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(2) (L5 pts) There are three children who want to halance 0n a 3.00-m long 10.0= kg board. The board is placed on & pivot point located meter from the left end ...

Question

(2) (L5 pts) There are three children who want to halance 0n a 3.00-m long 10.0= kg board. The board is placed on & pivot point located meter from the left end of the board. A 35.0 kg child sits at 0.500 m from the left end of the board. A 25.0 kg child sits at 2.50 m from the left end of the hoard . The third child sits at the left end of the board to balance the board on the pivot. (a) Make a free body diagram of the forces acting on the board. (6) Make diagrum labeling the torques acting

(2) (L5 pts) There are three children who want to halance 0n a 3.00-m long 10.0= kg board. The board is placed on & pivot point located meter from the left end of the board. A 35.0 kg child sits at 0.500 m from the left end of the board. A 25.0 kg child sits at 2.50 m from the left end of the hoard . The third child sits at the left end of the board to balance the board on the pivot. (a) Make a free body diagram of the forces acting on the board. (6) Make diagrum labeling the torques acting on the board about the pivot point. What is the Srd child '$ mass? (d) What i8 the force (magnitude and direction) applied by the pivot point to the board? Ifthe %5 kg child hopped off, which Wav do you think the board would begin to rotate (clockwise or counterclockwise)?



Answers

A uniform, horizontal, $0.20-\mathrm{kN}$ board of length $L$ has two objects hanging from it with weights of $300 \mathrm{~N}$ at exactly $L / 3$ from one end and $400 \mathrm{~N}$ at exactly $3 L / 4$ from the same end. What single additional force acting on the board will cause the board to be in equilibrium?
The situation is drawn in $\underline{\text { Fig. } 5-6, \text { where } F \text { is the force we wish to }}$ find. For equilibrium, $\Sigma F_{y}=0$ and so $$ F=400 \mathrm{~N}+200 \mathrm{~N}+300 \mathrm{~N}=900 \mathrm{~N} $$Because the board is to be in equilibrium, we are free to locate the axis of rotation anywhere. Choose it at point- $A$ at the left end of the board, since all the forces are measured (as to location) from that end in the diagram. Then $\Sigma \tau=0$, and taking counterclockwise as positive, $+(x)(F)\left(\sin 90^{\circ}\right)-(3 L / 4)(400 \mathrm{~N})\left(\sin 90^{\circ}\right)-(L / 2)(200 \mathrm{~N})\left(\sin 90^{\circ}\right)-(L / 3)$ $(300 \mathrm{~N})\left(\sin 90^{\circ}\right)=0$ Using $F=900 \mathrm{~N}$, we find that $x=0.56 L$. The required force is $0.90 \mathrm{kN}$ upward at $0.56 L$ from the left end.

Ground first. So here small M is the most off detained capital. M is the mass off the adult on envy is the miles off the world. Oh, you have small end to the 25 pages Capital and Hands 75 G's The Len Constable, the total land capital L It's given to the night he does on DDE. Mass. Of the board is 15. Gettys. My the pivot should be pleased so that the Net talk on the board is CEO or the board is invitational equilibrium. Let's say that F B is the upward force at the bigger point. Ah, but it exerts no talk about the bigger point because it passes through that point. Now we calculate the talks about this point. I'm assuming that positive, dark present clock on the clockwise direction No, in order to make this problem a little simple, as you know that for the first part, you have to ignore the marks off the boat. And in the second part of the problem, you have to consider envy. Uh, but to make this problem a little bit simply, I'm going to consider this mass and the life Oh, an equation for X, which is the distance off the pivot from the other ex. After diving X, I'll get an expression for ex basically having all these perimeters. Then for the first part where we have to ignore the moss, I can just plug in bait of the equal to zero because we're ignoring it on find the value effects. So let me just start Ah, doing it so again, that a clear idea of what I want to say wants to be a constituting. They're taking the talk about the favor point on the stark. The next talk should be CEO because, ah, the board is at equilibrium. So about this point, it's a water. The talks ah, likely have. So we have m G capital M G dimes. Thanks. That is giving. They found it, so that should be positive. Then we have a clockwise stock from months of the wood on the perpendicular. Distance from this force is hello, hello Over to mine sex. Plus it's so this is also ah, clock, one stock. So there's a bloke Let's start from the months off the child on this is MG times and so l minus X because that's the distance from the people and the start with sequel Placido. Now from here we can get an expression for X. It's indicates the position off the river. No X. It's because I do small m plus capital it and over too times Oh gosh Capital and less small lamb plus And we basically the d n a minute. That is the total mass. So now let's keep in mind this a question. So this is the question. It's going to the question one that we will be using for both first and second part. Now, for the first part, we have to ignore the minds of the board. So what they're going to do is sec nd to be equal to zero or simply ignore all the domes that have anybody in them. So finally they will get X to be Quito end overs capital m small am times with blood. The values into that, a question that we have here and we get extra the equal do to the point to five meter. Make sure you in you indicate what is X So basically this is the distance from the other. Now, for the second part, we have to consider it must be so. We will just use this a question and block all unplugged older values that we have. So in that case, let me just right that a question again that we got and this will come out. Toby, do point. I fooled me from the note. Now, why I found by I took this approach is ah, because this reduces Ah, lot of calculation that you might have done, uh, in case you took you choose toe sol for GNB separate so far, it would have right the topic question without taking the mask off the board. We have got this equation and then solve it. And then again for the second part, right? Another question for X taking the mass in the inter consideration. So the extract calculation that you might that you would have needed you know, that Raj is going now. But either way, you should give the same lesson as this

