Part A Using the dimensions given in the figure, find the forces exerted on the

Question

Part A

Using the dimensions given in the figure, find the forces exerted on the ladder when the person is halfway up the ladder.

Express your answers using two significant figures separated by commas.

f1,f2,f3 = ________ kN  

Part B

Find the forces exerted on the ladder when the person is three-fourths of the way up the ladder.

Express your answers using two significant figures separated by commas.

f1,f2,f3 = _______ kN  

please please make sure answer is correct, 3rd time asking same answer. will rate . Thanks

Explanation / Answer

Weight of person = 90 * 9.81 = 882.90 N

a)
Total weight of ladder and person.
= 882.90+ 100
= 982.90 N

All the weight is opposed by force f1 so
f1 = 982.90 N

The base of the ladder is (d) away from the base of the wall
d = root(4^2 - 3.8^2) By Pythagoras
= 1.24899 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
= 982.90 * 1.24899 /2 N.m

This is opposed by the force f3
so
f3 * 3.8 = 982.90 * 1.24899 /2
f3 = 982.90 * 1.24899 /(2 * 3.8)
= 161.5305 N

The only force to counteract this is f2
f2 = 161.5305 N

b)
The weight of the ladder and the person is the same, so
f1 = 982.90 N (again)

Moment clockwise about the base of the ladder due to the weight of the ladder
= 1.24899 * (1 / 2) * 100 N.m =62.4495

Total clockwise moment
= ( 1.24899 / 4) * (100 * 2 + 982.90 * 3)
= 0.3122475 * 3148.7
= 983.17370 N.m

This is opposed by f3
f3 * 3.8 = 983.173
f3 = 983.173 / 3.8
= 258.729 N

As before, f2 opposes f3

f2 = 258.729 N

All forces are in the directions shown in the diagram.

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