The three ropes in FIGURE EX6.1 are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with the tensions shown in the figure. What are the magnitude and direction of the tension T3 in the third rope?

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Hey, everyone in this travel, we have an object tied using three wires. The object is extremely small and lightweight and is in static equilibrium. And we're asked to use this given figure to calculate a third wire's tension in the magnitude and direction. We have a diagram and we have these three wires joined at a point. Yes. So we have the tension T three acting kind of up into the left. We have the tension T one acting up into the right making an angle of 45 degrees with the horizontal and that has a magnitude of 40 noons. And then we have the tension T two acting kind of down into the right, making an angle of 45 degrees below the horizontal with a magnitude of 60 noons. For given four answer choices, option A through D, each of them containing a different magnitude and direction. OK. So we have combinations of the magnitude being either 85 newtons or 72 newtons with the direction being either 169 degrees with the positive X axis or 79 degrees with the negative x axis. And we're gonna come back to those answer choices as we work through this problem. First thing we're gonna do is go ahead and draw a free body diagram. We're gonna draw it for the point where all of this is going on where all of these um wires are joined. So we have the tension T one acting up into the right OK. Magnitude of 40 newtons. And that's gonna be 45 degrees from the horizontal. And let me make that it's not quite horizontal. Is it? Let me make that more horizontal? There we go. That looks better. And we're gonna do the same with T two acting down into the right with a magnitude of 60 newtons. And again, at an angle of 45 degrees, we can break these tensions up into their X and Y components. OK. So we're gonna have the tension T one Y, the Y component acting upwards T two Y, the Y component of T two acting downwards and both of their X components are going to be pointing to the right. So T one, X and T two X and then we have T three and we are drawing it pointing straight to the left. But what we really want to figure out is what direction does it point. And so we're drawing it pointing to the left, but that's just um it's not set in stone of the, it's acting in this direction. So, hm, now we're told that this isn't static equilibrium if it's an equilibrium then the sum of the forces is equal to zero. So let's start there. We can say that the sum of the forces in the X direction is equal to zero right in the X direction. What forces do we have acting? Well, we have T one X acting in the positive X direction. We have T two X acting in a positive direction and we have T three X acting in the negative direction. OK. And let me just go back to our diagram. Let me label as T three X and we'll put A T three Y as well based on the figure we expected to be pointing to the left and upwards. So that's the direction that we're drawing those components in our diagram. But again, we'll see if that's true when we do the calculation. All right. So T one X plus T two, X minus T three X is equal to zero. We want to solve for the X component of T three. Having those components is gonna allow us to calculate both the magnitude and direction. This is gonna be equal to T one X plus T two X. Now, for T one X and T two X, we want the X components. So we're gonna look at the cosine of the angle. OK? Because that's gonna relate that adjacent side that we're interested in. And so we get 40 newtons multiplied by cosine of 45 degrees for T one X and then we're gonna add T two X which is going to be 60 newtons multiplied by cosine of 45 degrees. So the same process, they just have a different magnitude and this is gonna give us a T three X value of about 70.7 newtons. Hm. So we have the X component of our tension. We need to find the Y component as well so that we can calculate the magnitude and direction we're gonna go about this the same way. But we're gonna look at the, some of the forces in the Y direction now and the sum of the forces in the Y direction. Again, it's going to be equal to zero just like in the X direction because we are in static equilibrium. Let's take a look at our diagram in the positive Y direction. We have the Y component of T one, the Y component of T three. And in the negative direction, we have the Y component of T two. And so what we get is T one, Y minus T two YKT three Y is equal to zero. This tells us that T three Y it's going to be equal to T two, Y minus T one Y. Are you moving those to the right hand side? Now, these Y components are going to be related through sine. OK. The horizontal components were related through cosine because we were looking at the adjacent side. Now we're looking at that opposite side. And so we get that T three Y is going to be equal to 60 noons multiplied by sign of 45 degrees minus 40 newtons multiplied by Z of 45 degrees. And if we work this out, we get that, that Y component of T three is about 14.1 or two new. OK. And what we can see is that T three X and T three Y were both positive when we calculated them, what that tells us is that the direction we assumed that they were in is correct. OK. So the horizontal component does point to the left, the vertical component does point up, OK. Those directions we implied or reassumed are correct. OK. So let's go ahead and draw out this wire with our tension T three. We know it's going to point to the left and upwards. The vertical component is gonna be about 14.142 newtons and the horizontal component is about 70.7 nos. OK. That was our horizontal and vertical components of our tension T three. Now T is going to be the hypotenuse of our triangle and then the angle theta is going to be the angle from B at this point, the negative X axis upwards to B 10. So let's start with the magnitude. OK. We have a right angle triangle, we can use Pythagorean theorem. So the hypotenuse which is T three squared is going to be equal to the sum of the squares of the other two sides. So 70.7 Newton squared plus 14.142 Newton squared. This tells us that T three squared is gonna be about 5198 5486164, Newton square leaving lots of digits there. So we have no round off error and we get that the magnitude of our attention T three is about 72.1 news. All right. So we found the magnitude. Let's look at our answer choices for a second. So option A and C had that the magnitude was 85 newtons. OK. That's not correct. So we can eliminate those options. B and D had 72 newtons. So that is what we found. OK. So we're looking at either option B or D. Now we need to look at this angle of the, to get our final answer. So we can relate the through tangent because we have the opposite and adjacent sides. OK? So the is going to be the inverse tangent on the opposite side. 14.142 newtons divided by the adjacent side, 70.7 newtons. When we work this out, we get that data is going to be about 11.31 degrees. OK? All right. So if we were to go back to our answer choices, we would not see 11.31 degrees. And the reason for that is that we have to think about where this is relative to OK. So from the negative X axis, that's where we're measuring our angle. But if we imagined, let me do this in green, drawing our entire plan out our coordinate system and we wanted to measure from the positive axis, then our angle would go from our positive X axis in the counterclockwise direction to T three. And remember that we always measure counterclockwise when we're dealing with. And so the actual angle we are looking for which we're gonna call theta star is going to be equal to 180 degrees minus that theta value, right? So theta star is 180 degrees minus 11.31 degrees because 180 degrees would take us all the way to the negative X axis and then we're gonna subtract that little angle theta that's left over. OK. So theta star in the end is going to be 168.69 degrees from the positive X axis. All right. So now we have our direction as well as our magnitude. Let's go back to our answer choices. Now, option B, we had that this was 79 degrees from the negative X axis. That is not correct. But option D has 169 degrees with a positive X axis, which is what we just calculated. So the correct answer here is option D that third wire is going to have a tension of 72 newtons with the direction of 100 and 69 degrees from the X axis. Thanks everyone for watching. I hope this video helped you in the next one.

The three ropes in FIGURE EX6.1 are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with the tensions shown in the figure. What are the magnitude and direction of the tension T3 in the third rope?

Verified Solution

Key Concepts

Equilibrium of Forces

Vector Components

Trigonometric Functions