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Homework Help: Help to understand

  1. Nov 26, 2005 #1
    Hi , please help me to understand how to set up the equation for the torques.
    This is the context:

    A ladder of length l=5m and weigth wladder=160N rest against a wall. its lower end is at 3m from the wall.
    A man weigthing 740N climbs 1 m along the ladder.
    Nwall is the reaction ladder-wall.
    (counterclock wise direction is considered positive)
    The equation of the torque (equilibrium) about the bottom of the ladder is:
    Nwall*4 -160* 1.5-740*1*(3/5).

    I dont know how they get the equation especially the underlined coefficients.
    My problem is that I dont know which distance to consider when computing for the torque in this case: the distance in respect to x axis, y axis or along the ladder.
    I hope I was clear enough
    Thank you
  2. jcsd
  3. Nov 26, 2005 #2


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    Homework Helper

    Let A be the point where the ladder rests against the wall.
    Let B be the point where the ladder rests against the ground.
    Let C be the corner of the wall and the ground.

    I assume that the wall is frictionless and that there is friction between the ladder and the ground. You probably have a coefficient of static friction, yes ?

    The ladder and the wall and the ground form a triangle. This triangle is a 3-4-5 triangle. This simplifies some of the trig with cos@ = 3/5 and sin@ = 4/5, where @ is the angle between the ladder and the ground.

    About your torque equation
    Nwall is at A and is normal to the wall. Hence the perpindicular distance of Nwall from the ground, and hence also from the point B, is 4m, giving T1 = Nwall*4.
    The torque of wladder about B is the force wladder times the perpindicular distance from B to the line of action of wladder, which is 2.5m*cos@, giving T2 = wladder*2.5*(3/5) = wladder*1.5.
    The torque of the man about B is similarly given by T3 = 740*1*cos@ = 740*1*(3/5).
    Since T2 and T3 are clockwise torques, then they are negative.


    Ttotal = T1 - T2 - T3 = Nwall*4 - wladder*1.5 - 740*1*(3/5).

    When computing the torques, use either the x-axis or y-axis depending upon the direction of the forces.
    If the force(s) is parallel to the x-axis, take the distance along the y-axis, and vice versa.
    The distance should always be perpindicular to the line of action of the force.
    Last edited: Nov 26, 2005
  4. Nov 26, 2005 #3

    Thank you very much Fermat. so the key is to take perpendicular distance when calculating for the torque.
  5. Nov 26, 2005 #4


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    That's right. There should always be a right angle between the force and the distance measured.
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