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9 = 9.8 what is the normal acting on ml? whar is the friction experienced by m2? the acceleration of system is ? the tention in the cord is ?...

Question

9 = 9.8 what is the normal acting on ml? whar is the friction experienced by m2? the acceleration of system is ? the tention in the cord is ?

9 = 9.8 what is the normal acting on ml? whar is the friction experienced by m2? the acceleration of system is ? the tention in the cord is ?



Answers

Two blocks connected by a cord passing over a small, friction less pulley rest on friction less planes (Fig. P5.90). (a) Which way will the system move when the blocks are released from rest? (b) What is the acceleration of the blocks? (c) What is the tension in the cord?

In this problem. We have the following diagram and in part a we want to find or we want to be able to predict the side to which the blogs will site to do that first. Let's think about the forces acting on each of these blocks. First we have mass one over here and we'll draw its uh free body diagram. Of course we have gravity pulling it downwards. But as we know and inclines, we can decompose this force into two components, one that goes along the plane and the other one that goes perpendicular to the plane. This one We can call force fall. There's one we can call simply force of gravity with a capital jean just differentiated from this one. And here we have normal force from the plane on the block and these to cancel out because they have the same magnitude. Okay, And we have a 3rd final force force of tension from the string on the block. That's for blog one. Now, let's think about blogs number two, we have the exact same situation with gravity. We have the force of gravity. We have the component home gravity along the plane which we call force paul and the perpendicular component. And finally, well, almost finally we have normal force which councils out this component. And now finally we have the tension force. But notice that these forces are going to be specific to this mass. This one is going to be specific to this mass. It's all over here. We have one. I don't want Now if we want to be able to predict which side this will move, then we have to ignore the tension force for a second. And let's just think about these two forces. Force of fall one and fourth of all too, because those are the forces pulling, putting this block this side and the other one the other side. So they are competing To make the whole system slide to one side. So let's find them and compare their magnitudes and see which one is bigger. So first we have force of ball for blog number one. That is going to be the force of gravity which is must one times gravity. And since this is a component, we have to multiply it by the sign of that angle. Then we have four falls for First of All. Number two, Which is what must two times gravity times sine Number two. All right. And we have all we need to substitute these variables and find some values. This first one. It's going to give us 490 Newton's The 2nd 1 is going to give us roughly 300 92. Newtons. Clearly force of fall. Number one is greater than force of fall, number two. And this one points to the left. Therefore, we can predict that the system well slide to the left. Now we want to find the acceleration to elect in part B to do that. We can keep thinking just about these two forces and not think about the tension force in between for now. So we have Newton's second law which states that the total mass of the system times its acceleration is going to be equal to first force of fall. We're going to say the left is positive because we know that we're sliding to the left, so it's easier to make that side be positive minus first of all, Number two, which points in the opposite direction. Right? So if we want to find the acceleration, we solve for it by dividing everything by the total mass and the total mass of course is going to be must one class mass to And we found these forces up here, so we can just substitute them 490 392 100 kg 1st 50 programs And this is going to give us an acceleration of almost 0.65 m/s squared. That is her answering Now, in part C we have to find attention. So now let's stop ignoring that middle component, but we have been ignoring so far we can choose any of these two blocks because we know they're acceleration. So we can choose any of them to find attention for us. So let's choose block number one, let's use mutants Second law to find the tension force. So mass of block one times the acceleration is going to be equal to the force of fall of that block. Mine is the force of tension because that force of tension is pointing, of course, in the opposite direction. So we need to have that negative sign. Now, when we solve for the force of tension, we obtained that it is equal to the force of fall minus the net force. So now let's substitute the numbers. And this gives us Roughly 425 newton's. And if you apply the same logic to the second block, you should obtain the exact same force. So those are our answers for this problem.