Chapter nine problem 11. 75 kilogram adult sits at one end of a nine meter long board. His 25 kilogram child sits on the other end. Aye, Where should a pivot be placed so that the Board of Balanced ignoring the board's mass and B where should replaced if it has a mess of 15 kilograms and uniforms. So here is our free body diagram. And if we look, we can see that No, we have the 75 kilogram adult sitting at the left end with called Big Temps every five K G's your little and 25 G's your arms. All right. And then we also have the mass of the board. Right? But this is all this is changing. Yes, I'm gonna put that somewhere else just for you, m b. So I could be zero or it could be 15. This is basically the difference between part and part B, right? So let's talk about the free body diagram real quick. So the full length here, this is the full length of the of the longboard, a nine meter long board. So l is going to be 9.0 meters and the distance between. Well, we don't really know where the pivot point is quite yet, right? Because that's the question. Where should they be placed? If it's balanced, that's where we have this thing here that I put the variable pivot point. So because we don't know what it is, we kind of have to guess a little bit first and then with our guessing and with our educated guess this weekend, find no where exactly this place is depending on the masses. And so this is a very important there's a very important feature. So I want you guys to note this is the variable pivot point this X. It depends on the mass. Well, it depends on everybody, all the every mass in the problem, and you'll see that coming scene. But we have a variable, Matt, we have a variable pivot point. And what that means is, we could still we can talk about torques around this pivot point and then we can use these descriptions of the distance like X and then l minus exercise will never left from ex happy Subject X. And so once we do that, since we know all the masses and every since we know the masses on one of the length we should be able to find X with all of our givens. And we can't forget the pivot point itself. So I'm just I just drew this right under where my perfect point was. He really could be anywhere. Um, but this is going to be the forced due to the pivot point. And then you also have on board itself. So M b G's the weight of the board pulling down. But two MGs are the kid in the adults and F. P s, the pivot force and the distance between the pivot and the huh And the center of gravity is going to be l half minus x right, because the center of gravity is going to be in the center of the board. Halfway through the board, it's 4.5 meters, but then it's also going to be subtracted by well, wherever excess will defined his distance. Right? So we still don't know what. Actually, that's the best we're searching for. So set up the pivot with counterclockwise is positive so that there's no rotation. In other words, this means that or I eat. It means that the sum of the talks of zero and they were allowed to do Cem. Very nice stuff. So we have we set up the pivot right there. Right? Who sent this pivot right here than F P actually goes to zero. Right? What does an f p itself doesn't go to here, but the torque caused by any the torque caused by the pivot itself is not existent. Because to have a talk, he must have a distance and a force acting at that distance. Which the only thing we have in the first case in part a is just these two, right? You attempted to include this one, but in the first in a part, a part be part pay M B zero. So we're actually going to ignore it for now. Come back to later. But we can't ignore it from We just have big mg, little mg. And that's going to give us big mg times X. So this is exactly where we had big mg and ex distance away. Right from the pivot from the pivot. Too big Angie from the period. Also too little mgs l minus hex. So just gonna be a little empty times l'm ice picks. And then when you do all the outer before this, you have rights and equal zero, of course. And then you just if we put all the exits on one side and put all the put everything else on the other side and you should get about 2.3 meters 2.25 to be exact for part B, we have to just include the mass of the part B. If you include the mass, and that just means one more factor, right? So if we go back to our be arguing here from the pivot points to the ER to the center of mass of the board, it's just gonna be l half minus x. And so when we include this one, she's gonna be NBG multiplied by el half minus six. Which is exactly what I have here. So same process is a little owns a little bit longer just because it has more terms after you distribute it. But I promise it's not that hard. Um, you just bye. Distribute all your terms and then separate X only on this, like just like 1/2 here. X x things only on the left and then everything else on the right. Then you can factor out ex divide and you should end up with the next pages result which we do here you can pull out. You can also factor g out of both the top in the bottom of the nominator denominator. And if you just plug in those masses, big M Beghin was Ah, what was that? 75? Little in was 25 M B is now 15 kilograms. Way have to include that in there. And this should basically change the position to about 2.5, so it just moves 0.2 after.

