A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable, as shown in FIGURE EX7.21. It is found that $theta sub 1$ = 20 degrees when the center rope is adjusted to be perfectly horizontal. What are the tension and the angle of rope 3?

Question

A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable, as shown in FIGURE EX7.21. It is found that $theta sub 1$ = 20 degrees when the center rope is adjusted to be perfectly horizontal. What are the tension and the angle of rope 3?

Diagram showing forces on a mobile with weights of 8 kg and 12 kg, angles, and tensions labeled.

Accepted Answer

Hey, everyone in this problem, we have an intersection with two traffic lights installed a one for traffic that weighs 12 kg and the other for pedestrians that weighs eight kg. Ok. So we have a figure that shows these two traffic lights hanging and they're connected using highly flexible and lightweight cables. We're asked to determine the tension and angle fatal one of the first cable if the center cable is adjusted to be perfectly horizontal. So we're given the diagram here of our two traffic lights. We have five tensions shown in the diagram T one up to T five. We're looking for T one and we're also looking for the angle beta one. Now we're given four answer choices. All of them have a different answer for the tension in newtons and for the angle in degrees. And we're gonna come back to those answer choices as we work through this problem. Now, we're looking for that tension, we're talking about forces here. So let's go ahead and start by drawing free body diagrams. And in this case, we have a lot to do. OK? We have four free body diagrams to draw because we have a free body diagram for each of our traffic lights. And then we also have a free body diagram for these knots. The left knot that connects T three T four and T two and the right knot that connects T two T one and T five. So let's start with the first traffic light and we're gonna use black and we have the tension T four acting upwards. We have the force of gravity acting downwards. And I'm gonna say that this is F G eight. Hey, that eight kg traffic light. So the force of gravity for that one and those are the only two forces acting on this particular object. OK. This tropically now we're gonna take up into the right as positive for all of our diagrams. We're gonna add that in now. OK? So now we're gonna move to the next traffic light. We are gonna draw this one in bolu our next traffic light. We have very similarly tension acting upwards. In this case, it's the tension T five. We have the force of gravity acting downwards and we're gonna call this F G 12. Is it the force of gravity for the 12 kg traffic light? And those are the only two forces acting here. Next, we're gonna move on to the knots. So we're gonna start with the left knot and we're gonna draw that in red with the left knot. Now what's not has the tension force T four acting downwards? It has a 10 force T two acting to the right and acting up into the left is the tension force T three. And this tension force T three makes an angle of 25 degrees with the horizontal. OK. So we can break that tension force T three up into its horizontal component, T three X and it's vertical component T three. Y. Similarly, we have the right knot. We're gonna draw this one in green and it is just like that left knot. In this case, we have the tension T five acting downwards. We have the tension T two acting to the left, we have the 10 to 1 acting up and to the right. OK. And that makes an angle of theta one. Oops, sorry. This is not T two acting up into the right, this is T one acting up into the right and that makes an angle of theta one with the horizontal. OK. Where T one and theta one are what we are looking for. And again, we can break that tension T one up into its components, T one X for the horizontal component and G one, Y for the vertical component. All right. Well, let's start by thinking about our forces. If we want to find T one and theta one, we can look in either the Y direction or the X direction of our green free body diagram. You'll see that in order to find T one, we're gonna need T two and, or T five. And we don't know either of those yet. So let's start with T five because that's gonna come from our blue diagram. OK. Which is a little bit simpler. We only have two forces acting. One of them is a force of gravity which we know we can calculate. So let's start there. OK. So we're gonna start with the sum of forces in the Y direction for our ball blue diagram. And I'm gonna write all of these equations in the corresponding color through the diagram. So we can keep track of everything we're doing because there are gonna be a lot of calculations we're working through in this problem. So in the positive Y direction, we have T five in the negative Y direction, we have the force of gravity for this 12 kg um traffic light. So we have T five minus F G 12. And I've written that, that some of the forces is equal to zero. And that's because these traffic lights are an equilibrium, they are not moving. OK? So this is gonna be equal to zero that tells us that that tension force T five that we're looking for is equal to the force of gravity of the kg traffic light A, we call it, the force of gravity is equal to the mass M, we're gonna call it M 12 multiplied by the acceleration due to gravity G. And if we substitute in our volumes, OK. Again, this