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Torque Forces

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A weightless ladder at 7m long rests against a frictionless wall at an angle of 65degrees to the ground. A 72kg person is 1.2m from the top of the ladder. What horizontal force at the bottom of the ladder is required to keep it from slipping?

    d= 7m
    m(man) = 72kg
    r(man) = 1.2m
    Theta = 65

    2. Relevant equations
    T = r(Fa)sin(theta)

    3. The attempt at a solution
    I know that this is a torque question because of the nature of the ladder positioning. First off, I started by finding the distance of the man from the pivot point at the bottom of the ladder, which is:

    To get the angle of the force that the man is exerting down on the ladder, I used the Rule of Sum for the C-pattern angle. Since the bottom angle is 65degrees, I subtracted that from 90 to get 25degrees, the angle of the force acting on the ladder.

    Afterwards, I calculated his torque value that is going CW:
    T = r(Fa) sin(Theta)
    = 5.8(72*9.8) sin(25)
    = 1729.56N-m

    At this point, I was thinking that since the ladder has to be in staticity to prevent it from slipping, the Torque forces were all equal. Thus:
    T(CW) = T(CCW)

    From here, I plugged in the values for the equation and attempted to solve for (Fa) of the ladder. Since the force at the bottom is now in regards to the pivot point at the top of the ladder(?), the distance (r) is equal to 7m. The angle of the force would also then be 65degrees:

    T(CW) = T(CCW)
    1729.56 = r(Fa) sin(theta)
    1729.56 = 7(Fa) sin65

    However, when I solve this equation for the (Fa) value, all I get is around 180N. Everyone in my class got 272N (approx.), and I am not sure how they did that. I am sure that I am missing something extremely simple, but I have spend over 45 minutes on this problem and I cannot figure out how to do it. Any hints on what to do after finding the Torque(CW) is much appreciated :smile:

    EDIT: Even after I fixed my answer (which was not 1279N-m), I still couldn't get 272N for the horizontal force. I am going to go crazy from this problem.
    Last edited: Nov 1, 2007
  2. jcsd
  3. Nov 4, 2007 #2


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    Homework Helper

    I'm not sure how you're getting 180 N. I describe the approach to this solution a little differently to my students, but what you are finding is the magnitude of the reaction force from the wall exerted against the ladder, which is equal in magnitude to the static frictional force on the foot of the ladder.

    I find that the torque exerted by the person's weight is

    72 kg · 9.81 m/sec^2 · 5.8 m · sin 25º = (72 · 9.81 · 5.8 · 0.423) N·m = 1731.3 N·m .

    I also find that the torque exerted by the reaction force is

    (F) · 7 m · sin 65º = (7 · 0.906 · F) N·m = 6.344 F N·m .

    This gives F = 1731.3 / 6.344 N = 273 N .

    You are doing this problem right! On the other hand, I'm at a loss to explain how you were getting 180 N (if you back-figure from that, you'd have to be using the sine of some angle as larger than 1). I can only surmise that you were repeatedly making some input error...

    (One tip I always give to students is that, before you start a calculation involving trig functions, you should always check the mode setting to make sure you're in degrees or radians, depending on which units you intend to be using. I'm not sure that explains the problem here, though.)
    Last edited: Nov 4, 2007
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