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Balance of Torque forces (scrambler machine)

  1. Sep 29, 2015 #1

    I'm trying to understand the physics on a Scrambler; and am having difficulty with balance of torques equation.
    In below image a vehicle is moving to the right while breaks are applied.

    The balance of torque is defined as:
    h * F1 + h * F2 + b * N1 = (B-b) * N2


    I understand the normal forces; and the direction of friction force; but, having difficulty with balance of forces.
    Please let me know if my interpretation of diagram is correct; and if I am missing anything.

    1) Force direction
    When we apply the break, the torque force is to the right. So, am I right to say that we are going to use the friction force magnitude that is pointing to the left; but, the real torque direction is still clockwise against an imaginary vertical wheel arm.

    2) Expression: h * F1 + h * F2 + b * N1 = (B-b) * N2
    Am I correct to say that we are assuming a rigid arm of height "h" in this problem; thereby, ignoring the fact that we have an axis in the center of the wheel?

    3) Expression: h * F1 + h * F2 + b * N1 = (B-b) * N2
    I understand the location of horizontal center of mass; but, are we ignoring the vertical center of mass for this problem? I know friction force is k * N, which considers the mass; but, I can't help think that the center of mass is not considered along height h (vertically).

    4) Expression: h * F1 + h * F2 + b * N1 = (B-b) * N2
    Why are we adding these forces? Are we to visualize one giant torque system (regardless of distance between the wheels)?

    5) Equation: h * F1 + h * F2 + b * N1 = (B-b) * N2
    Why is (b * N1) added to the left side of the equation. I guess we add this torque because the rotational direction is clockwise (same as when breaking is applied); and since the downward direction force is same as normal force, we add it to the left side of our equation.

    6) Expression: h * F1 + h * F2 + b * N1 = (B-b) * N2
    Am I correct to state that this expression goes to the right side of the equation because it is opposite direction of all other forces.

    7) Am I correct to say that the equation would remain the same, even if we lower the horizontal axle and center of the mass such that it is the bellow the wheel axis. I guess only torque to the center wheel changes; but, the equation remains the same.

    Thanks in advance.

  2. jcsd
  3. Oct 4, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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