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Torque Problem: Weight of box hanging from Pivoting Boon

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A crate with a mass of 181.5 kg is suspended from the end of a uniform boom. The upper end of the boom is supported by the tension of 2323 N in a cable attached to the wall. The lower end of the boom pivots at the location marked X on the same wall. Calculate the mass of the boom.
    [​IMG]

    2. Relevant equations

    L - length of boon
    Ff - friction
    T - tension
    Wb - weight of boon
    W - weight of box

    net forces in x direction = Ffx - Tx
    net forces in y direction = Ffy + Ty - Wb - W
    net torque = Wb(L/2)cos(theta) + WLcos(theta) - TLsin(theta)

    3. The attempt at a solution

    This is what I did exactly: I set up the above equations and then worked with the torque equation setting the tension equal to the weights. I canceled out the L's and reset up the problem so that it was equal to what I wanted to find, Wb. Upon doing that I got this equation: (2Tsin(theta) - Wcos(theta))/cos(theta) = Wb. I eliminated the cos in the denominator and got 2tan(theta) - W = Wb and from there I dervied theta from the graph making a right triangle with the boon as the hypotenuse I took the tangent of 5/9 and got 29.1 for theta and filled that into the equation. However, obviously that didn't get me the right answer, so I'm wondering what I'm doing wrong. Help would be appreciated.
     
  2. jcsd
  3. Oct 24, 2007 #2

    Doc Al

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    Staff: Mentor

    OK. I assume that what you call Ff is not friction, but the force that the pivot exerts on the boom.
    This is the one you need. But I don't understand what angles you are referring to. Please specify the angles.

    You didn't state it, but these expressions all equal zero. The only one you need is the torque equation.
     
  4. Oct 24, 2007 #3
    sorry here's where my theta is in respect to the graph:
    [​IMG]
     
  5. Oct 24, 2007 #4

    Doc Al

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    That's the correct angle to use when calculating the torque due to the cable tension--the last term in your torque expression. But that's not the same angle you need for calculating the other torques.
     
  6. Oct 24, 2007 #5
    I understand what I'm doing wrong now. I mistakenly used the same angle to derive all the lever arms when the angle to use to find the lever arm of the tension is larger. I instead drew a line across from the coordinates (6,2) to (6,11) and found the angle across from the 90 degree angle and added that to theta to get an angle alpha to find the lever arm for the tension. Is that correct? Although I'm pretty sure I've got it straight I'm still not getting the right answer.
     
  7. Oct 25, 2007 #6

    Doc Al

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    I'm not quite sure what you mean. If I draw a horizontal line from point (2,6) to (11,6), it forms the top of a right triangle whose hypotenuse is the boom. The angle in the right corner of that triangle is the one you need to use in order to find the moment arms of the vertical forces (using the expressions in your equation).
     
  8. Oct 25, 2007 #7
    ^ yea I that's the angle I used to find the angle to find the lever arms for those vertical forces. I added that angle to the angle above in the lower right corner that you would get in the right triangle above that has the tensions as the hypotenuse.
     
  9. Oct 25, 2007 #8

    Doc Al

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    I don't understand why you would add that angle to anything.

    If you're still getting the wrong answer, it will be helpful to mark the angles you're using on the diagram. (Call them theta_1, theta_2, etc.)

    Then you can show what you're getting for each moment arm and torque, and show how you solved for the mass of the boom. (While your equation is in terms of the weight of the boom, you will have to convert to mass to get the final answer.)
     
  10. Oct 25, 2007 #9
    I wanted to try to explain without having to spend time amending the graph in photoshop, but I guess it just got confusing so heres my line of thinking:
    [​IMG]
    as you can see I renamed theta as just the angle you would use to find the lever arms for the weights. Now from my understanding the angle you use to find the lever arm for the tension would be the sum of theta plus the angle above it, as a labeled in the graph so that whole angle that I arced out would be the angle you use to find tension's lever arm.
     
  11. Oct 25, 2007 #10

    Doc Al

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    This looks good. (I assume your angles are with respect to the horizontal.) If you are still not getting the correct answer, show what you did step by step.
     
  12. Oct 25, 2007 #11
    ok, I'll give it to you step by step:
    1) upon realizing my mistake of using the same angle to derived all the lever arms, I amended the graph to the one shown above and found the apropriate angles.

    2) since the picture is place on a graph, drawing lines across from (2,6) to (11,6), I created two right triangles. I counted up 5 spaces and across 9 spaces to find the legs of the bottom triangle and found theta by taking the inverse tangent of 5/9 which came out to be approximately 29.1 degrees for theta

    3) for the top triangle using the same methods in the previous step I counted up 4 spaces and across 9 spaces and and took the inverse tangent of 4/9 which came out to approximately 24.0 degrees for phi

    4) I deduced that theta would be used to derive the lever arms for the vertical forces down for the weights, and in the case of the tension the angle to be used to find the lever arm of that force would be the sum of phi+theta. phi+theta = approximately 53.0 degrees, this new angle of phi+theta I will call alpha

    5) I original derived a torque equation before I calculated the angles:
    net torque = Wb(L/2)cos(theta) + WLcos(theta) - TLsin(alpha)
    I rearranged the equation so that it would equal my unknown
    (2Tsin(alpha) - Wcos(theta))/cos(theta) = Wb

    4) Filling in my numerical values I got this:
    2(2323)sin(53.02) - 1778.7cos(29.05)/cos(29.05) = Wb

    the 1779.7 is the mass of the crate (181.5) times 9.8, and the extra decimal places for the angles I put in for the sake of accuracy

    5) Completing that math I got approximately 2471 and divided that by 9.8 to get the mass and got approximately 252.2, however this answer is incorrect so where did I go wrong?
     
  13. Oct 25, 2007 #12

    Doc Al

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    Good.
    You should also have a 2 in front of the W:
    (2Tsin(alpha) - 2Wcos(theta))/cos(theta) = Wb
     
  14. Oct 25, 2007 #13
    ah, you know that's probably what I'm doing wrong, thanks again.
     
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