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Point) block Dl mass Mt 225.0 Kg sits on an Incrinud plang and connuctod buckot with PUley Tha coctlicient of static masa 0s$ string Oves Iniction butwreen the bloc...

Question

Point) block Dl mass Mt 225.0 Kg sits on an Incrinud plang and connuctod buckot with PUley Tha coctlicient of static masa 0s$ string Oves Iniction butwreen the block and the plare macelos} und Inctionloss 60". Mass Mz Can bo 0.74, and Iha angle botwcon the pland und the horzontal changed by adding Inttm taking awa} tand Irom tho bucketM1M2What Mumtium Uaua Mz lor whioh the "yatem wIll remain rent? (Hint; The body Mi tunds Movu Up G0 static Iricion down |What Ine minutlum valuo of M; I

point) block Dl mass Mt 225.0 Kg sits on an Incrinud plang and connuctod buckot with PUley Tha coctlicient of static masa 0s$ string Oves Iniction butwreen the block and the plare macelos} und Inctionloss 60". Mass Mz Can bo 0.74, and Iha angle botwcon the pland und the horzontal changed by adding Inttm taking awa} tand Irom tho bucket M1 M2 What Mumtium Uaua Mz lor whioh the "yatem wIll remain rent? (Hint; The body Mi tunds Movu Up G0 static Iricion down | What Ine minutlum valuo of M; Ior which Iho syotor will rumnoid F09l? (Hm: Ihe body MI londe Movu own alalic Iriction Up ) Supposu Ino cooll ciunt kinotic Iriction botwoon Iho block and Ino Incured plano MUa 0,57 im 7301 , Ko; *t Iha mnon tuda ucculufallon ol M ? (int: Tnu ditucion 0i molion nuads bu ducldod Ivet Inon Lnom the d roction = wnotic Irction )



Answers

In the system shown in Fig. $5.54,$ block $A$ has mass $m_{A}$ , block $B$ has mass $m_{B},$ and the rope connecting them has a nonzero mass $m_{\text { rope }}$ The rope has a total length $L,$ and the pulley has a very small radius. You can ignore any sag in the horizontal part of the rope.
(a) If there is no friction between block $A$ and tabletop, find the acceleration of the blocks at an instant when a length $d$ of rope hangs vertically between the pulley and block $B .$ As block $B$ falls, will the magnitude of the acceleration of the system increase, decrease, or remain constant? Explain. (b) Let $m_{A}=2.00 \mathrm{kg}$ $m_{B}=0.400 \mathrm{kg}, m_{\mathrm{rope}}=0.160 \mathrm{kg},$ and $L=1.00 \mathrm{m} .$ If there is frition between block $A$ and the tabletop, with $\mu_{k}=0.200$ and $\mu_{s}=0.250$ , find the minimum value of the distance $d$ such that
the blocks will start to move if they are initially at rest. (c) Repeat part $(b)$ for the case $m_{\text { rope }}=0.040 \mathrm{kg}$ . Will the blocks move in this case?

