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# A uniform aluminum beam 9.00 m long, weighing 300 N, rests ISBN: 9780321675460 31

## Solution for problem 12E Chapter 11

University Physics | 13th Edition

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Problem 12E

A uniform aluminum beam 9.00 m long, weighing 300 N, rests symmetrically on two supports 5.00 m apart (?Fig. E11.12?). A boy weighing 600 N starts at point A and walks toward the right. (a) In the same diagram construct two graphs showing the upward forces FA and FB exerted on the beam at points A and B, as functions of the coordinate x of the boy. Let 1 cm = 100 N vertically, and 1 cm = 1.00 m horizontally. (b) From your diagram, how far beyond point B can the boy walk before the beam tips? (c) How far from the right end of the beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip?

Step-by-Step Solution:

Solution 12 E Step 1: In the question,we need to construct a graph showing force F and F on the beam at A B point A and B From this graph we need find how far beyond the point B can the boy before the beam can tip In the third part we need to find the the distance at which the support B has to be placed such that the boy can walk tipping Data given Length of the beam L = 9.0 m Force exerted by the boy on the beam W =B600 N Weight of the beam W = L00 N Distance between the supports d = 5.0 m Step 2 : We shall find distance of the boy from point B We shall assume that the boy is verge tipping , hence the force at point A will be F = 0 A And the torque at point A will also be 0 that is T = 0 Thus we can write as F = W × ((2.0 m + d) (4.5 m )) + W x A L B A Note L/2 = 4.5 m Substituting values we get F = 300 N × ((2.0 m + 5) (4.5 m )) + 600 N × x A B 0 = 750 N + 600 N × x B 600 N × x B 750 N xB= 750 N/600 N xB= 1.25 m From point A x = 7.0 m + 1.25 m = 8.25 A The force on point B is F B 600N × 1.25 m FB= 750 N

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##### ISBN: 9780321675460

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