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Three masses are attached as follows_ assuming no friction force: ml] 19.0 kg; mz 11.0 kg; m, 5.0 kgM)eWhat is the acceleration of the blocks?b) What is the tension...

Question

Three masses are attached as follows_ assuming no friction force: ml] 19.0 kg; mz 11.0 kg; m, 5.0 kgM)eWhat is the acceleration of the blocks?b) What is the tension in the string at point A? What is the tension in the string at point B?4) Look at the diagram from question 3. Ifthe FtOn mz is 35 N and the Ft On mz is 18 N, find their acceleration:

Three masses are attached as follows_ assuming no friction force: ml] 19.0 kg; mz 11.0 kg; m, 5.0 kg M)e What is the acceleration of the blocks? b) What is the tension in the string at point A? What is the tension in the string at point B? 4) Look at the diagram from question 3. Ifthe FtOn mz is 35 N and the Ft On mz is 18 N, find their acceleration:



Answers

As shown in the figure, two masses, $m_{1}=3.50 \mathrm{~kg}$ and $m_{2}=5.00 \mathrm{~kg},$ are on a frictionless tabletop and mass $m_{3}=7.60 \mathrm{~kg}$ is hanging from $m_{1} .$ The coefficients of static and kinetic friction between $m_{1}$ and $m_{2}$ are 0.60 and 0.50 , respectively. a) What are the accelerations of $m_{4}$ and $m_{2} ?$ b) What is the tension in the string between $m_{1}$ and $m_{3} ?$

In this problem, it has given that three blocks of 10 cases six KC and four Kz are connected with the help of a string and the force of 10 newton is applied on the four KZ bloc and the forces pulling uh in pulling nature. So we have to determine what will be the acceleration of each block. So we know that when we will pull the four kids blocking right hand side direction, so this entire system will move in right hand side direction, so they will all have the same acceleration. So we can assume the entire system, we can assume the entire entire block system as a single mass. So for that, if we convert this in an equivalent mass system, so an equivalent mass, or if we consider the entire 10 Casey for Casey and six KZ as a single mass system, so the total mass will be 10 plus six plus four. That is 20 K Z. So basically, this system will reduce to an equivalent system in which a 10 Newton force is applied on a block of 20 Kadima's. So what will be the acceleration? So actually, reason will be equals to force divided by mass. That is 10 divided by 20, which is 0.5 m/s squared. So this will be the acceleration for each of the block. Therefore, acceleration of each block Is equals 2.5 m/s squared. So, this is the answer for part of the problem. Now, in part B, it is saying that what are the are tencent forces that are present in in this case or what are the trends and forces that is present in between 10-K C and six KZ and six KZ. And for Casey. So for that, let us assume that that tencent force between 10-K C and six KZ is steven. So for the same string there will be even with the same for steven will be applied on six KZ block also and far between the sixties and four KZ. The tencent forces let us 82. Now, we will drop free body diagram for each of the blocks. So, what will be the free body diagram for 10-K C block? So, there will be a tencent forcing for in horizontal direction. There will be a normal reaction, let us say anyone is acting in vertically a poor direction and there will be weight of the block. That is 10 in two G in downward direction. All right. So, and what will be the free body diagram for second six KZ block? So, there will be tencent force even in left side direction, and he too will be in right hand side direction, normal force, and two will be in vertically a poor direction. And weight of the block will be downward direction masses six. Therefore, MG will be six and two G. What will be the primary diagram for the four K. C. Block? So there will be forced F is equal to 10. Newton is applied in forward direction. There will be a backwards force of T to newton And there will be a normal rates, an entry that will act in vertically upward direction, and M3G, that is four in two G, will be the weight of the blog that will act in downward direction. Now, in previous part, we have calculated that acceleration of each blockage .2m per second squared, and this acceleration in the right direction. So this is the acceleration of each block. E as we can see that the acceleration is completely in the horizontal direction. Therefore, we have to apply Newton's second law of motion in horizontal direction. So if we apply Newton's second law of motion for the first case, so we will write the equation as even will be equals to mass of the first block. That was 10-K Z Into Axle reason a. and acceleration was 0.5. So 10 into 0.5 will give five Newton. Therefore, what was the 71 was the tencent force between block one between 66 Kz block and 10-K C block. So this will be the force that will be acting between six K C block and 10-K C block. So this represents tencent port between 10-K Z. And six Gezi blocks. Now, if we go to write equation of motion for the third keys. So what we can right f minus T. Two must be equals two M two into a. What is the M two? M two is the weight of the 3rd and 3rd mass. So better to write em tree since there is no notice. And is given that which one is blocked too and which one is blocked? Three. So let me write it as three. So basically here M three's four KZ Okay. The forces given as 10 Newton And we have to find the value of T two. M 3 is four and acceleration is 0.5. Therefore, T two will be equals to 10 minus four in 20.54 into 0.5 is two there for 10 minutes to will be eat So 10cent between the uh, four KZ block and six KZ block will be eat newton. So T two is eat newton. So what is the key to? It is the tens and fours between six KZ and four KC Block.