is a 12 kg traffic light. So the mass is 12 kg, the acceleration due to gravity is 9.8 m per second squared. And we get that this is equal to 117. newtons. And that is T five. All right. So we've got T five and I'm just gonna put a little check mark on the diagram be beside T five. So that we know that we've already found it right now. We can go back to the Y direction of our green diagram. OK. Where we have that tension T one that we're looking for. OK. In the Y direction, we have the tension T five and we have the Y component of T one. So let's go ahead and write up that equation now that we know T five and see if we can find out any information about this tension T one that we're looking for. So we have at the sum of the forces in the Y direction again is gonna be equal to zero. This is an equal agreement. It's not moving in the positive Y direction. We have the Y component of T one Y in the negative Y direction. We have T five. So we get T one, Y minus T five is equal to zero. That tells us that T one Y which we can break down using T one in our angle theta OK. This is the Y component is gonna be the opposite side of our angle. So it's gonna be related through the sign. So we have T one sign of theta one and we can move our T five to the right by adding it. So T one sine theta one is going to be equal to T five, which we just found to be 170 sorry, 117.6 nodes. Now we have two unknowns in this equation T one and theta one. So we can't solve for either of them yet. What we need to do is use the X direction to also write an equation in terms of T one and theta one, then we'll have two equations, two unknowns and we'll be able to solve. OK. So I'm gonna put a star beside the equation you've just written because we're gonna come back to that one. So now we wanna look in the extraction of our green diagram in order to find T one X and theta one or to have only those variables, we need to know T two. OK? Where else does T two appear? Well, T two appears in this red diagram in the X direction. OK. Now, in the X direction, we also have T three, the X component of T three. Now we don't know T three yet either. Well, how can we find T three? Well, it also appears in the Y direction. We don't know that Y component. But in that Y component, we also have T four, right? Where does T four appear? Well, T four appears in our block diagram and we can calculate T four because the only forces acting are T four and the force of gravity on that eight kg traffic light, which we know how to calculate. OK. So we're going back multiple steps here. We're gonna first find T four. That's gonna allow us to find T three. That's gonna allow us to find T two, which will then allow us to write this equation for the X direction of T one. OK. So that's why I've written these colors because we're doing a lot of steps here. We're working with a lot of different equations. So we're working with the black diagram, we're gonna write in black and we're looking at the sum of the forces in the Y direction which we know is equal to zero again. OK. In the positive Y direction, we have the tension force T four in the negative Y direction. We have the force of gravity of that eight kg traffic light, we get T four minus F G eight is equal to zero. This tells us that the tension force T four is gonna be equal to the mass of that eight kg traffic light multiplied by the acceleration due to gravity. OK? That's that force of gravity. And if we substitute in our values, we have eight kg multiplied by 9. m per second. Squared, which gives us a 10 T four of 78.4 newtons. OK? All right. So going into our diagram, we're gonna give a checkmark to that one as well. So we know that we found it, we know we have it if we need to use it and we're gonna switch back over to this red diagram. OK? We know T four. Now, in the Y direction, it's gonna allow us to find T three. So let's do that. We have the sum of the forces and we're looking in the Y direction again, this is equal to zero in the positive Y direction. We have T three Y in the negative Y direction, we have T four, T, three, Y minus T four is equal to zero. We can move T four over to the right hand side and we can break T three up, OK? This is a Y component. We know it's gonna be related through sign of the angle because it's the opposite side. And so we can write T three, Y as T three multiplied by stein of degrees. And this is gonna be equal, moving T four to the other side to positive T four, which we just found to be 78 0.4 newtons. This tells us that T three is gonna be equal to 78.4 noons divided by sine of 25 degrees. And we are gonna leave that just like that. We're not gonna calculate that, that value and round just yet, we're gonna try to leave the rounding until the end. So we don't end up with more rounding error than we need. All right. So we have T three. Now, now we can get to the X direction of our red diagram. OK? We have T three. We can calculate T two and remember that's what we want so that we can work with the X direction of our green diagram. So we're getting closer and closer, working through this bit by bit. And this is what you have to do. When you have multiple free body diagrams, you just kind of got to work backwards. OK? Figure out what it is you're trying to find and how you can get it. So now we're looking in the X direction of our red diagram, the sum of the forces in the X direction is gonna be equal to zero. OK. These