Physics

A 56.5kg, 2.4m uniform ladder leans against a frictionless wall. A 95.5kg person is standing up 0.8m from the bottom of the ladder. The ladder makes an angle of 57 degrees with the horizontal. The coefficient of static friction between th eladder and the ground is 0.15. What is the force of the wall on the top of the ladder?

WORK:
F-Ff = 0
F-usN=0
F=usN = (0.15)(56.5+95.5)(9.8) = 223.44N

However the answer is 382.4N

WHY??

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  1. Careful, this is a torque problem. There are three sources of torque we need to solve for. The torque the person applies on the ladder. The torque the ladder applies on the wall, and the torque the wall exerts on the ladder.

    The torque exerted by the person:
    T1 = (mg(0.8))sin(33)

    The torque from the ladders weight:
    T2 = (Mg(L/2))sin(33)

    The force the wall exerts:
    FLsin(57) = -(T1 + T2)

    It's important to keep in mind here that since we have chosen the point where the ladder meets the floor as our pivot point there will be no force exerted on the floor by the ladder or vice versa.

    Plug in values and solve for F

    F = cos(57)(mg(0.8)+(Mg(L/2))) / Lsin(57)

    Plug in values and prosper.

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