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 = 4.7 m,

a weight of

WL = 66.0 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.)

(b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.)

Explanation / Answer

(a) If the vertical projection of the 4.7 m ladder is 3.75 m, then the horizontal projection is

d = (4.7² - 3.75²) m = 2.83 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 * 66N * 2.83m - ½ * 90kg * 9.8m/s² * 2.83m

N2 = 357.712 N

Summing the horizontal forces, observe that f1 = N2 = 357.712 N

Summing the vertical forces, observe that N1 = 66N + (90kg * 9.8m/s²) = 948 N

(b) Now M = 0 = N2 * 3.75m - ½ * 66N * 2.83m - (3/4) * 90kg * 9.8m/s² * 2.83m

N2 = 524.116 N

and therefore f1 = N2 = 524.116 N

Vertical load unchanged, so N1 = 948 N

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