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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 weighs 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?

a. $360 \mathrm{N}$b. $171.0 \mathrm{N}$c. $2.7 \mathrm{m}$

Physics 101 Mechanics

Chapter 11

Equilibrium and Elasticity

Section 3

Solving Rigid-Body Equilibrium Problems

Kyle P.

April 26, 2020

NO FBD

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

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So we know that the Wallace frictionless Ah and so defined the maximum frictional force. We simply just need to find the normal force exerted by the floor. So we can say that the sum of forces in the UAE direction equals zero and this would say and to equals Don't be so boring. Clothes, W C M. And this is equaling 1 60 plus 7 40 So 900 Nunes. So this would be the magnitude of the normal force exerted by the floor on the ladder. So we can say that the maximum friction I was going to be equal to the coefficient of static friction. This would, of course, give us the maximum friction because thie coefficient of kinetic friction is always less than the coefficient of static friction times. So it's music best times and some too. And this would be point for Times 900 and this is giving us 316 yoon. So this would be your answer for party. This should be the maximum frictional force on DH. This occurs between the point of contact of thie ladder and the floor for Part B. They wanted to find, and someone so this would be the normal force that is exerted by the wall on the ladder. So we can say that the sum of torques equals air again. The ladder is in rotational equilibrium. This would equal 1.5 meters times 160 Newtons plus one meter times three over five times 740 Nunes and then minus for 0.0 meters times ends of one. And we simply saw for and someone so and someone is going to be equal to four times 360 better my qualities for a 1.5 times 100 60 plus three over five times 7 40 And this is going to be divided by four meters. So for and this is giving us 171 Newton. So this would be the magnitude of the normal force between the batter and the and the wall. And lastly we have part. See, we still have rotational equilibrium. However, now that just wants to find X, so we can say that four meters times 360 Nunes is going to equal 1.5 meters times 116 unions plus X meters times three over five times, 749 years. And we find that acts is going to be equal to four times 3 60 minus 1.5 times 1 60 divided by three over five times 7 40 And this is giving us 2.70 meter. So this would be your value for X. This would be your answer for part. See, that is the end of the solution. Thank you for watching.

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