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(20%) Problem 4 block with mass of m = 39kg rests on frictionless surface and subject to two forces acting on it; The first force directed in the negative X-directi...

Question

(20%) Problem 4 block with mass of m = 39kg rests on frictionless surface and subject to two forces acting on it; The first force directed in the negative X-direction with magnitude of F 9.5 N. The second has magnitude of 22.75 N and acts on the body at an agle 0 = 17 measured from horizontal, as shown: Jeme Rodney = [email protected]

(20%) Problem 4 block with mass of m = 39kg rests on frictionless surface and subject to two forces acting on it; The first force directed in the negative X-direction with magnitude of F 9.5 N. The second has magnitude of 22.75 N and acts on the body at an agle 0 = 17 measured from horizontal, as shown: Jeme Rodney = [email protected] F m



Answers

A 2.5 $\mathrm{kg}$ block is initially at rest on a horizontal surface. A
horizontal force $\vec{F}$ of magnitude 6.0 $\mathrm{N}$ and a vertical force $\vec{P}$ are
then applied to the block (Fig. $6-17 ) .$ The coefficients of friction for
the block and surface are $\mu_{x}=0.40$ and $\mu_{k}=0.25 .$ Determine the
magnitude of the frictional force acting on the block if the magnitude of $\vec{P}$ is (a) $8.0 \mathrm{N},(\mathrm{b}) 10 \mathrm{N},$ and $(\mathrm{c}) 12 \mathrm{N} .$

For this problem. On the topic of force and motion, we are told that a 2.5 kg block is addressed on a horizontal surface. A force F of magnitude six Newtons acts on the block horizontally. And a vertical force pr then applied to the block has shown in the figure. The coefficient of friction for the block and surface are U. S. 0.4 and U k 0.25 We want to find the magnitude of the frictional force acting on the block. If he has a magnitude of eight newtons, B, 10 newtons and c 12 newtons. Now we can choose X to be positive right words, and why it be positive upwards. And we can apply Newton's second law in the horizontal direction. We have the applied force F minus the frictional force. Little F. Is equal to M. A. In the vertical direction. We have P plus the normal force that the flow exerts on the block, minus the weight of the block. Mg was equal to zero. Since there's no net motion in the vertical election, we have F. Being six newtons and M 2.5 kg. And so for part A. If P is equal 28 newton's, then we can see that the noble force F. N. Is equal to 2.5 kg times 9.8 m per square second minus P, which is eight newton's. And so the normal force has magnitude 16.5 newtons. And so the static frictional force Fs max is equal to us times F. N, which is 6.6 newtons. And so this is larger than the six newton right would force so the block does not move. Which so let the acceleration be zero in the and into the first. For equations above, we get the frictional force F to equal to the blood force P, which is six newton's in part B. We have the applied force P. To be 10 newtons, which gives us the normal force F. N. To be 2.5 kg times 9.8 m per square second. And so the normal force and this is minus P, which is minus 10 m. So the noble force is 14.5 newtons. And the maximum static frictional force F. S. Max in this case, new S. Times F. N. Is 5.8 newtons. This is less than the six nutrient right would force. So the block block does move. So we're not dealing with static but we're dealing with kinetic friction. And so the kinetic frictional force F. K. Is equal to the coefficient of clinic fiction mu K times the normal force FM. Which gives us a frictional force of 3.6 newtons. For part C. We have the applied force p equal to 12 newtons. This gives us the normal force F. N. To be 12 0.5 newtons. And so the maximum static frictional force is equal to five newtons, which is less than the six newton Reitman force. And so again the block moves. And we have kinetic friction, and so the kinetic frictional force f. K, is equal to the coefficient of kinetic friction times This normal force FN, which gives us the frictional force, the 0.1 newtons.

