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Find the tension $T$ in each cable and the magnitude and direction of the force exerted on the strut by the pivot in each of the arrangements in $\textbf{Fig. E11.13.}$ In each case let w be the weight of the suspended crate full of priceless art objects. The strut is uniform and also has weight $w$. Start each case with a free-body diagram of the strut.

(a) $ T = 3\sqrt{3}w/2 $, $ F = \sqrt{43w/4} $, $ \theta = 37.6º. (b) $ T = 4.1w $, $ F = 5.39w $, $ \theta = 48.76º.

Equilibrium and Elasticity

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University of Michigan - Ann Arbor

Numerade Educator

University of Washington

University of Winnipeg

and it's problem. We're gonna talk about static equilibrium. So let's suppose we have a stretch and a block that's supported by this strut. And the threat itself is supported by rope and a pivot. Okay. And here we have a lot. This is pretty much the set up off the problem in our exercise. But let me give you a brief overview of how we generally solve these types off these types of problems. I'll let say that this angle here is state of, um so for an object to be in static equilibrium, we need that the some off the forces over the object be equal to zero. That must be true. And also the some off the torques that act on the object must be, which is Europe. The first condition guarantees that the center of mass of the object will move the second conditions. Condition guarantees that the torque, uh, I'm sorry. Every object will not rotate around the period. Okay, so now we can go on to our problem. And before going on, we just need to remember that. Yeah, actually, the Torkham Tom is equal to and across are where r is the distance between the point where the forces being applied and the central rotation. Um and so the magnitude of the Torkham route Nazef times are times a sign off theta where theta is the fourth, the angle between the force being applied and the line that connects the center of rotation to the point where the forces we apply. Okay, um, so now we can move on to our actors. What we have are two setups, and in the first one, I'm gonna call it A. We have the stretch exactly as before. So actually, not even going to use the same, uh, drawing. So for Anglo thing, we have being, well, 30 degrees. We have that The weight of the block is w The weight off the strut is also W. And the tension on the string is t. So what I'm gonna do is to draw force diagram on the strip. So here, at this point, we have the wait force coming from the block. Wait. Forced W At this point, we also have the tension in particular to the weight force. We have the weight force acting on the street itself. It acts on the center of mass of the structure. And finally, we have the force of the pivot exerts on the strip. You point somewhere in this direction. Okay? I'm just drawing a general direction, um, to the northeast. And also I'm going to write that w I'm going to set up a organisms such that the y axis points that Teoh the north. The acts access points to the right. So why W is equal to minus W J T is equal to minus t I where that were in jail are the magnitude of the forces eso. What we need to do is to find both the tension magnitude of attention t and also the direction and magnitude of the Force f. And here in my drawing, it seems like the force necessarily is pointing in the direction of the stretch. Since that's not true, let me just be a little clearer. Let's say that points in a general direct. So we have two equations. The first one that I'm gonna apply is the equation for the torque. So the some of the talks is zero. Notice that the strut if it were true, it rotate, it will rotate around pivot. So, uh, the force of the pivot exert. Some of the strut will not be counted in the torque equation because, uh, the distance between the point where the force is applied and the centralization is zero eso in this, uh, angle here is 30 degrees. This music, this oneness 60. And by the right hand rule, we can see that the direction of the torque coming from the tension on the ropes is equal direction is in the Z direction, even though this direction is coming out of the US and I'm gonna consider this direction, the positive direction, the negatives, the direction I'm going to consider the negative direction. So the torque coming from T eyes equal to tee times the length of the rope we call it L T L times the sign of 60 the torque that is exerted by the weight force coming from the block. I'm gonna call it how that we want is equal to omega L Times sign of 60 actually appear. I wrote that the son of 60 Street have been signed up 30 times, right, And there's a minus. Sign here because by the right hand rule, you can see that the, uh Tor coming from the weak force is pointing in the negative direction. The Tao that will you chew coming from the weight force on the strut is equal to W L over two. Since it it, uh, it acts on the central mass time sign of 60. So the some of the torques is t ell times sign off 30 minus w three w l over to sign of 60 and this is equal to zero eso the else can be canceled out so t is equal to three w over to the sign of 60 is the squared of three over to while the sign off 30 is one hand So t is three square to three that we over achieved. This is the magnitude of T That's the first Ansari needed. The second one, um, is the total force of the pivot. No notice that the some of the forces in this case is equal to the force After the pivot exerts those attention was two times w this must be which is zero. So the force is miners. Tension minus two w. If you remember, the tension is equal to minus T in the extraction. So if we put a miner's That's t I was to W J Um, no, Using that t is the square root of 33 times square of three divided by two I plus t j So magnitude of the force is w Times Square, the square root off, um, three every nine times three over four plus four Because squaring, uh, each accomplish component and then take the square root eso This is equal to the square root of 43/4 times W This is the value off the force. That's the magnitude of the force. And the angle theta will be the our attention off the why component divided by the X component. And this is equal to 37.6 degrees. And this concludes the first item on how we can move on to the second one be. And the set up we have in that item is the threat right here. Then we have an angle 45 degrees. The rope is making an angle of 30 degrees with the ground, and the block is just as before. Okay, um so before doing anything else, I'm gonna write again That, uh, W is minus w J T is equal to minus the attention the magnitude of attention times uh, the sign of sexy degrees because noticed that this angle here a 60. Okay, um so actually, let me just be a little more clear. I'm going to ride. This is the call. Sign up. 30. Mine is a cool side of 30. I plus call sign was a sign of 30 j. Okay, eso What we have to do first is to apply the total turkey is equal to zero. Uh, notice that the angle that the the rope does with the strut is equal to 15 degrees, and we can show that I'm gonna zoom in here. This is the situation. Notice that the struck that rope makes an angle of 30 degrees with the ground here isn't angle 45 degrees. So this means that, uh, in order for this to be a triangle, we need that this thing will be 15 degrees so that the sun is 90. Yeah, eso The torque exerted by the rope is equal to the force T times the sign of 15 times l The torque exerted by W the force The first verse The gravitational force on the, um, block is equal to w ell sign off 45 because this angle here must be What do you find? Tell them what you choose. Physical too. Uh, w l sign of 45 divided by two since it's applied here. So the total torque is t ell sign on 15 minus three that we will to send 50 45 and this is equal to zero. The ELCA sold out and we're left with tea is able to three w over to sign off 45 divided by sine of 15. And this is equal to 4.1 that will you. This is a magnitude attention. Now I have to find the direction and magnitude off the force F so from a four, you can see that the force of his mind as the some of the tension was too. W Now the tension is minus. T gonna write it This minus t times call sign of theater. Uh, so close. How 60 I plus sign of 60. Actually, it should be 30. I'm sorry. Outside of 30 I what time of 30 j? That's what I said before. Minus two that will you J The minus minus is console out. We have 4.1. That will you Time school sign of 30. I was sign of 30 j plus to the blue Jay. So this is equal to 3.55. Ah, that will I plus 2.5 w j was to the blue Jay, and this is equal to 3.55 w line plus four point off five w j Sophie S O. The magnitude of the force is equal to the square root of 3.55 squared was four point 05 squared on. This is equal to 5.39 mutants. Actually. About this there should be a that we're here. So 12, 5.39 w This is the magnitude and the direction of the force data is the arc tangent off 4.5, divided by 3.55. And this is equal to 48.76 degrees. And this is the conclusion of our exercise