Problem

\bulletA good workout. You are doing exercises on…

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Problem 57 Hard Difficulty

$\bullet$ You are asked to design the decorative mobile shown in Figure 10.69 . The strings and rods have negligible weight, and the rods are to hang horizontally. (a) Draw a free-body diagram for each rod. (b) Find the weights of the balls $A, B,$ and Find the tensions $S_{1}, S_{2},$ and $S_{3}$ in the strings. (c) What can you say about the horizontal location of the mobile's center of gravity? Explain.

Answer

a) See drawing
b) $24 \mathrm{N}]$
9 $\mathrm{N}$
$= 8 \mathrm{N}$
32 $\mathrm{N}$
c) The center of gravity must be somewhere directly below $S_{1},$ otherwise the mobile would tilt.

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Physics 101 Mechanics

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Video Transcript

Hello. Number 57 talks about a mobile that you see in the picture there. And you have some various masses hanging from them and, uh, some different things going on here. So I'm gonna drown a little bit of a picture and more as to sell for several numbers here. So first off, we know we have an upward s one right here that were asked to sell for, and we have a downward s two. This downward force on the mobile is also the upward force right here. And this is string three downward here. Aboard. Here. Where has to solve for all three of those? Well, so that three different objects. We have objects. See? Right here. Object to be right here. 16 18 Object rule right here. An object? A right here. We've got all this in equilibrium drawn on there, the different distances. So what? We're going to work out our kind of three different rods for all three of the rods were gonna look at the network, which is zero and the Net Wife forces, which are CVA. Okay, so the star may as well start up here. Okay? And look at this one you know the net torques here are zero. And if we look at that, that would tell us that s one is equal to If we're going to use this rotating point right here, it would be C times 0.0 a. And if we also look at the forces here, we would see that s one is us too. Let's see. We're gonna jump over here and look at this middle one. Same thing are torques zero use to the point here that be we would see that as to equals as three. No, wait. No note Forces being zero on this one would tell us that as two this big was as three. No, don't want to put our pivot point on this one right here, because this is our only known value. Go and stick our pivot point here. I don't want to cancel out that six. We need its help. So here, Arnett torques zero. So if we measure everything from a we see, that s three. His point of late is equal to six times 0.12 Hey, we can actually solve something right here. We can solve for us three. That tells us It's nine. News that one. Okay, go back to net Forces zero. We can see that as three is six waas a Now we just saw for us three. So it helped us out with a Is three. Okay, now, kind of working backwards. Since we know as three here, let's go in and summit in here and help ourselves out as two terms. Point of three is nine point away. Gives us an s two of 24. Our attention should be getting bigger. As we move up. We're holding more. So this is 24. No, it's B plus as three. Mine. We just saw for bay 15 minutes. Okay. Then we could move up a level there throwing what? We know. We just learned as to that we can do this one. Get you gonna look here? I think we're gonna strange up something a little bit here. Let's change our private boy two right here. Change these up to pivoting from C and using us to hear you. Just help us out a little bit. Is we know what has to is. We change our pivot point menus as one and it knowing it's 0.8 Ana Portnoy to its point of six. And that's equal to us, too. Times point away helps us out a little bit here because I know us to make itself for us one. So it's one we know. Six. 24. Turn away. Use me as one. Been 32 natures. Now that helps me out here. 32 as to 24. Yeah. Then is eight noons It was about everything that

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