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Homework Help: Test question today

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data
    The image is a crane.
    For the diagram given, find the tension in the cable and find the moment around point A.
    The length of the crane the dark bar is 12 m.

    http://img249.imageshack.us/img249/679/exqjm6.jpg [Broken]

    2. Relevant equations
    sigma m = F *D

    3. The attempt at a solution

    I had this question on a test today and am very curious to know the answer.
    if someone strong in physics please solve it.

    I got 51 something Kilo Newton for the reaction at A.
    For the tension i got 46.08 Kilo newton i believe.

    thank you.

    EDIT: In my reaction/moment calculation I included the force of the cable. that being said i forgot to account for the y direction of the cable. should the cable have been accounted for?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 19, 2007 #2


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    Here's what I get:

    let T be the force of tension..

    let B be the force in the bar...

    Bcos30 - Tcos20 = 0

    B = 1.085T

    Bsin30 - 35 - Tsin20 = 0

    plug in B = 1.085T

    I get T = 174.6N

    Then the torque about A:

    10kN*6cos30 + 35kN*12cos30 - Tsin20*12

    = 10kN*6cos30 + 35kN*12cos30 - 174.6sin20*12
    = -300.9 kNm

    so the reaction has to be +300.9 kNm ie 300.9 kNm clockwise

    I was a little thrown off by the problem... because I initially just tried to get T by making the torque about A to 0... then when I saw the question asking for the moment about A I changed...

    Not very confident with how i did this problem.
  4. Oct 19, 2007 #3
    If someone as intelligent as you are not confident then I am sure my class took a strong beating during this test.

    Btw the weight hanging is 25KN i should have wrote it neater.

    Yup totally blew the last page of this test. I even got the tension wrong..which was the easier question.

    He gave us 50 minutes to solve 4 questions. By the time i got to the 4th had very little time left.

    but listen in your moment calculation why didnt you account for the x component of the tensin in the string?
  5. Oct 19, 2007 #4


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    Thanks I appreciate that. But in this question, I don't feel confident at all because I changed the way I solved the problem after reading the second part of the question. That's never a good sign. ;)

    The assumption I'm making here that there's a reaction force acting at A along the axis of the beam... along with a reaction moment at A. I don't feel good about this assumption that the net reactional force acts along the axis...

    yeah, I used 25kN... The 35kN... is the 25kN + the 10kN of the beam...

    I was using the component of tension perpendicular to the beam... ie Tsin10. then the torque due to that is -Tsin10*12 (minus sign indicates counteclockwise). The component parallel to the beam is Tcos10... but this creates 0 torque...

    You can also divide into the x (horizontal) and y(vertical)-components... magnitude of x-component of tension would be Tcos20. torque due to that is -Tcos20*(12sin30). The torque due to the y-component of tension is Tsin20*12cos30... net torque would be Tsin20*12cos30 - Tcos20*(12sin30) = Tsin(20-30) = Tsin(-10) = -Tsin10. comes out to the same answer...

    I would have blown this question on a test too.
  6. Oct 19, 2007 #5
    Go back and look at the moment about A and ask yourself "Is the boom arm rotating about A?" If not, what does that say about the moments?
  7. Oct 19, 2007 #6


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    Oh... I was making this whole problem way more complicated than it was... for some reason I was fixated on the answer to the second part not being 0.

    so to get tension:

    10kN*6cos30 + 25kN*12cos30 - Tsin10*12 = 0

    T = 149.6 kN ?
    Last edited: Oct 19, 2007
  8. Oct 19, 2007 #7
    do you mean T cos 20? because the triangle of the given 10 degrees is not a right triangle.

    149.6N doesn't SEEM right because how could the tension be so low compared to the other forces?
    Last edited: Oct 19, 2007
  9. Oct 19, 2007 #8


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    I blundered... it should be 149.6 kN.

    I'm taking the component of tension perpendicular to the arm... that's Tsin10
  10. Oct 20, 2007 #9
    Thanks ok I understand what you did there.

    Can someone clarify for what TVP45 meant about the boom bar please?

    Why should the moment/reaction be 0 again?
    Last edited: Oct 20, 2007
  11. Oct 20, 2007 #10
    Since the boom bar is not rotating about A, the moments about that point must sum to 0. That is a necessary condition for static equilibrium.

    That is a fairly simple equation to solve.
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