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Question about torque and ladders.

  1. Apr 14, 2006 #1
    A ladder is against a frictionless wall, elevated at angle theta above the ground.

    I have a question about distance components. We find the torque pushing the ladder to the ground by using horizontal distance components: the distances of the weights acting on the ladder multiplied each by cosine of theta. What is strange is that we use a vertical distance component when finding the force exerted by the wall onto the ladder: distance from the ground to where the ladder touches the wall multiplied by sine of theta.

    Why don't we use cosine as we did for the others?

    Sorry if I am unclear, I really don't know how to put it another way.
     
  2. jcsd
  3. Apr 14, 2006 #2

    Doc Al

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    Staff: Mentor

    In each case you need to find the perpendicular distance from the axis to the line of the force. For vertical forces, the perpendicular distances are horizontal. To find the horizontal component of a distance along the ladder, cosine is used. But for horizontal forces, the perpendicular distances are vertical--that requires the sine.

    Another way to think of it is to examine the angles of the right triangle formed by the ladder. Realize that if two angles (theta1 & theta2) add up to 90 degrees, then sin(theta1) = cos(theta2). (And vice versa.)
     
  4. Apr 14, 2006 #3
    I understand that, but why would we use both vertical and horizontal components in the same torque equation?
     
  5. Apr 15, 2006 #4

    Doc Al

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    Staff: Mentor

    Because the forces due to gravity act vertically, while the wall force acts horizontally.

    Why don't you tell me the definition of torque that your book uses. (There are several equivalent definitions of torque.)
     
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