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Homework Help: Stuck on these boom problems

  1. Mar 11, 2005 #1
    A wrecking ball (weight = 5400 N) is supported by a boom, which may be assumed to be uniform and has a weight of 3800 N. As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32°, and the angle between the boom and the horizontal is 48°.

    what i have done which i guess is wrong is.
    L*5400 - Lsin(32)Ft + 1/2Lsin(48)(3800)=0
    and then i got rid of the L's
    and got 5400 - sin(32)Ft + 1/2sin(48)3800=0

    but this was wrong, if you could just explain to me how to do this it would really help a lot thanks.
  2. jcsd
  3. Mar 11, 2005 #2


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    you have posted only the background info, not the actual problem (whatever it is)! Also...I don't see where "L" is defined...so I have no idea what that is either.
  4. Mar 11, 2005 #3
    it wants to know the force of the cables tension, the length isn't defined but we can get it out right?
  5. Mar 11, 2005 #4


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    Okay. So the problem is asking you to find the force due to tension in the support cable. When I said that L wasn't defined, I didn't mean that the problem didn't specify the length, I meant that you introduced the symbol "L" without stating what it was. In fact, it is still unclear what it is. Remember that we can't see the picture, so it's hard to figure out what's going on.

    Your equation seems to be trying to sum the torques around the point where the boom is attached to the tractor. So your terms seem to be

    L*5400 = the torque on the boom (around that point) due to the weight of the wrecking ball (so, I'm inferring that you meant that L is the length of the boom). I don't think this term is correct, because you want to multiply the force by the moment arm, or lever arm, which is the perpendicular distance from the point of application of the force to the pivot point. That is not just the distance along the lever. The force is downward, so this perp. distance is the horizontal distance.

    L/2*sin(48)*3800 seems to be the torque on the boom due to its own weight. It is in the same direction as the torque above. It has an error too. We want the moment arm, or the horizontal distance in this case. Is that given by Lsin(48)?


    Now this one is very confusing. It seems that you're trying to calculate the counter torque provided by the support cable, (ie the torque that must prevent the boom from actually falling). But if so...then why did you use L? Isn't L the length of the boom? It's the force in the cable we're talking about now. And even if you had used the right length, can you see again why sine is not correct?
    Last edited: Mar 11, 2005
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