Free by diagrams party in the first block over here on the surface. So you have normal force going upwards. You have mass times grab me the weight going down and then you have the force attention being pulled on. Then for second block, you have force of tension being pulled upwards and then you have mass two times gravity weight being pulled down. As for the free body doctors since two blocks are connected, the maxims of their accelerations will be the same. So what a one. The extraction equal a to in the y direction the T equals the overall acceleration. So if we combine forces equations in second law, we get this is on for part de now. So force attention equals and one times a that's two times gravity. My eyes force attention equals mass to times acceleration. So rearrange subs to you get massive too Times gravity, my eyes, massive block One times acceleration equals mass to times acceleration, the neck business and to times a being and one times a plus and two times a equals and to times she so you re range and solve for a celebration equals gravity times. That's what to over and one plus into forced tension came equals mass and one times a equals g times m one and to over and one plus. And to this is for your part a free, wide diagrams over here. And then we saw four acceleration terms. Love Newton's second law, all right?

In this exercise, we have these two blocks A and B that are arranged as in the picture. Both blocks have mass. I am. And the coefficient of kinetic friction between Block B and the surface is Omega K A mu K. And our goal is to find based on on this information, what is the acceleration of both blocks? So let's draw the force diagram for each block individually for first for block A. So this is blood A. We have the tension acting upwards, and we have the the force of gravity, the weight force backing downwards. And the fourth equation for this block will be the weight force minors. The tension is equal to the mass of the block times the acceleration. Now the weight force in this case is M G. The tension is Oh, no. So this is the equation which I'm gonna call equation one four block a. Then for block B right here we have the tension that's acting to the left and we have the friction force F f attacking to the right and the sum of forces t minus. The friction force is equal to, According to Newton's law, the mass times acceleration. Now the friction force is equal to mu K, times MGI, and this is equal to m A. This is equation too. Now I'm gonna some equation one with equation two. Okay, so let me just write them again. Here we have in equation one m g minus T is equal to m A in equation to have t minus u k m g is, um a how if I some these two equations we obtain m g one minus mu K is equal to two m a. So the acceleration A is equal to G divided by two times one minus mu k, which is the exploration we're looking for.

We know that the applied force causes a torque which gives the pulley and angular toleration. We know that the applied force Berries with time. And if this is a case, so will the angular acceleration. We can say that, Ah, in order to find the angular velocity, we can integrate the variable acceleration to stall for this. And then we can then say the speed of a point on the rim is the tangential velocity of the rim of the wheel. So we can set this up and say that Ah, Sigma Tau equaling. Ah, the radius times the force tension. And this would be equaling the moment of inertia multiplied by the angular acceleration. So we find that the angular acceleration would simply be equaling the radius times of force. Tension, which is again burying with time. And this would be divided by the moment of inertia from the definition of the angular acceleration in your acceleration is gonna be equaling the derivative of the angular velocity with respect to Time T. So we can then say that we can integrate from Brother Omega initial to Omega final times d omega, and then this would equal the integral from zero to t of the radius times the force tension divided by the moment of inertia d t. And so we know that the moment of the angular velocity rather is equaling the linear velocity divided by our by the Radius. And this would be equaling the moment of the initial angular velocity plus r divided by I multiplied by the integral from zero to t of the force tension DT. Given that the force tension is the only variable that actually depends on time. So we can then say that the initial angular velocity zero so we can eliminate that. And we can then say that the angular velocity would be equaling the radius, divided by the moment of inertia multiplied by the integral from zero to t of the force tension d t. And we're simply going to then, um, we can simply say that the tangential velocity would be equaling. So the final velocity times the radius and so this would be equaling the radius squared, divided by the moment of inertia, Time of the integral from zero to t of f sub t. Now this would be equaling 3.0 T minus 0.20 t squared times D t. And so we can then solve and say that this is gonna be the 10 gentle Ah, velocity would be equaling r squared, divided by the moment of inertia multiplied by three over to T square minus 0.20 divided by three t cubed and the units Newton seconds. And we can then say that the velocity, the tangential velocity AT T equaling 8.0 seconds. This would be equaling the radius of 0.330 meters, divided by the moment of inertia to be 0.3 85 kilogram meter squared. And then this would, uh, the radius would be squared on, then multiplied by three over two times 8.0 seconds. Quantity squared minus 0.20 divided by three times 8.0 seconds. Quantity cubed. And then again here the the units here would be new 10 seconds and we find that the tangential velocity AT T equaling 8.0 seconds would be approximately equal to 17.5 meters per second. This would be our final answer. That is the end of the solution. Thank you for one


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