A uniform ladder stands on a rough floor and rests against a frictionless wall a

Question

A uniform ladder stands on a rough floor and rests against a frictionless wall as shown in the figure. Since the floor is rough, it exerts both a normal force N_1 and a frictional force f_1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N_2 on the ladder. The ladder has a length of L = 4.4 m, a weight of W_L = 59.5 N, and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90 kg is standing on the ladder, determine the following. (a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.) N_1 = N N_2 = N f_1 = N (b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.) N_1 = N N_2 = N f_1 = N

Explanation / Answer

According to the given problem,

(a) If the vertical projection of the 4.4 m ladder is 3.75 m, then the horizontal projection is

d = (4.4² - 3.75²) m = 2.30 m

If we sum the moments about the base of the ladder, we can ignore the friction and normal forces at the floor. The sum of the moments must be zero, or the ladder would be rotating. Then
M = 0 = N2 * 3.75m - ½ * 59.5N * 2.3m - ½ * 90kg * 9.8m/s² * 2.3m
N2 = 288.93 N

Summing the horizontal forces, observe that f1 = N2 = 288.93 N

Summing the vertical forces, observe that N1 = 59.5N + 90kg * 9.8m/s² = 941.5 N

(b) Now M = 0 = N2 * 3.75m - ½ * 59.5N * 2.3m - (3/4) * 90kg * 9.8m/s² * 2.3m
N2 = 423.96667 N = 424N
and therefore f1 = 424 N
Vertical load unchanged, so N1 = 941.5 N

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