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A uniform aluminum beam 9.00 m long, weighing 300 N, rests symmetrically on two supports 5.00 m apart ($\textbf{Fig. E11.12}$). 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 $F_A$ and $F_B$ 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?

a. The graphs are given in Figure 11.12b. $x=6.25 \mathrm{m}$ when $F_{A}=0,$ which is $1.25 \mathrm{m}$ beyond point $B$c. $1.50 \mathrm{m}$

Physics 101 Mechanics

Chapter 11

Equilibrium and Elasticity

Section 3

Solving Rigid-Body Equilibrium Problems

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

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So we have the free by diagram for apart A here on DH. They're asking us to find the force with respect to X. So we can say that some of forces for the sum of the torques from the point A would be equal to 600 Newtons times X plus 300 Nunes times 2.5 meters but and then minus the force of B times, five meters And we can just say here here is a 600 to enforce. And then here we can say somewhere it was the 300 noon force and we can say at this distance from here it's going to be 2.5 years. So essentially here we're going Teo solve for force of be because we know that this is going to be zero because this is in rotational equilibrium. So we can say that the force of B is going to be equal to 120 x plus 150. So as you as you move farther and farther away from the point, eh? We know that force of B is going to be increased. Let's do the sum of torques. We know Ah, from point B and this is going to be equal to zero because again, we have rotational equilibrium and this would be equal to the force of a times five meters minus 300 Newtons, times 2.5 And then again, no minus 600 Nunes And here it would be five minus acts. So I would be the distance from B and we when me say when we use this, we can say that force of a equals 750 minus 126. So in order in order graph this, we simply need thio weakened labour weaken plot these both in the same graph and essentially the force will be on the Y axis and then the ax will, of course, be on the X axis And this would these would be your two equations for the line. So we want to graph this. It would look something like this. So this would be the graph of force of B and the force of a ah graft on the same graph again, your force in Newton's is your Y axis and your accent meters would be your X axis. Now for part B they want us to find the ex with which the bar tips so we can use any. We can use the formula for the force of a or the form of the field force would be. I'm going to use the formula for the force of A and I'm going to say force of a equal 750 minus 120 x, and we're going to set this equal to zero, and this is going to give us acts equaling 6.25 meters now. This essentially means this would be a distance from Point A, and this critical distance would essentially mean that this is going to be the distance of from point A before the bar tips. So essentially, it would explain, equals 6.25 meters. If it does not equal 6.25 meters, then essentially the bar is going to tip and it won't be in rotational equilibrium anymore. Now here they're saying that for part, see, this would be your free body diagram for part C, and they're saying that for part C, let's just say that force of a equals zero Newton's. So let's say the Sigma T Sigma Tau sort of sum of torques evaluated at point B will be equal to zero. So again, this isn't Rotational Eagle, and this would be equal to ass of a seven meters plus 600 noone's times Why minus 300 Nunes times 4.5 meters minus y. And we find that why is going to be equal to 1.5 meters and this would be the location of point B and this will be again from right and of being so this would be the distance such that we have a rotational equilibrium if the force of a suddenly were to break. So this term right here is going to be equal to zero. That is the end of the solution. Thank you for watching.

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