Question

A 21.7 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.29. The angle between the horizontal and the ladder is 0. The person wants to climb up the ladder a distance of 1.2 m along the ladder from the ladder's foot. 2.9 m WWW +01 1.2 m 121.7 kg u= 0.29 XXXXX

Answer 1

The free body diagram is given as follows: L = 2.9m Lsin e 10 W *d cos e Here, the static frictional force between ladder and the floor acts towards right, and normal force exerted by the wall (N) acts towards left. The net force along horizontal direction is, Fe = f-N, Here, f is the static frictional force between ladder and the floor, and n, is the normal force exerted by the wall. From Newton's second law of motion, the net force along horizontal direction is, F = ma From above two equations: f-N = ma f-1) = m (0) f-N,=0 1. f = No The net torque about the point of contact of the ladder and floor is, mg (1.2m)cos8 -f(2.9 m)sin 8 = 0 Here, f is the horizantal normal reaction from the wall. f = umg = (0.29 mg

Thus, mg (1.2m)cos -(0.29) mg (2.9 m) sin 0 = 0 tan =1.42 0 = tan-(0.70) =54.90 = 550 At equilibrium the net torque at the point of contact between the ladder and the floor is zero. (Fxr)=0 W(d cos 6)-(N.) (L sin ) = 0 w (d cos 6 )= (N) (I sin ) Wd cose) L sino Wd cos) L sin Wd N . sin e cos e Wd L tan Substitute values: (21.7 kg) (9.8 m/s) (1.2 m) (2.9 m tan 55° = 61.6 N (21.7 kg From expression, f =N, umg =61.6 N 4. 61.6 N mg 61.6 N (21.7 kg) (9.8 m/s) -0.289 = 0.3