Okay, let's take a look at the $35 for each block. So they freed by government for the hanging block is something like this. Which means that the hang of hope or experience a graph to you from the Earth and the ground, these evil MSG images a massive hanging Bob, and you also experience attention. It was going upward in the direction of an air force on the hanging block. Should be going downward is because the hanging block is trying to move down. Okay. And the everybody that was full of block on inclined plane should be something that is, and the angle between the inclined plane and the grounds. 45 degree. Any block yourself experience of gravity from the Earth, which is, um a g m I am eyes the mass of a block from inclined plane, and it experienced attention that was, uh, moving upward. But along the direction of the inclined plane and only, uh, political direction, you each friends a normal force there was moving upward and which is perpendicular to the inclined plane. And lastly, you also experienced the friction. There was allow the inclined plane, but in opposite direction when compared with the direction of attention and in the air Force off the block on an incline plan should be moving upward, but a long incline play. Okay, so you've reconsidered. Okay, Upward direction on an incline plan is the deposit X axis and the direction that is perpendicular to the incline planes in upward direction is the Y axis. So therefore can decompose the gravity which you will have the s component of the gravity and the y component with gravity. OK, As you can tell, no more force is equal to the gravity times cause I'm 45 degree which is equal to oh, I m i g because I 45 degree. And we also know the friction in his case sixties kinetic friction. So it Ziegel Toby coefficient of kinetic friction passed a normal force, which is you go to meal times M I. G cause I 45 degree and an air force only vertical direction that was perpendicular to the incline plants is equal. Zero is because it blocked in the moving up or down. He only moved along the inclined plane. Okay, so therefore we can say that the net force equation for the block on in climb were just net. I is equal to the tension minus the friction. An M minus name. Ask component over the gravity. OK, which is going which is also going to the opposite direction when you compare with that direction of attention, which is, um, Jane sign for inviting me, which is you go to is the question we put the t minus mu n minus McGee. Sign for my degree. If we expend it will have t minus mule m i g because I 45 degree minus MRG sigh or if I'd agree. And we also know that the net force off the block on Inclined Plan is equal to the mass off the block on an inclined plane has the acceleration A and is or you go to t minus. Um, I gee, because I 45 degree, the I minus MRG sign 45 degrees. So it seems that we don't know how much tension is. Remember, the block on incline plan was connected. Ah, with uh, hanging block to real strength. Okay. Which means that if we take a look at everybody that were funny hanging block, we can have the therefore see questionable hanging ball. Is there definite H music? Green pencil? I have no h. You see? You go to T minus t. OK, because it's going Donald were okay, This is equal to Mohd finest he and then we'll have attention. T is equal to MSG minus f in their age. Which is he going to MSG minus, um age eight. Okay, because then their force on hanging glass is equal to the mass of the hanging above has the acceleration and attention finally hanging Block is you go to the tension on a block off inclined play. Okay, is because they're connected with each other and there in the sand. Sister. Therefore Okay. Along these inclusion here, back into here and we'll have and my a equal to m h g minus. Um, my j Yeah. Minus mule. Am I G cause I 45 degree and then minus am I signed 45 degrees. Okay, I am convinced of the m e J two left side. No have. And my a plus I m a j is able to msg minus you. Am I Jane? Course I 45 degree minus. Am I sigh 45 degree. Okay, So now I can determine acceleration off this system, which is you can take a out. We'll have eight times in my plus image is equal to MSG minus, um, I g cause I 45 degree minus. Am I signed for by degree, you know, have a which is the acceleration. So you go to MSG minus, Um, I oh, j cause I 45 degree minus. And why so everybody me divided by and my plus a mate And from the question we know the Mbai, which is, uh, mass of the block on an inclined plane is 2.0, kilograms. And the mess overhanging block image is able to three point. I'll kill where a coefficient of a kinetic friction is given a 0.1 I and we know the acceleration of gravity, you see, go to 9.8 meter a second square. We know, cause I'm 45 degree and signed 45 degrees right now. So cause I or if I agree, is you go to So for if I'd agree, which is square to over to. Okay, So if you're plugging these values back into the equation here we'll have. Acceleration is equal to, um three point. Oh, you were kinds nine point a meter per second square minus Joe 0.1 I times 2.0 killer worms times Now I'm going eight meter per second square and end Times Square to over to minus to a bono kilograms times 9.8 meters per second square 10 times square to over to and it's holding here, divided by 2.1 kilograms plus 3.1 kilograms and this will give us acceleration is equal to 2.6 meter But seconds were okay Simple question Be things when we know the acceleration of on a question A Now we can determine attention on us on a string OK, was his tea is equal to remember Fun question a worry Uh, this the equation pretension which is right here. Attention is evil toe msg minus Emma Jay Gray. That's right down. So its able to msg minus amazing A Which can we go toe image G minus a. We know the mass overhanging block image is 3.1 kilograms 10 times G minus eight, which is the XRs on ground D minus the acceleration off the block, which is not going a university square minus 2.6 University Square. And this will give us attention is about 21. Only six new times. If we run into the tens digit, we'll have 22 newshounds. I'm sorry. No intense digits. Integer here. So is about 22 new times. Okay, so this is the answer for this question. Thank you.

