A symmetrical ladder of mass M = (mass 15.7 kg) leans against a smooth, friction

Question

A symmetrical ladder of mass M = (mass 15.7 kg) leans against a smooth, frictionless wall so the top of the ladder is height h = 4.4 m above the floor, and the bottom of the ladder is distance d = 1.36 m from the base of the wall. Since the floor is also frictionless, a horizontal wire connects the bottom of the ladder to the wall so the ladder does not slip. a) With no one on the ladder, find T, the magnitude of the tension in the wire. N b) Suppose the wire will snap when the magnitude of the tension is T = 175 N. Find x, the distance a man of mass m = 83.7 kg can climb up along the ladder before the wire snaps.

Explanation / Answer

Tan = h/d = 4.4/1.36 = 3.235

= atan(3.235) = 72.8 deg

Length of ladder L = (h^2 + d^2) = 4.605 m

a)

Balancing moments about the base of ladder, Rw*h = W*(d/2) where Rw is reaction from wall.

Rw = W/h*(d/2) = 15.7*9.81/4.4*(1.36/2) = 23.8 N

Balancing horizontal direction forces, T = Rw = 23.8 N

b)

Balancing moments about the base of ladder, Rw*h = W*(d/2) + mg*(x*Cos)

Putting Rw = T = 175 N

175*4.4 = 15.7*9.81*(1.36/2) + 83.7*9.81*x*Cos72.8

x = 2.74 m

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