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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

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Comments

Assignment Expert
29.07.15, 11:12

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

Assignment Expert
28.07.15, 14:29

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

abc
27.07.15, 08:24

the first answer is wrong. cos50/sin50 is not tan(50). Its cot50

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