A uniform 95-kg flagpole of length 8.4 m is being erected by p...

Question

A uniform 95-kg flagpole of length 8.4 m is being erected by pulling on a rope attached 2/3 of the way to the top (Fig. 12–94). When the pole is inclined at 35° and the rope makes an angle with the ground of 18°, what is the tension in the rope?

Accepted Answer

Welcome back. Everyone in this problem. A uniform 87 kg column is held still by a cable that is attached to it at a point that is three quarters of the length from the ground. The other end of the cable is attached to the ground which makes an angle of 33 degrees with the horizontal given that the length of the column is 16 m and the angle between the column and the ground is 83 degrees. Calculate the tension in the cable for our answer choices. A says the tension is 1.1 multiplied by 10 to the negative two newtons. B 9.1 multiplied by 10 newtons. C 2.7 multiplied by 10 squared newtons and D 5.7 multiplied by 10 squared newtons. Now, if we're going to figure out the tension in the cable, let's make sure we understand all of the forces that are acting here, both our translational and rotational forces in our diagram. So I'm going to reproduce this diagram on the side. Just try to look a bit simpler. OK? For us to do our free body diagram here. OK? And for it, we know that we have our attention T and we know the value of the angles. I'm not going to put those values in yet. But what do we know so far? Well, we can let L be the length of the column. OK. If that's the case, then we can consider the equilibrium condition of torque above the 0.0 here to find an expression for the tension T. Now, since the column has uniform mass distribution, we can assume the weight acts at the midpoint of the length. In other words, we could say the midpoint of the length is here and the weight will be acting vertically downwards. So our weight is acting at a distance of half L from the foot of the column. Now let's think about our torque here for that length, half L. Then to find our torque, we can multiply that force by its perpendicular distance. And in this case, that perpendicular distance would be going in this direction. OK. So far OK. We already know that we have our 83 de degree angle here in our diagram. So this is being affected by the angle that we can, we can use here as Phi, OK. And thus the component of the way that is acting here is MGS. Well, let me write it about it. OK, MG Sine Phi. OK. So the torque due to the weight is going to be the product of uh our distance, which is a half of L multiplied by MG sine phi. So let's make a note of that here. So the torque due to the weight equals M sorry, half L multiplied by MGSP. What else do we know? Well, we also know that our cable with the tension, it is at a distance of 3/4 of L from the foot of the column. OK. So in other words, we know that this distance is going to be 3/4 of L. So we, we're going to be concerned with the perpendicular force for that distance which is going to be coming out here. OK. And if we say up here that this is our angle theta, then that component is going to be T sine theta. OK. So that's the product of T sine theta and 3/4 of L is going to be the torque due to the tension. So in other words, the target due to the tension is 3/4 of hell multiplied by. Well, let me put that here, 3/4 of L multiplied by T sine theta. OK? No, with that in mind, then like we said at the equilibrium position, the net network is equal to zero. OK. So at equilibrium add equilibrium equilibrium, we can see them that or net torque which is going to be equal to the torque or torque due to the weight minus the torque due to the tension is going to be equal to zero. That tells us then that the torque due to the tension will equal the torque due to the weight. So 3/4 OK? Of LT sine theta. OK? Is going to be equal to a half of L oops, my apologies there. Yes, it's gonna be equal to a half of L. OK. Multiplied by MG sine phi. OK. Well, no, we can simplify here because if we can cancel L from both sides and solve for T, then that means T is going to be equal to two mg sine phi. OK. Divided by three sine theta. So we are on our way to find the tension. All we need to do now is to figure out the values of Phi and theta because we already know the mass and we can take G the acceleration due to gravity as 9.81 m per second squared. So first, let's try to find our angle, Phi. Now again, if we go back to our diagram, OK. Notice that here, we knew that this angle between the column and the ground is 83 degrees. OK. So if I were to represent that here, we know that this anger is 83 degrees. OK. Notice here it forms a right angle with, well, the weight and our horizontal, the ground forms a right angle. So our angle between the column and the weight would be equal to 180 minus 90 plus 83. OK? And that is going to give us seven degrees. Let me actually let me point the toad from the side here. Well, I think we can all see it. So that would give us seven degrees. Ok. Now, if you notice here, we have a straight line on which Phi and our seven degree angle is and angles on a straight line some to 180. So what that tells us then is that Phi ok, is going to be equal to 180 minus seven degrees, which equals 173 degrees. Next, what do we know about our Anger Theater? Well, here again, notice that we have a right, well, not a right angle, but we have a triangle formed. OK? And we know that this angle is 33 degrees. Now notice that our angle here, if we call our angle at the top B of this triangle, OK, then notice that B B is a, well, our 83 degree is an exterior angle and some of our remote interior angles would be equal to our exterior angle. In other words, B plus 33 degrees must be 83 degrees. So since B and 33 make 83 degrees, then B is going to be equal to 83 degrees minus 33 degrees, which is 50 degrees. And that is useful because again, we have a straight line here. OK. So if B is 50 degrees, then theta must be 100 and 80 minus 50 degrees. OK. So that means theta equals 180 minus 50 which is 130 degrees. Now that we have a value for a theta and we have a value for Phi, we can substitute it to solve for attention because this tells us that our attention is going to be two multiplied by the mass 87 kg multiplied by acceleration due to gravity multiplied by the sign of 173 degrees, all divided by three multiplied by the sine of 130 degrees. Now, when we put, when we put this uh expression into our calculator, then we find that we get our attention t to be equal to 90.518 newtons. If we write it to two significant figures like the rest of our answers, then it's going to be 9.1 multiplied by 10 newtons. Thus, the tension in the wire is 9.1 multiplied by 10 newtons. And when we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

A uniform 95-kg flagpole of length 8.4 m is being erected by pulling on a rope attached 2/3 of the way to the top (Fig. 12–94). When the pole is inclined at 35° and the rope makes an angle with the ground of 18°, what is the tension in the rope?

Key Concepts

Torque

Trigonometry in Physics

Equilibrium of Forces

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