A cleaner is trying to get to a window on the side of a house by climbing a unif

Question

A cleaner is trying to get to a window on the side of a house by climbing a uniform ladder that is 5.0 m long and weighs 180 N. The painter, weighing at 800 N, stops a third of the way up the ladder. The bottom of the ladder rests on the concrete drive-way and and leans across, in equilibrium, against a vertical frictionless wall. The ladder makes an angle of 53 degrees with the horizontal, form a 3-4-5 triangle.

Please solve using torques about an axis at the top of the ladder if possible:
A) Find the normal and friction forces on the ladder at it's base
B) FInd the minimum coefficient of static friction needed to prevent from slipping
C) Find the magnitude and direction of the contact force on the ladder at the base

Explanation / Answer

If the system is equilibrium, the net force acting on the system is zero(Fnet = 0).hence torque = 0
Taking F = 0 and = 0 components,
Fx = 0, fk = n1 -------------------- (1)
Fy = 0 , n2 = 800 N + 180 N = 980 N -------------------- (2)
= 0 ,n1(4.0)-180(1.5)-800(21.0)+n2 (0)+fs(0) = 0 ------------------- (3)

therefore n1 = 268 N and from equation (1)

fs = 268 N

(b) (s)min = fs/n2 = 268/980 = 0.27

(c) the components of the contact force ,fs = 286 N and n2 = 980N

the resultant force F = (286^2 + 980^2)^1/2 = 1020 N

the direction = Tan^-(980/286) = 75 degree with harizontal.

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