Problem 11.87. So there's a bar that we're trying to measure the mass of. So capital M is the mass. This bar is four meters long and we're told that it has holes and stuff in it. So it's ah, its center of mass is not in the middle. In fact, the center of mass of the bar, we're told, is 1.88 beers from its left edge. So we then are told that we place the bar when we determine this by by placing it on a knife edge. And then we're told that we placed the bar, uh, slightly further to the right and then suspend a mask from one end and then another mass that we move along and measure where the center of mass of the whole system moves, too. And we have this table here, and there are a few things we need to do. So the first thing is to draw free body diagram, and this will help illustrate what it is exactly that we're talking about. So we have our bar. Mass one is suspended 0.2 meters from the left hand side. The knife of it is here one and 1/2 meters from, uh, from the left hand side. Recall the normal force it exerts F p. The center of mass of the bar is over here at M G. On this is, um, 1.5 meters from the air. It is It's 1.5 meters from left hand side. This is 1.88 meters and then, uh, this is the second mass and the distance from the left hand side to the second mass is X and removed. We pick different masses for this and find out where we have to position it to balance the bar on the So here's a free body diagram. Next we want the plot or not. We want to use the equilibrium condition to find the our X. Here is a function of the second mass. So this way over here has a mess of two kilograms and we're using the access for the access were using the pivot. This acts 1.3 meters from the pivot. We have the mass of the bar and its weight on the acceleration of gravity to make this its weight. This is one point. It's eight meters right now 1.88 meters from the left hand side. Uh, so it's 0.38 meters from the pivot. This will have second masters attracted to from it and then X minus 1.5 meters because exes from the left hand side. But we're measuring cross. If it this all needs to equal zero. And so we just solve this for X as a function of em too. But it will be a function of all these things, but we the rest of the constant, um and we get one over M too. I actually this is, uh, make a little more room for this. X is going to be sorry. One over M too. Times two kilograms. Times 1.3 meters. Um, hunts 0.38 meters plus 1.5 meters. So now I want to plot. Uh, X is a function of one over em too. Um, this is an inverse kilograms. This is the meters, so point to interest kilograms. It's two meters and then it one inverse kilogram, 3.5 meters here would be our best fit line. Obviously, you should plot and your plot all of the values that were given. But this is just a sketch of what it and looking like on then. However you like to do these things, you, uh, make a best fit lion. And from there, uh, you will find that m the mass of the bar is 1.59 I feel like art scene. So now, um, from doing the best fit line, we also find that the Y intercepts. So this right here is at 1.5 meters. And so this makes sense because as one over m goes to zero or one over him, to be specific goes to zero. This is the same is saying that M two goes to infinity. And so, as as this becomes larger than all of the other masses involved, you're going to have to recall this is the location of the night EJ. And so you're going to have to, you know, basically put this giant mass as close as possible to the pivot as it gets larger and larger so that the, uh, the center of mass of the system will be over the pivot. So this makes sense because it just gets bigger. It has to get closer to the pivot so that it's the moment arm of the torque that it produces a smaller because the force is increasing.

From what we have given. Mhm the boss to the Y axis and exit system, Center of Mars of the torso, neck and head. You configured on the Y axis at a 0.0 point 89 Some of these Well, this is Yeah, there's a distance is given zero point Teammate Peter at the center of mass upper leg, located on the X axis at a 0.0 point one centimeter. 0.17 m to the right of the origin. Yeah, this distance is zero. And said if this is supposed treatment, this is an and the lower leg and see total mas 9.90 at the center of the 0.43 m. Who that off? And the 0.26 m below the origin. So here this is this distance 0.1 women. And this distances giving you like 26 This is the stances. 0.26 Yeah, this was empty. So and money is given for the kitty for the kitty and bamboos. So pjsc connects for Yeah. The mask of the cabinet is 17. Katie, this entity is a horrible I am 0.90. So first we will write the coordinates of and the coordinators and when will be excellent, Raymond, this is equal to zero on 0.39 Mhm. This is for you. For them to court here this extra common way. This 0.1 step. Oh, my computer coordinate for entry. This is next three, Whitey. It's just 051705 to 6 x coordinate. So this will be for 0.16 Yeah. 26 This is equipment you look like four weeks and like what it is like this zero point 26 Yeah. And when it is for people, influence 70 These things. So we will be here to fight X and Y coordinates of the center of Mass. So for the X coordinate well, right and one. Excellent. Let's put it ST listen to the next city invariably anywhere. Let's end recently. So and when is for demon extraordinary? 0 to 17. 80? It's for nearly 0.17 m. That's empty is the name or getting your next for your head to the point moving and total masses 14. 79.9. So the exported it when we called. Yeah, 17 0417 This is 9.90 point 43 They ran away for differentness 17. This with like this is important. 0.10 5 m. This is the X coordinate of central Hamas. Broth will fight for you. England like this in the wake of Let's Empty might be ready to write and one less and less empty. And one is 45 minutes. 0.39 right? And who is 17 years by 20? Yes, and he is 9.90 here. White, please, Brian A zero for impulses. The weight of 70 Let's provide for So this is important. It's not even multiplied. 49. This 70 to zero. Let's when we play with 0.26 Seriously apartment this 17 plus I. So this is a quarter zero come into one night 76 suite at the System y coordinate of Central

