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Torque Help

  1. Dec 6, 2006 #1
    I don't understand these concepts very well. Could someone please take the time and help run through a few of these problems. I don't want the answers I wanna know how to get them. Any help would be appreciated, thanks.


    1. Two builders carry a sheet of drywall up a ramp.

    Assume that Width = 2.10 m, Length = 3.10 m, Theta = 19.0°, and that the lead builder carries a (vertical) weight of 165.0 N (37.1 lb). What is the (vertical) weight carried by the builder at the rear?

    I know the net Torq=0 since it is not rotating. I also know torque=r*F8sin theta. Other then that I don't know how to set up the problem. I don't know what I am suppose to set equal to each other.

    2.A 33.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal.

    If the angle between the beam and the cable is q = 62.0° what is the tension in the cable?

    I took 1/2 the weight of the beam and divided it by the cos 62 and it gave me the right answer. I have no idea why this is. Could someone explain?



    3.A 24.4 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of q = 22.2o with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)

    I got Fh=1/2mgsin theta. Thats comes out to be 45.2 N and the answer is 41.87N. I think I am doing something wrong :-(


    4.A window washer with a mass of 91.0 kg stands a distance, D = 0.600 m, from the left end of a plank of length, L= 2.30 m, with a mass of 15.0 kg. The plank is hung on two cables as shown. Find T2, the tension in the right cable.

    For this one I did the Horizontal forces T1+T2=m1g+m2g

    Also did torque1=T1(.6) and torq2=-T2(1.7) sooo T1(.6)=T2(1.7) T1=2.83T2

    substituted 2.83T2+T2=m1g+m2g
    3.83T2=m1g+m2g
    T2=271.23N
    But answer says its 301N, Haha yeah definitely not understaning something.
     
    Last edited: Dec 6, 2006
  2. jcsd
  3. Dec 6, 2006 #2

    OlderDan

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    Welcome to the forum. Please read this

    https://www.physicsforums.com/showthread.php?t=94379

    Let's get one problem done before working on 4 of them. There must be an error on your first problem statement. Wat does W = 2.10 m mean? Please post an attempt to solve the problem.
     
  4. Dec 6, 2006 #3
    Updated the 1st one and wrote down all i understand about the problem.
     
  5. Dec 6, 2006 #4

    OlderDan

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    You have enough information to calculate the torque the front builder applies about any point in the drywall. You can pick any point you want since there is no rotation. I'll tell you for this one that a good point to use is the middle of the drywall. Find the torque from the front builder, and then find the torque fom the rear builder and the force needed to produce that torque.
     
  6. Dec 6, 2006 #5
    Then wouldn't I have the same force? torque= r*F*sin theta I get .50463*165=83.264N the 83.264N= 1.55*sin 19*F and i get the same F as the original.
     
  7. Dec 6, 2006 #6
    I have also put that I know how to get the answer to #2 but don't know why it is. torq= 1/2mg and T=Torq/cos theta?
     
  8. Dec 6, 2006 #7

    OlderDan

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    They are the same.
     
  9. Dec 6, 2006 #8

    OlderDan

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    Calculate the torque about the wall hinge. mg acts at half the length of the beam. Resolve the tension into horizontal and vertical components. The horizontal component contributes no torque. The vertical componet does. Take it from there and get your answer.
     
  10. Dec 6, 2006 #9
    It says the answer to #1 is 265N and not the 165N and I can not figure out why that is.
     
  11. Dec 6, 2006 #10
    I just edited what I tried to do for #3
     
  12. Dec 6, 2006 #11
    Just edited #4 on what I tried to do.
     
  13. Dec 6, 2006 #12

    OlderDan

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    I cannot make any reasonable assumption that would lead to their answer. I thought maybe in spite of their (vertical)s they were assuming the lead builder was applying his force perpendicular to the plane of the drywall, so the trailing builder would have to support more weight, and also apply a horizontal force. This assumption does no make nearly enough difference to get up to 265N. My guess is they made a typographical error and the answer is 165N. It certainly is 165N if both builders are applying only vertical forces.
     
    Last edited: Dec 6, 2006
  14. Dec 6, 2006 #13
    So the torque equation looks something like this?
    L*T*sin theta-L/2*mg=0

    L*Tsin theta=L/2*mg

    Ls cancel leaving

    T*sin theta=1/2*mg

    T=mg/2sin theta?

    I get the right answer like this. Just wanna make sure that is the right way to go about this.
     
  15. Dec 6, 2006 #14

    OlderDan

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    It is the right way. You can interpret LTsinθ as the vertical component of T acting at a distance of L, or T acting along a line that is Lsinθ from the pivot point. Either way, the sinθ comes from the cross product of the position vector from the point of rotation to the point of application of the force

    Torque = r x F
     
  16. Dec 6, 2006 #15

    OlderDan

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    This problem is what I though they might be doing in #2 (but they were not). If you take the pivot at the wall, the torque from the full F must balance the torque from the weight. If the beam were horizontal, F would be ½mg. But the beam is not horizontal, which reduces the torque from the weight and therefore reduces F. The horizontal forces from the pivot and the cable must be equal and opposite. If you get F correct, then Fsinθ will give you the correct answer.
     
  17. Dec 6, 2006 #16

    OlderDan

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    You left out the weight of the plank in the torque equation. Since you do not know T1, it would be good to caclulate torque about its point of application, the left end of the plank. Then youget an equation for T2 directly. I get 306N, but did not round off until the last step.
     
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