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A 9.00-m-long uniform beam is hinged to a vertical wall and held horizontally by a 5.00-m-long cable attached to the wall 4.00 m above the hinge ($\textbf{Fig. E11.17}$). The metal of this cable has a test strength of 1.00 kN, which means that it will break if the tension in it exceeds that amount. (a) Draw a free-body diagram of the beam. (b) What is the heaviest beam that the cable can support in this configuration? (c) Find the horizontal and vertical components of the force the hinge exerts on the beam. Is the vertical component upward or downward?

a. The free-body diagram is shown in Figure 11.17 , with $\vec{T}$ resolved into its $x$ - and $y$ - components.b. $533 \mathrm{N}$c. $267 \mathrm{N}$

Physics 101 Mechanics

Chapter 11

Equilibrium and Elasticity

Section 3

Solving Rigid-Body Equilibrium Problems

Van S.

May 24, 2021

The vertical component of the hinge force is actually downward

Fv = mg - T(sin53º) = 54kg (9.8m/s^2) -1000kg*m/s^2(sin53º) = -266 N

University of Washington

Hope College

University of Winnipeg

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problem. 11.17. We have a beam attached to a wall, uh, by a hinge that's supported by a cable being nine meters long. The maximum tension that the cable can support before it breaks is one killer Newton, and so we'd like to do is draw free body diagram of the forces acting on the beam. We want to find what the maximum weight of the being could be before the able break. And then we want to find the X and Y components of the force that the hinges exerting on the beam. And then we want to know whether the vertical component of that is positive or negative. So or another way of putting It isn't directed upwards or downwards, because whether it's positive or negative will depend on whether we put our Y axis going up or down. So to begin with, you are free body diagram, so we have our being halfway along it. The weight will be acting, Um, that's to the left of that will be the tension and the cable, which has its X and Y components and then the force from the hinge or have proponent H sub X and then we'll have the why excess going downwards. I will say that the H supply is downwards. And if we're wrong about this, then we just get a negative number and then we know it's actually part of the other. And as usual, we're gonna call the counter clockwise the direction of positive rotation. And another thing we should do before we go too much further. Figure out. Actually, you know what is the the distance between the hinge of where the cable attach is which we could call l and so this is this is fairly straightforward. We know that l squared us 16 meters squared is equal to 25 meters squared 6 25 minus 16 as nine. And so Ellis three meters. Uh and then furthermore, the, uh, angle fated that this makes is going to be the dark side, uh, four meters divided by five meters. Is it over? I thought muse, and so fade is value is 53.1 degrees. Good. So the maximum lee we confined using our equilibrium conditions for the tour tableaus in the sob. And this gives us that attention. The magnitude of the tension which will said equal to the maximum one killer Newton, because you want to know what the maximum weight times the sine of the angle. Because only the vertical component of the tension exerts a tour as it's a cross product. And if a court Forrester you know, vector in general is directed in the same, the same, uh, direction is another vector. You're taking the cross product of its its the value of the cross biotic zero. This will be times R l, which is three meters. It's the weight of the beam. Times are four and 1/2 meters. If we call the entirely for this big l l over too zero. And we're not that we don't have any forces from the hinge because we're setting this equal equal. We're using this as the axis about which were computing the ports. And so then solving this for the lead we have TL science data over l over too is equal to the weight and again using a one killer Newton. Maximum tension gives us the maximum weight, which is going to be 533 nutrients now in part. See, we want to find the components of the force that him is exerting starting new page for this. So here will use our other equilibrium conditions that the sum of all of the x and Y components of the forces acting on, uh on the B and have to be zero because it's not accelerating or we don't want it to be accelerating. So for X, this gives us that the vertical component of the force from the hinge minus the the horizontal component of attention in the wire has to be equal to zero. And so then you just add this to both sides. And so this is going to be our one hell of Newton tightens a coastline that 53.1 degrees and so him is exerting a force of 600 Newtons. So the right for why we call that we have if survive in the negative, wider action which may or may not be true and if it's not true, will get a negative number and they don't know it's actually upwards. So we have the horizontal or vertical rather component of attention. Uh minus like component of the hinge force minus the late all add up to zero than adding this to both sides. Supply equals just the difference between the vertical part of the tension and the weight, and this has a value of 267 Newtons, which means that the horses, in fact outwards.

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