are all in equilibrium. They are all not moving in the positive X direction. We have the tension force T three X. And you know what? That is not in the positive direction that is in the negative direction T two is in the positive direction. That's my mistake. So we'll put a negative in front of T three X and then we have plus T two is equal to zero. That tells us that the tension force T two is gonna be equal to the X component of T three X, OK. Just like we wrote the Y component with sine, we're gonna write the X component with cosine because it's the adjacent side. So this is gonna be equal to T three multiplied by cosine of degrees. And we know what T three is, OK. We found that T three is 78.4 newtons divided by sine of 25 degrees. And so T two is gonna be 78.4 newtons divided by sine of 25 degrees multiplied by cosine of 25 degrees. Now sine divided by cosine is equal to tangent. OK. You can recall that from trick and so here we have cosine divided by sine. And so what we really have is tan in the denominator. So we have T two is equal to 78.4 newtons divided by the tangent of 25 degrees. OK. So now we know T two and T three. If we look back at our red diagram, we know all of these components. T two, T three, the X and Y components. OK. All right. Now that we know T two, we're getting back to this green diagram. Again, we wanna find T one and theta one right now, our star equation, we have one equation with two unknowns. So we need to use the X component to write another equation with two unknowns. So the horizontal component of this equation, we have that the, some of the forces in the X direction is equal to zero. OK. In the positive X direction, we have the tension T one X K, the X component and then negative X direction we have T two. So T one X minus T two is equal to zero A T one X, we can write as T one multiplied by cosine of theta one. OK? Because it's the adjacent side, this is gonna be equal to the tension T two, which we know is equal to 78.4 newtons divided by the tangent of 25 degrees. So now we have two equations. We have our equation star which is T one sine theta one is equal to 117.6 newtons. And we have this other equation which we're gonna call equation two which is T one coast theta one is equal to 78.4 newtons divided by tangent of 25 degrees. Two equations. Two un notes, there are a couple of different ways we could solve this. We could solve this using the substitution method, the elimination method. What we're gonna do is we're gonna go ahead and divide these two equations. If we divide these two equations, the T one term is gonna cancel out and we're gonna be left with just the one as an unknown which will allow us to solve. Think. So let's give ourselves some more room to work this out. I want to leave that equation. Up. So we can see it, we're gonna take equation star and divide it by equation two. Hm. So we get T one sign beta one equals 117.6 new. On the left hand side, we divide by T one cosine of beta one K, the left-hand side of our equation two. And on the right hand side, we divide by 78.4 newtons divided by tangent of degrees. All right. This looks messy right now. But remember that everything on the right hand side is just a number. OK? We haven't simplified that tangent yet because we wanted to avoid rounding error, but it's just a number. All right. Simplifying on the left-hand side, I mentioned this before, but these T ones are gonna cancel, they're gonna divide it. We have sine theta one divided by cosine of theta one which is just tangent of theta one, right. And this is gonna equal, we have 78.4 newtons divided by tangent of 25 degrees in the denominator. So we can actually write this as our 117.6 newtons multiplied by tangent of 25 in the numerator. And all of that is divided by 78. newtons. And if we take the inverse tangent of everything on the right hand side, we get that theta one is equal to 0.97 degrees. OK? So we have our data one value. Now, that's one piece of the puzzle we were looking for. Next is to calculate that tension. Now, we have equation star and we have equation two, we can use either of those to find that tension. Now that we know theta one. So let's use equation Star, we're gonna substitute theta one equals 34.97 degrees into that equation. We got team one multiplied by sign of 34.97 degrees is equal to 117.6 newtons. And to solve for T one, we're gonna divide by sign of 34.97. We get that T one is equal to 205. newtons. And that's that first tension, the tension of the first cable that we were looking for. So now we have the tension, we have the angle, we have everything we needed for this problem. Let's go back up to our answer traces and see what we have. OK? And we can see that the correct answer corresponds with answer choice. A hey, our tension was approximately 205 noons and the angle is about degrees. Thanks everyone for watching. I hope this video helped see you in the next one.

A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable, as shown in FIGURE EX7.21. It is found that θ1\theta_1 = 20 degrees when the center rope is adjusted to be perfectly horizontal. What are the tension and the angle of rope 3?

Verified Solution

A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable, as shown in FIGURE EX7.21. It is found that $theta sub 1$ = 20 degrees when the center rope is adjusted to be perfectly horizontal. What are the tension and the angle of rope 3?