Question #107263
A uniform ladder 10m long and weighing 59N rets against a smooth vertical wall. If the ladder is just on the verge of slipping when it makes a 50 degree angle with the ground, find the coefficient of static friction between the ladder and the ground.
1
Expert's answer
2020-04-01T10:20:49-0400

As per the given question,

Length of the uniform ladder(L)=10m

Weight of the ladder =59N

Angle between the ladder and the ground =5050^\circ

Let the reaction on the bottom of the ladder N1N_1 and reaction on the wall N2N_2

Now, taking the torque about the center of the ladder

N1×L2sin50μN1L2cos50=N2×L2cos50N_1\times\dfrac{L}{2} \sin 50-\mu N_1\dfrac{L}{2}\cos 50=N_2\times \dfrac{L}{2}\cos 50

N2=N1tan50μN1(i)N_2=N_1\tan 50-\mu N_1--------(i)


Now, balancing horizontal force

N2=μN1(ii)N_2=\mu N_1-------(ii)

From equation (i) and (ii)

2μN1=N1tan502\mu N_1=N_1\tan 50

μ=tan502=0.59\mu=\dfrac{\tan 50}{2}=0.59

μ=0.59\mu =0.59


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