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 N1 and a frictional force f1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N2 on the ladder. The ladder has a length of L = 4.5 m, a weight of WL = 52.0 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.) N1 = 831.5 Incorrect: Your answer is incorrect. Draw a careful diagram which includes all forces acting on the ladder. See if you can apply the first condition of equilibrium to vertical forces acting on the ladder in order to obtain N1. N N2 = 831.5 Incorrect: Your answer is incorrect. Determine a good point about which to sum the torques and then see if you can write a statement of the second condition of equilibrium about that point that will allow you to determine N2, the normal force exerted on the ladder by the wall. You may need to use your knowledge of trigonometry to determine the angle between the ladder and the floor. N f1 = N (b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.) N1 = N N2 = N f1 = N

Explanation / Answer

L = 4.5 m,

WL = 52.0 N,

d = 3.75 m

m = 90 kg

1) (a) If the vertical projection of the 4.5 m ladder is 3.75 m, then the horizontal projection is
d = (4.5² - 3.75²) m = 2.48 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 * 52.0N * 2.48m - ½ * 90kg * 9.8m/s² * 2.48m
N2 = 504.21 N

Summing the horizontal forces, observe that f1 = N2 = 504.21 N

Summing the vertical forces, observe that N1 = 52.0N + 90kg * 9.8m/s² = 934 N

(b) Now M = 0 = N2 * 3.75m - ½ * 52.0N * 2.48m - (3/4) * 90kg * 9.8m/s² * 2.48m
N2 = 1701.25 N
and therefore f1 = 1701.25 N
Vertical load unchanged, so N1 = 934 N

2) What is the 47º w/r/t ? Try
F * 0.47m * cos47º = 65 N·m
but it might be sin47º instead

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