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Nonamiya [84]
3 years ago
13

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

s 160 N. The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing 740 N climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder.
(a) What is the maximum friction force that the ground can exert on the ladder at its lower end?
(b) What is the actual friction force when the man has climbed 1.0 m along the ladder?
(c) How far along the ladder can the man climb before the ladder starts to slip?

Physics
1 answer:
DiKsa [7]3 years ago
5 0

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force =\mu \left ( N\right )=0.4\times \left ( 740+160\right )

=360 N

cos\theta =\frac{3}{5}

sin\theta =\frac{4}{5}

(b)Moment about Ground Point

740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

N_1tan\theta =1140

N_1=171 N

N_1=f=171 N

(c)

740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

Here maximum friction force can be 360 N

Therefore N_1=360 N

Where x is the maximum distance moved by man along the ladder

360\times 5\times \frac{4}{3}=740x+160\times 2.5

740x=2000

x=2.702m

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