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Torque and static friction

  1. Nov 26, 2005 #1
    Romeo takes a uniform 10-m ladder and leans it against the smooth wall of the Capulet residence. The ladder's mass is 30kg, and the bottom rests on the ground 2.55m from the wall. When Romeo, whose mass is 70 kg, gets 88 percent of the way to the top, the ladder begins to slip. What is the coefficient of static friction between the ground and the ladder?

    For this problem i attempted to find the torque of each of the masses (romeo and the ladder) add them together and with the equation f = u*N find the coefficient of static friction.

    I found the torque by finding the angles of the ladder. I found that the angle the ladder makes with the ground to be 75.226. Therefore the torque of each of the masses will only be acting on the sin of 14.77 degrees. (this may be where i mess up but i don't know what else to do)

    so i took these equations: mass(romeo)*9.81*sin(14.77)*distance from axis
    mass(ladder)*9.81*sin(14.77)*distance from axis

    70*9.81*sin(14.77)*8.8=torque of romeo = 1540
    30*9.81*sin(14.77)*5= torque of ladder = 375

    sum the torques to find the total force working against the ground to get 1915 N.

    The normal force i found to be the mass of the total system times 9.81 so it is 981 N.

    So i took 1915/981 to get the coefficient of static friction to be 1.95. this is not right, where did i mess up? thanks
  2. jcsd
  3. Nov 26, 2005 #2


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    You're taking the moments around what part of the ladder? I was unclear what you meant by torque of the ladder. There is no ladder torque. There may be a torque due to gravity and the two surfaces acting on the ladder.

    I would recommend using the top of the ladder as the pivot point. Take the moments with respect to this point. There will 4 forces contributing to the torque from this point.

    1. gravity due to ladder's mass
    2. gravity due to Romeo's mass
    3. normal force at ladder base: hint - Fnet(y) = 0 = Fgrav(Romeo+ladder) - Fnorm
    4. static friction at the base of the ladder

    Now from here, all you have to do is carefully find the right effective distances these forces are acting from to produce the torque.
  4. Nov 27, 2005 #3


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    Taking torques around the bottom of the ladder (like you did) is OK.
    Don't forget the (unknown) horizontal Force by the wall,
    or else you won't be able to obtain Sum of Torques = 0 !

    (The sum of torques [Newton*meter] is NOT a FORCE applied to the ground!) (The friction Force must cancel the Force by the Wall.)
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