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Simple torque problem

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data
    A man ties one end of a strong rope 8.00m long to the bumper of his truck, 0.500 from the ground, and the other end to a vertical tree trunk at a height of 3.00m. He uses the truck to create a tension of 8.00x10^2N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.

    2. Relevant equations
    torque = Fxd

    3. The attempt at a solution
    I know this is an easy problem. I can't picture what's going on. Is the tension at an angle?

    Can someone guide me please?
  2. jcsd
  3. Aug 3, 2010 #2
    Last edited by a moderator: May 4, 2017
  4. Aug 3, 2010 #3


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    It is a very good and nice drawing!

    The torgue depends on the angle the force encloses with the vector r pointing from the pivot to the point where the force is applied. If this angle is α and the force acts at distance r from the pivot the torgue is

    τ=F* r *sinα.

    Try: You can not rotate the door with a force parallel with it. The force needs to have a perpendicular component.

    So you have the force of tension acting on the tree at 3 m hight. This force points toward the car. Calculate the angle it makes with the tree.


    Attached Files:

  5. Aug 3, 2010 #4
    Why is the tension going in that direction?
  6. Aug 3, 2010 #5


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    The cord pulls the tree towards the car, but pulls the car towards the tree. You want the torque acting on the tree.

  7. Aug 3, 2010 #6
    with respect to the car, the tension points away from the car like the one I drew (if the tension is great enough the rear wheels will actually be lifted off the ground).

    with respect to the tree, the tension points away from the tree like the one ehild drew (similarly if the tension is great enough, the tree will be uprooted).

    Use the one ehild gave instead. I misread the question and thought that you are looking for the torque acting on the car :tongue:
  8. Dec 18, 2011 #7
    Would somebody mind explaining why you use cos in this problem?

    I think I have it right (please correct if wrong, i always get confused with this!)

    So you take the cos-1 ( adj/hyp) because you know one side is 90.

    Now there is info that would allow you to use sin opp/hyp at the remaining angle near the truck.

    What makes you choose that top angle for theta?

    I understand the next part: That you plug that theta inton rFSinTheta but am unsure why you use the imaginary or dotted triangle values? I see why you are using the 3 but am unclear why that information is needed.

    If anybody could shed some light I would greatly appreciate it, thanks!
  9. Dec 18, 2011 #8


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    I do not understand your post. There is no side of 90 (in what? m, cm? . ..) I do not see any dotted triangle.

  10. Dec 19, 2011 #9

    ^ link to image.

    So you have the hyp (8) going from bumper to top of trunk at 3m high.
    You then have a "dotted" line going from bumper to bottom of trunk and you have a right triangle. (Triangle 1)

    Triangle 2:You also make a triangle by going from bumper to top of trunk. Then height of bumper to meeting point at trunk (instead of leg being 3, it is 2.5).

    Now I think you solve by:
    1) Finding Horizontal Component:

    I used the top angle (same for both triangles) as my reference angle.

    So I used cos-1 = adj/hyp = 2.5/8 = 71.8 = theta.
    Now to get the right answer in the book I multiplied theta by 3.00 (the full height of the of the trunk) and got the final answer of 2279.8J.

    My questions:

    Why can you use cos in the non perpendicular torque equation ( T= rFSintheta).

    Why is the top angle the "correct" reference angle?

    Why do you need to use both triangles (imaginary ones), i see the difference in them with the leg being 2.5 and 3 but I am not making the distinction of why you use both instead of just one of them?

    Any help to get my thought process going would help a lot. I got the right answer, and think my math is correct but wouldn't of known this is the right answer without looking in the book.
  11. Dec 19, 2011 #10


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    You can not use it. Nobody said it could.

    Look at the definition of torque. The component of force normal to the axis of rotation F sin(alpha) has to be multiplied with the distance of the point of attack from the axis (3 m).
    You use the right triangle with leg 2.5 m and hypothenus 8m to get the angle between the force and tree: alpha = cos-1(2.5/8)

    You use 3 m and the angle alpha to get the torque. It is not a triangle.

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