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Question about Torque

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    A nonuniform bar 4 meters long has a weight of 70 newtons. The bar is on a fixed pivot at its center of mass which is 1 meter from the heavy end. If a 300 newton weight is placed 0.4 meters left of the center of mass and a 100 newton weight is placed 1.0 meters right of the center of mass,
    (a) What are the clockwise and counterclockwise torque values?
    (b) What is the magnitude and direction of the net torque?
    (c) where must a 200 newton weight be placed to establish rotational equilibrium?

    2. Relevant equations

    Torque=Force*distance (If it's perpendicular, which it is)

    3. The attempt at a solution

    I don't know weather the fact that the bar is 70 newtons matters at all. Assuming it doesn't, here's what I have:

    (a) Clockwise: 100 N. Because 1.0m*100N=100N
    Counterclockwise: 120 N. Because 0.4m*300N=120N

    (b) Magnitude: I really have no idea here.
    Direction: Counterclockwise... ?

    (c) Assuming everything else is correct, I would say it needs to be 0.1 meters right of the center of mass. That way both sides are 120N

    All that assumes that the weight of the bar doesn't matter. Am I even close?
  2. jcsd
  3. Feb 6, 2008 #2
    You are right in assuming the weight of the bar does not matter. This is because gravity is constant, and the pivot is at the center of mass. Intuitively, in most cases, you can think about the center of mass as a balancing point. Thus, without weights, the bar is balanced.

    From the question, I can't tell which weight is on which side, but assuming your clockwise and counterclockwise are correct, then:

    (a) You're right, although you need to make sure your units are Nm, the units for torque, not N, since you are multiplying force (N) by distance (m).

    (b) Two torques in opposite directions cancel each other out, just as two people sitting on a see-saw can cancel each other out because they rotate in opposite directions. The convention is to give counterclockwise torque a positive sign and clockwise torque a negative sign perhaps because angles in math are done that way.

    The net torque is the final rotation of the bar after all cancellations. In fact, there is already a cancellation: the two ends of the bar, which cancel each other and keep the bar balanced because the pivot is at the center of mass.

    In your case, you have a 120 Nm torque and a -100 Nm torque, so what is your net torque? The magnitude of that is just the absolute value, just in case clockwise had won out and your net torque was negative. You have the direction right.

    (c) On the money...
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