Consider a 3 m long ladder leaning against a wall at an angle of 70? from the ho

Question

Consider a 3 m long ladder leaning against a wall at an angle of 70? from the horizontal, as shown in the figure. The wall has a very slick surface, so there is no friction between the ladder and the wall. The ground, however, has a coefficient of static friction of ?s = 0.236. The ladder has a mass, m, of 9.85 kg. A maintenance worker, with mass M = 91.5 kg, is climbing the ladder. At the time when he is 2/3 of the way up the ladder: a. Find the frictional force (in Newtons) exerted on the ladder by the ground. Positive forces point to the right and negative forces point to the left. b. How far up the ladder can he climb? If he can climb all the way to the top, just enter 3 m. c. What is the minimum angle (in degrees) that the ladder can make with the horizontal that will allow the worker to get to the top of the ladder?

Explanation / Answer

a.

the wall exerts force of F

F sin70 * 3 = Ww (2) cos70 + Wl (3/2) cos70

F ((70)s) = 91.5*9.8*2/3 * ((70)o) + 9.85*9.8/2 * ((70)o)

F = (220.97)/(0.939693)

F = 235.15 235 N

N = Ww + Wl = 91.5*9.8 + 9.85*9.8 = 993.23 N

f_max = N = 0.236 * 993.23 = 234.402 N

234.4 < 235.15 ; therefore

f = 234.4 N

b.

N = Ww + Wl = 91.5*9.8 + 9.85*9.8 = 993.23 N

f_max = N = 0.236 * 993.23 = 234.402 N

F = f_max = 234.402 N

F sin70 * 3 = Ww (x) cos70 + Wl (3/2) cos70

234.402*(sin(70))*3 = 91.5*9.8*x*(cos(70)) + 9.85*9.8*3/2*(cos(70))

x = 1.993 m

c.

234.402*(sin(x))*3 = 91.5*9.8*3*(cos(x)) + 9.85*9.8*3/2*(cos(x))

x = 1.3277 radians = 76.1 degrees

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