For this problem. On the topic of static equilibrium, we're told that a seesaw has a land of five m and a mass of 50 kg. A girl of mass, 45 Kg sits on the left hand of the seesaw, And a boy of mass, 60 Kg sits on the right end. We want to determine the position of the perfect point that will allow static equilibrium to occur. Now. For static equilibrium, the pivot needs to be placed at the boy girl board system, center of mass. We'll take the girls position to be the original the coordinate system and assume the board's wait accented center and over to. So the distance X. As we can see, the figure can be calculated by using the center of mass equation. Now the total mass of the system and total is able to the mass of the seesaw, that's the mess of the boy and be less the mess of the go OMG. And in the center of mass, X is equal to one over the total mess times the position of the mass, center of the seesaw, X. M. Times M plus XB times MB. For the boy less the girls position X. G. Times he goes mass E. G. And we know that X. G. Is equal to zero since we use the left hand as the origen for the coordinate system. So this becomes one over the total mess into half mm. Oh plus the mass of the boy, M. B. Times ella. So if we substitute our values into this, this becomes one over the total mess, which is 1/50 kg plus 60 kg plus 45 kg, Which is 100 and 55 kg into a half times the mass of the vessel, which is 50 kg and the length of the seesaw, which is five meters plus the boys mess of 60 kg and his distance five meters. And so we get the position of the center of mass. X. To be two 0.74 meters from the left end.

All right. We have a drawing looks like this. Okay. M. one is greater than M. two. The tables frictionless that P. Is the contact force between the two blocks. Okay. So there's the uh free body diagram, what's the net force on the system? Well, the net force on the system is just F. As you can see above see The Net Force on M. one. Well, force net on M. One. Mhm. Is gonna be F minus P. Net forcing them to. It's just pee the X. Component component of Newton's 2nd Law. So for M. One F minus P equals M. One A. Mhm. And then P equals M. Two A. F. So I need to solve for A. And P. Well Um in the 2nd 1 A equals P. Over em too. So substituting that into the first F minus P equals M. One P. Over em too. So F equals just rearranging this. I'm one over M. two plus one times P. Um M. One M. Two over M. Two plus one times P. Okay good. So I need A. And P. In terms of M. And F. So wait a minute. P. Is therefore going to be F over M. one over M. two plus one. Okay. Now, I also need A. A. Is P. Over em too. So A. Is gonna be F. Over. I am too um one over M. Two plus one. So simplifying this, Distributing the M. two mm. Yeah. All right. How would this change If the forces applied to M. 2? Well, the force so I think what they mean is applied in the opposite direction to um to so because I was gonna say only M. Two is gonna move if it's if it's applied to M. Two in the same direction. So I think you're saying this, well the acceleration is not going to change at all. However, P is just gonna be F over M two over M one plus what? Yeah. Using symmetry, we just replace two and 1 is the force larger or smaller and why? Well, we know that M one is greater than M. Two. So this is a large number, larger number. And so this going to be a smaller number. If you divide by a smaller number. If the denominator is a smaller number then the whole thing becomes larger. Soapy. In part G is greater than P. At the beginning of the problem. Yeah. So larger


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