First calculating the center of mass of X coordinate. So here I can use the formula M one X one plus M two, X two plus M three, X three by M one plus M two plus M three. So just putting the value in this expression, I can write the expression is 41 multiplication zero plus 17, multiplication 170.17 plus 9.9, multiplication 0.43 Hold by 41 kg plus 17 kg plus 9.9 kg. Which is equal to 7.147 x 67.9, which is equal to 0.11. Meted now calculating the center of mass for Y coordinate. So I can use the formula M one Y one plus M two, Y two plus emptory Y three by M one plus M two plus M three. So just putting the value in this exhibition, I can write 41 multiplication 410.39 Plus 17, multiplication, Jiro plus 9.9. Multiplication minus 0.26 Bye 41 plus 17 plus 9.9. Which is equal to 0.20 m. So this is our final answer.

My discussion. We're looking at a similar example to example seven. I told 5.7, uh, in which we have a We have two blocks off mess resting on the table and you have in external force acting from the left and pushing the blocks together. Now hear this. No idea. The blocks in contact with each other's We have a contact Force P Now the only difference between this and 5.7, it's that there is friction that we have to account for and not it. There's two different kind off, two different coefficients or friction depending on the block. So M one has coefficient off kinetic friction off you. One rest for rental is mutual. I'm gonna draw first of all the individual diagrams for each other blocks. So they started. Look, one m one. So what experiences is the external force? Of course, coming from the left, you know, so extra inserts the fictional force because of you Assume that it's gonna move towards the right. Right? So it s fictional force. Opposing is motion. If and you have gravity every every normal force. Finally, we have tea Contact force, actually contact force from M two will be sighing for someone from the right hand side toasty left. So a more accurate representation like this Course it dissecting from the itch nights from the right, It on the other end for him too. The smaller block again experiences fiction this well And from the left inside you experienced e pushing force from RM one. Basically it's the contact for speak and also the mess. I mean the weight and the normal force the usual forces. So this for him too. Not a funny net force on the system off two blocks, right? So you're considering the entire block. Here s ah, a single system. So what are the Net forces acting on? The system will be for sex on a force f as well Esti fictional forces you want from me when I'm you. Now we know that the fictional forces sexually related to the coefficient like by the normal force. So the total fiction we have a new one and the normal force from the M one. It's just equal study. He wait right to bed. And so all the forces in the Y direction this m one time street with us The fictional force for the second block and to times two. Therefore the net force considering the right side STD positive X direction then the net force would just be f minus away. The daughter frictional force. Next, you want to find what is the net force for m one and forceful and to respectively So, using the free body diagrams we can very easily find mighty Why direction their balance don't 19 or more and the rich So the only thing s t ah horizontal forces along the X direction. So again, positive direction is twisty rights. So the Net Force 40 and one b f minus p my nasty fictional false you one m one g for the m two the case off into the net Force B P minus a fictional force which is mewtwo m two g. So what we do, right? All the u turns second law for these equations for M one and m two. And it was a Loy's. Basically, based on the acceleration and the acceleration to get from the Net force equation, we just divide by the mess. So, for your 1st 1 first equation for m one e nets acceleration. Just if 0.1. It's Peter of I m one minus the one. You you are heading for the 2nd 1 creation. The net acceleration for him too. We just invite this by and to No, because the both off the blocks are moving as a single object by them. We gotta they must share the same acceleration. So actually, both off this have the same acceleration, which you can't just called it a Now we got to find what it's a You think these two equations So, uh, it's quite simple. First, you on a substitute for a GOP and the younger one p using the second equation a wish your we're gonna do is gonna find in terms off p. He scores to eight times two plus you two and two. So substituting it to this question here you have If the m one minus p that p iss, it's toe a to plus you two and two g voted by and one when I see one g close to a. Not to simplify this expression would be again modifiable sites by m one. And I'm gonna bring the term with a two d my hands site so you can get a It's just ef minus you one time you to m two G minus one and one g Put it by in one plus and no, we can find Ah, what iss de apply force. So you want to find the contact force, Pete, And we have this p or you. So what we're gonna do is just suffered you in our A do this equation nights or desist e expression for people.


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