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.
1
Expert's answer
2015-07-28T07:30:06-0400
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
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
Dear Customer, please accept our apologies for this inaccuracy.The answer is edited now, so please check an updated version. Thank you!
Dear Customer, please accept our apologies for this inaccuracy.The answer is edited now, so please check an updated version. Thank you!
the first answer is wrong. cos50/sin50 is not tan(50). Its cot50
Leave a comment