So the figure that we can we have with us is this. This is just for reference so that you can have a much better understanding. Just 35 degrees. This is 35 degrees. There is a fully here. There are two Mosses, one off 3.50 Katie on the other off a ticket. Now what happens here is since this'll one This block has a larger mosque. So the mosque one is eight k G and master is 3.5 kids. No sense m one is greater than m two. We expect this block to move down the plane. So we expect this to move down the plane now The acceleration for both the block should have the same magnitude. So if the acceleration for this is a two on for this, it is a one. We consider that a one equal to eight to why? Because they are joined together. There is joining you. No What? We can write this. You can write that Sigma F one equals to m one a one which is equal. Shoot M one g sign 35 degrees lusty which is equals to tee. It is equal to M 18 because he is same football and stigma effort equals shoot em to eight to equal. Shoot M two g sign 35 degrees minus t which is equals to em to a So this is the falls that we have no suppressing the units. We can get minus 3.50 Find us off 3.50 into 9.80 Sign 35 degree. Yeah, lusty. You're putting the substituting the values dusty equals two equals weekend. Put it us am to value. We can put it that is 3.508 Similarly, you can put it us eight in 29.80 into sign 35 degree minus T equals soup 88 no vote. If you consider this to be equation number one and this trip equation number two, what we can put us and we're adding one plus two. We get 45 plus 45 Newton minus 19.7. Newton equals to 11.5 kg into eight. Therefore, in the a part and we substitute for a or acceleration, we get it us 45.0 Newton minus and just have a clear idea? Yes, minus 19.7 Newton upon 11 point five kitchen. So in this case, we get the value. You have to see the calculations of it. So here we get the value off A s 2.20 m per second squared. So this is our absolute wish. Yeah. By substituting the value off a in the question off t we get it us when we see the value for the tension EPA at this minus 19.7 Newton lusty equals to 3.50 k g into 2.20 m per second squared equals two 7.7 neutral. When we consider this so the tension would be 27.4 new. So this is the required attention.


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