    R_y + Tsin60 - w = 0

    T_1 = -L/2 * W * sin130 (counter clockwise is +)
    T_2 = L*T*sin160

    Are these two tensions right?

    If so, I'll have three equations

    1. R_x + Tcos60 = 0
    2. R_y + Tsin60 - w = 0
    3. L*T*sin160 - L/2 * W * sin130 = 0

    Then I assume solve for t
     
  7. Oct 2, 2011 #6
    Rx and tcos60 must be in opp dir for equilib
     
  8. Oct 2, 2011 #7
    Try to use acute angles. i cannot follow what you mean by the above equations.
     
  9. Oct 2, 2011 #8
    Thanks for the help,

    I got T = 2300

    Now for part b though, I would think the coefficient should be u=f/N

    f is R_x
    N is R_y

    R_y + Tsin60 - W = 0
    R_y = W - Tsin60
    ......= 210*9.8 - 2300*sin60
    ......= 66.14157

    R_x - Tcos60 = 0
    R_x = Tcos60
    .....= 2300*cos60
    .....= 1150.000

    u = f/N
    u = R_x / R_y
    u = 1150 / 66.14
    u = 17.38736 to 2 sig figs... = 17

    but it says its wrong?
     
    Last edited: Oct 2, 2011
  10. Oct 2, 2011 #9
    How did you get the value of T?
     
  11. Oct 2, 2011 #10
    LT*sin160 - .5*L*W*sin130 = SUM OF TORQUES = 0
    LT*sin160 = .5*L*W*sin130

    T = .5 * W * (sin130)/(sin160)
    The website said that was the right answer
     
  12. Oct 2, 2011 #11
    yes tension is Ok
     
  13. Oct 2, 2011 #12
    Did I do everything right to find the coefficient of kinetic friction then? wWhere did I go wrong in that solution

    thanks
     
  14. Oct 2, 2011 #13
    Are you sure the weight of the beam is 210kg?
     
  15. Oct 2, 2011 #14
    Yes, I think the book didn't use the rounded answer for the tension force when they calculated the normal force. Hence, they would get an answer around like 18.5 or so
     
  16. Oct 3, 2011 #15
    The coefficient of friction CAN have a value greater than 1. But a value close to 17 appears to me to be unusually a high value. But I checked your work again and could not find any mistake, neither in Physics nor in Mathematics.
    Of course I could have made a mistake myself!
     
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