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A uniform ladder 5.0 m long rests against a frictionless,

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 10E Chapter 11

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 10E

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?

Step-by-Step Solution:

Solution 10E Introduction We have to first draw the free body diagram of the ladder with man. Then we have to calculate the maximum frictional force that ground can exert on the ladder. Then we have to calculate how far the man can go upto ladder before the base of the ladder starts slipping. Step 1 The free body diagram of the ladder is shown below. Now, since the latter is not moving in any direction. The sum of the vertical and horizontal forces will be zero. Now equating the vertical forces we have N 1 160 N 740 N = 0 N =1900 N Now, we know that the coefficient of friction is = 0.40, Hence the maximum possible frictional force is f S,max= N 1 (0.40)(900 N) = 360 N ………………………………………..(1) Step 2 (b) When man climbed 1.0 m along the ladder, then the horizontal distance of the man from the point of contact B is x = (1.0 m)cos Now from the figure we can see that the angle is given by 1 = cos ( 5 m Hence the we have x = (1.0 m) 3 m = 0.6 m ( 5 m Since the ladder is not moving in horizontal direction, we have fs N 2 0 f s N ……2……………………………………………………….(2) Now taking the moment about B, we can write that (160 N)(1.5 m) + (740 N)(0.6 m) N2(3.0 m) = 0………………….(3) N = 228 m 2 Now, using equations (2) and (3) we can write that f = N = 228 N s 2 Hence the actual frictional force is 228 N when the person climbed 1.0 m along the ladder.

Step 3 of 3

Chapter 11, Problem 10E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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