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Homework Help: Objects in equilibrium beam problem

  1. Apr 16, 2014 #1
    Problem: A uniform beam of length 4 m and mass 10 kg, connected to the side of a building by a pivot hinge, supports a 20-kg light fixture,as shown in figure 4.43. Determine the tension in the wire and the components of the reaction force at the pivot. ImageUploadedByPhysics Forums1397700079.438119.jpg

    Equations: sum of forces =0
    Sum of torques=0

    Attempt at problem: Since the sum of the torques must be zero the torque of the cable must be equal to the torque of the weights of both objects so T( 4 cos 45)= 20( 3 cos 45) + 98( cos 45). Where T is the tension in the rope. The correct answer is T= 160
  2. jcsd
  3. Apr 16, 2014 #2


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    You are in general not calculating torques correctly. Its force times perp distance or use the cross product rule for torques. Identify your pivot point and show your free body diagram .
  4. Apr 17, 2014 #3
    ImageUploadedByPhysics Forums1397748499.173913.jpg
  5. Apr 17, 2014 #4
    Why are you using cos 45 when calculating the torque of the tension?
  6. Apr 17, 2014 #5
    Because the distance of 4 is along the length of the beam and you must find the actual x distance from the pivot point by using cos 45.
  7. Apr 17, 2014 #6
    No you don't. Look again. there are two different angles in your picture.
  8. Apr 17, 2014 #7
    I know the second angle is 60 degrees I just don't understand how to use it.
  9. Apr 17, 2014 #8
    Create an imaginary line connecting the pivot to the cable at 90° to the cable forming a triangle. use that triangle to find the distance used in the torque formula. The equation you used implies that triangle as an angle of 45 ° somewhere in it but that's not correct. It's a fairly simple trigonometry/geometry problem.
  10. Apr 17, 2014 #9
    Can you send me a diagram? I'm having trouble visualizing the triangle.
  11. Apr 17, 2014 #10
    Draw a line perpendicular to the cable making sure it passes over the pivot.
  12. Apr 17, 2014 #11
    ImageUploadedByPhysics Forums1397765906.695928.jpg
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  13. Apr 17, 2014 #12


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    i don't see any perpendicular line . It would run from the side T to the pivot, at a right angle to T. The force T times that perp dist to pivot would give the torque caused by T about the pivot. Or try cross product rule it might be easier if you are at all familiar with it.
    Incidentally, your torques for the applied weight forces are wrong also. In one case you used mass instead of weight, in the other , you forgot to include the moment arm length. Might be a careless error on those ?
  14. Apr 17, 2014 #13
    Should I have 98(2 cos 45) + 196(3 cos 45) for the applied weight? If so what makes the distance measurement for T different since it is on the same beam? I also have never heard of the cross product rule.
  15. Apr 17, 2014 #14
    Originally I tried to break down T into its X Y components so T sin 60 (4 cos 45) + T cos 60 (4 cos 45) - 98(2 cos 45) - 196( 3 cos 45)= 0. Is this more near what I should be doing?
  16. Apr 17, 2014 #15


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    Yes that's correct for the applied weight torques. Note that torque is force times perpendicular distance . The perpendicular distance from the tension force to the pivot seems to be stumping you. It just requires some geometry and trig. Now you could break up T into its x and y components and sum the torques of each component which you tried to do, but you didn't calculate the components correctly because you used the wrong angle . I see the basic trig and geometry is troubling you. On the other hand, you handled the geometry and trig pretty good when doing the weight torques. Hmmm. Well' heck there are so many ways to calculate torque try the cross product T (r)(sin theta), where T is T , r is the distance from the point of application of T to the pivot ( the length of the beam), and theta is the angle between T and r. But better to practice up on your basic geometry first.
    Last edited: Apr 17, 2014
  17. Apr 18, 2014 #16
    Actually for the first two terms you should have either

    T sin 15° (4 cos 45°) + T cos 15° (4 sin 45°)

    (because the angle between the cable and the horizontal is not 60°)


    T sin 60° (4 cos 90°) + T cos 60° (4 sin 90°)

    (If you decide to break the force into components parallel and perpendicular to the beam.

    Also note that one of the cos 45° was replaced with the correct sin 45°.

    But none of that is necessary and both of these expressions reduce (through simple trigonometry) to the expression that is more easily obtained by using the triangle I mentioned in earlier posts.
  18. Apr 18, 2014 #17
    Thanks! I understand what I did wrong. And thanks to everyone who helped me with this problem. I thank you all for your great amount of patience!
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