# Answer to Question #13047 in Mechanics | Relativity for regina

Question #13047

a uniform ladder weighing 330N is leaning against a wall. the ladder slips when the angle with the floor 50degrees. assuming the coefficient of static friction at the wall and the floor are the same,obtain a value of coefficient of static friction.

Expert's answer

The normal force F1 is:

F1 * L * sin 50 = W * L/2 * cos 50

F1 = (1/2) * W

* cot(50) = 0.4195 * W

The normal force at the base equals the weight W, since

the other

contact point is smooth and frictionless. If it is just about to

slip,

the maximum static friction force at the base (Ff,max) equals

the

normal force at the top, which I called F1.

Ff,max = mu * W = F1 = W/2

cot(50)

mu = F1 / W = (1/2) cot(50) = 0.4195

F1 * L * sin 50 = W * L/2 * cos 50

F1 = (1/2) * W

* cot(50) = 0.4195 * W

The normal force at the base equals the weight W, since

the other

contact point is smooth and frictionless. If it is just about to

slip,

the maximum static friction force at the base (Ff,max) equals

the

normal force at the top, which I called F1.

Ff,max = mu * W = F1 = W/2

cot(50)

mu = F1 / W = (1/2) cot(50) = 0.4195

## Comments

Assignment Expert29.07.15, 11:12Dear Customer, please accept our apologies for this inaccuracy.The answer is edited now, so please check an updated version. Thank you!

Assignment Expert28.07.15, 14:29Dear Customer, please accept our apologies for this inaccuracy.The answer is edited now, so please check an updated version. Thank you!

abc27.07.15, 08:24the first answer is wrong. cos50/sin50 is not tan(50). Its cot50

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