Question

A uniform ladder rests against a frictionless wall at an angle of θ = 54.0° with respect to the ground. If the weight of the ladder is WL = 240 N, the length is L = 7.0 m, and the coefficient of static friction between the ladder and the ground is μs = 0.55, determine the distance a man with a weight of WM = 875 N can climb along the ladder before the ladder begins to slip

. ______m

Answer #1

The net vertical force of the system should be zero for equilibrium

so, R - (m+M)g = 0

=> R = 240 + 875 = 1115 N

this is the vertical contact force between the floor and the ladder.

the net horizontal force on the system should also be zero.

So,

F_{friction} - S = 0

u_{s}(1115) - S = 0

=> S = 613.25 N

this is the horizontal contact force between the ladder and the wall.

now, for the ladder to not slip while the man is climbing, the net torque on the system should also be zero.

So, taking moments about the point of contact of floor and ladder, we get:

R(0)cos54 + 240(L/2)cos54 + 875(x)cos54 - 613.25(L)sin54 = 0

=> 0 + 493.74 + 875(x)cos54 - 3472.907 = 0

=> x = 5.7925 m

so, the man can climb 5.7925 meters along the length of the ladder until it begins to slip.

Please upvote if the solution provided was helpful :)

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