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Torque up a ramp

  1. Mar 9, 2008 #1
    Two builders carry a sheet of drywall up a ramp. They exert vertical forces at the lower corners of the sheet. Assume that W = 2.00m, L = 3.10m, theta = 15.0 degrees, and that the lead builder carries a weight of 123.0N (27.7lb). What is the weight carried by the builder at the rear ?

    [tex]\tau[/tex] = F[tex]\bot[/tex]*r

    I know that there is no rotation of the drywall, and it is a thin, rectangular plate. The lever arm of the leading builder is (L/2) * cos(15). The torque of this builder is thus 123*cos(15)*(1/2)*(L/2) = 184 N*m.

    This means that the torque of the other builder must be the same. However, I can't figure out what the lever arm of this other builder is; is it the same as the leader? It looks like the corners produce different lever arms at an angle, but I'm not sure how to prove that. I thought that the lever arm would be shorter by a distance of W*sin(15), but this is wrong. What am I missing?
     
  2. jcsd
  3. Mar 9, 2008 #2

    tiny-tim

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    Hi physicsgrouch!

    You seem to have calculated the torques about the centre of the bottom of the sheet. :frown:

    If you want the torque of the force of gravity on the sheet to be zero, you must calculate the torques about the centre of gravity of the sheet (or about any point on a vertical line through it).

    Try again! :smile:
     
  4. Mar 9, 2008 #3
    So...the triangle of the ramp has a similar triangle inside the drywall. The cosine of the angle is W/(length of line through the rectangle at its center of mass). This is similar to another triangle, with the base of the formed trapezoid as the hypotenuse, meaning that the lever arm of the front builder is...3.8637 m?
     
  5. Mar 9, 2008 #4
    I set up a trapezoid (with an area of half that of the rectangle itself) and got that the lever arm of the lead builder is 1.282 m. So his torque is...152.32 N*m?
     
  6. Mar 9, 2008 #5
    Wait, sorry, I mean 1.238 m. So that makes the lead builder's torque...152.32 N*m...and the second builder's force to counter the torque is...211.34 N?
     
  7. Mar 9, 2008 #6
    Crud...I messed up again...I mixed up the base of the trapezoid and the bases of the trapezoid...

    I FINALLY got 174.414 N as the other builder's force.

    I'm really sorry; I'm very confused myself.
     
  8. Mar 9, 2008 #7
    What should my answer be?
     
  9. Mar 10, 2008 #8

    tiny-tim

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    Use coordinates, not Euclid!

    Hi physicsgrouch!

    (I've just woken up!)

    I'm completely confused by your trapezoid. :confused:

    All you need to know is the vertical distance from the centre of mass to each corner.

    In other words, the x coordinate.

    (You seem to be treating this like a geometry problem, using the methods of Euclid - that'll work, but it's far too slow, just use Descartes' invention, the coordinates!)

    So just go along half the bottom edge, and get the x coordinate of that length, and then go from there to the c.o.m, and get the x coordinate of that length.

    Then you add for one corner, and subtract for the other.

    That's all it is! Have a go … :smile:
     
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