A straight ladder that has a length L 36 ft. is placed against a building that i

Question

A straight ladder that has a length L 36 ft. is placed against a building that is 75 ft. tall. The ladder forms the desired angle with the wall, the base being L/4 ft. from the wall. Assume the ladder weighs 100 lbs and the person climbing on it weights 200 lbs. Assume that the person's center of gravity acts 3 ft. vertically from the point of contact on the ladder rungs. The coefficient of friction between the ladder feet and surface it rests on is 0.45. a) How far up the ladder can the worker go before the ladder slides out at the bottom? b.) If the worker stands 3 ft. from the top of the ladder, how much force is required to tip the ladder away from the wall, assuming the force is acting horizontally against the top of the ladder? Neglect vertical forces at the wall. (Draw a free body diagraph) and (show all your work.)

Explanation / Answer

Given,

length of the ladder, L = 36 ft

distance from the wall = L/4 = 9 ft

height on the wall = (362 -92 )1/2 = 34.86 ft

forces on the system

weight of the ladder = 100g

weight of the person = 200g

Normal reaction of floor N2

Normal reaction of wall N1

frictional force of floor Fs

Frictional force of wall Fw

1,2 act vertically downward whereas 3 and 6 act vertically upward

4 and 5 are the only forces in horizontal direction

equating the vertical and horizontal components

100 g + 200 g = N2 + Fw ..........(1)

N1 = Fs ............(2)

co-efficient of friction = 0.45 hence

Fs = 0.45 N2

Fw = 0.45 N1 = 0.20 N2

substituting the values in (1)

300g = 1.20 N2

N2 = 250 g

N1 = 0.45 N2 = 112.5 g

Fs = 0.45 x 250 g = 112.5 g

Fw = 0.45 N1 = 50.625 g

Let the person is at a distance d from the wall

equating the CW and CCW moments about top of the ladder

d x 200g + 4.5 x 100g + 112.5 g x 34.86 = 250 g x 9

d <0

indicates the person can go up to the top of the ladder without slipping.

when the worker is 3ft from the top of the ladder

distance from the wall d= 3 x 9/36 = 0.75 ft

let f be the force to be applied at the top of the ladder then

take moments about bottom of the ladder

f x 34.86 = (9-0.75)200g + 4.5 x 100g

f = 60.24g = 1928 lbf

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