A 2.5 m long, 20 kg ladder leans against a wall, making an angle of 15° with the

Question

A 2.5 m long, 20 kg ladder leans against a wall, making an angle of 15° with the vertical. Assume that the ladder can slide without friction against the wall, but the coefficient of static friction between the ladder's feet and the ground is 0.6. A 75 kg person stands on one of the ladder's rungs, 1.0 m from the bottom end.

a) What is the magnitude and direction of the torque that the person's feet exert on the ladder, about the point of contact between the ladder and ground?

b) What is the magnitude of the normal force that the wall exerts on the ladder's upper end?

c) What is the magnitude of the friction force acting on the ladder's feet?

Explanation / Answer

(A) torque = r x F = r F sin(theta)

= 1 x (75 x 9.8) sin15

= 190.23 N m clockwise

(B) torque due to weight of ladder.

torque = (2.5/2) (20 x 9.8) sin15

= 63.41 N m clockwise

torque due to normal force = L N cos15 counterclockwise

net trque have to be zero.

190.23 + 63.41 - 2.5 N cos15 = 0

N = 105 N ........Ans

(C) in horizontal,

Fnet = f - N = 0

f = 105 N

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