Aladder of weight 30 lbs rests against the floor at point b and the wall at point a. how far up the ladder can a man of w = 200 lbs climb before the ladder is on the verge of slipping, given: l = 14 ft, θ = 50 °, μs floor = 0.4, μs wall = 0.25

Explanation:

The weight of the ladder will act from its middle point downwards. Reaction force will act at A and B. Friction force will act on the ladder at point B .

Reaction force at B = 230g . ( 200 + 30 )g

Friction force at B ( R ) = 230g μ

This will be equal to reaction force S  at point A on the wall.

S = 230gμ

Taking torque of all forces about point B

Moment due to weight of ladder and weight of man

30g x 14 / 2 cos50 + 200g x L X  cos50 where L is required length

= 135 g + 128.55 x gL

= 4320 +4113.6 L ( g = 32 )

This torque will be balanced by torque of reaction force S.

= S x 14 cos 50

= 230gμ cos 50 x 14

.4 x 230 x g x .6428 x 14

= 26493.64

Now for equilibrium

= 4320 +4113.6 L = 26493.64

4113.6 L = 22173.64

L = 5.4 ft .

i think the answer would be b

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