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 = 4.2 m,

a weight of

WL = 66.5 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.)

(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

Weight of person = 90 * 9.81 = 882.9 N

a)
Total weight of ladder and person.
= 882.9 + 66.5
= 949.4 N

All the weight is opposed by force f1 so
f1 = 949.4 N

The base of the ladder is (d) away from the base of the wall
d = root(4.22 - 3.72) By Pythagoras
= 1.98 m

Consider the base of the ladder as a pivot. Since the total weight of ladder and person are halfway up the ladder their weight will provide a turning moment about the base will be
= 949.4 * 1.98 /2 N.m

This is opposed by the force f3
so
f3 * 3.7 = 949.4 * 1.98 / 2
f3 = 949.4 * 1.98 / (2 * 3.7)
= 254.03 N

The only force to counteract this is f2
f2 = 254.03 N


b)
The weight of the ladder and the person is the same, so
f1 = 949.4 N (again)

Moment clockwise about the base of the ladder due to the weight of the ladder
= 1.98 * (1 / 2) * 66.5 N.m

Moment clockwise due to the person three quarters of the way up the ladder
= 1.98 * (3 / 4) * 949.4 N.m

Total clockwise moment
= (1.98 / 4) * (66.5 * 2 + 949.4 * 3)
= 1475.7 N.m

This is opposed by f3
f3 * 3.7 = 1475.7
f3 = 398.37N


As before, f2 opposes f3

f2 = 398.37 N

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