In this question, we're trying to determine the force of friction acting on a block. So first let's draw our free body diagram for our situation. So we gotta force after the right the force of gravity down force of friction opposite that force. And then up we have some force P. And our normal force. Yeah. Okay. So what we can say in this situation up and down has to be balanced to the block is and flying up into the air, breaking down to the table. So please p. Plus the normal force is equal to the force of gravity and left and right. We don't know if they're balanced or unbalanced yet but we really don't care. All we know is that this force is six mm. So we want to figure out the force of friction. So a couple other pieces of information that we need. So the mass of the block Is 2.5 kg. The coefficient of static friction is .4. The coalition of kinetic friction This .25. Mhm. Okay. Yeah. So now we are using three different values for P. To determine our normal force. Let's move this equation around here, I'm gonna subtract peter both sides. So the normal force is equal to F. G minus P. Yeah. Yeah. Mhm. Okay, so let's calculate the force of gravity first. So force of gravity equals 2.5 kg. The force of gravity is mass times gravity Times 9.8 meters per second squared. So our force of gravity acting on the block is always going to be 24.5 mittens. Okay, so now let's calculate our three normal forces for the three situations. So in part A P equals eight newtons. So when we plug into our equation here, F.G. This is going to give me a normal force. Uh 16.5 Newtons in B. He is tendons. This gives me a normal force Of 14 5 Newtons. And see here's 12 movements, which gives me a normal force of 12.5 mittens. Okay, so now how do we figure out the forces of friction? So in each part we'll start with part A. We could have the force of friction static which is less than or equal to the coefficient of static friction comes normal force or the force of friction, kinetic is equal to the coalition of kinetic friction times normal force. So the way that we're going to know which one to use is we're going to start with static friction. Okay. And if static friction, when we calculate here, if the force of friction is less than or equal to a number that's bigger than six because that's our force here, We know it's not going to move. Static friction has to be less than six in order for the back to block to be moving. So let's first calculate for part A. So we plug in we get four sub static friction is less than equal to 0.4 and our normal force which is 16.5 Millions. So this comes out to force of friction is equal to Yeah. All right. Not equal to his less than equal to 6.6. Newton's Okay. So since static friction can be bigger then are pulling force of F. That means that the force of static friction is only going to balance it out. So our static friction in part A. Is six newtons. We don't have to worry about carrie correction. So part of our friction is going to be six. Newtons just exactly balancing out are pulling force of F. So part B. Same thing. We're gonna start with static friction less than equal to 0.4 times 14.5 Newtons. And this comes out to remember Yeah, Which only comes out to 5.8. So, since our number is less than six, we have to switch over to kinetic friction. So when we plug in for kinetic friction, That's .25 times are 145. Newton number force. So our force of friction kinetic In this scenario is going to be 3625, which will earn a 36 newtons. Mhm. And part C. Mm. Well, since our force of friction was less than six in part B, it's going to be even less than that. So, we can skip the force of static friction in this part and go right to kinetic. So .25, 10- 12.5 Newtons. And this comes out to are force of friction. It is 3.1 newtons. So in part A it's static friction exactly equal to six. The block is not moving at all, and then part B and C. It's kinetic friction, and we have our two values here.

Three blocks apart. A is simply asking us to draw the free body diagram, so this it would be in with this would be the answer for per day. We just have to late rather draw the mass and label of the forces that are acting on that mass. Now. For Part B, however, we need to find the acceleration of the total system. So we can say that the sum of forces would be equal to the force minus the force of Block AM Block B plus the force on block a block B and block A minus the force of Block C on Block B plus the force of Block B on Block C. And then this would be equal to F because all of these would cancel out due to Newton's third Law. And so we can say that the sum of forces would then be equal to the sum of the masses. Time's A which is giving us that the acceleration would simply be the force divided by the sum of the masses. So this would be your answer for a part B now, for a part. See, they want us to find the net force on Block C Whether the net force on each block So we can say that the force the net force on block A would be equal to the force times massive a divided by the total mass um massive A plus Master B plus mass subsea on B It would follow the same pattern So this would be the force times amsa be This would be divided by M of a plus B plus m sub sea and then for C, This would, of course, be forced times the mass of C divided by mass again. So this would be those three answers for apart. See, Now Part D is asking us to find, um ah, essentially the expression Oh, are the magnitude for all of the ah intermediate forces. So essentially efs of a B s, A b a B c. So we can say that for d, we can say that force of block B on sea would be equal to force C net and this would be equal to again the forced times m subsea divided by that total mass. And at this point, we can say that this is going to be equal to force the BC due to Newton's third Law of Motion. Now we have to apply it to block A. So we can say that force minus force of a B ah would be equal to, of course, force a net. We've we know that this is going to be equal. Uh, rather weekend actually say that. Force A B equally, of course, foursome be a with equal force minus. And then that force a net so we'd be forced times I'm sub a divided by the total mass. And after this, we can ah substitute. And so we can say that the force of a B equal in the force force of B A. This is giving us force times the sum of the mass of Block B and Block C divided by again that total mass. So at this point for a party, they actually want us to find the force in the net force on each mass and the acceleration. So the acceleration is quite easy. We found the expression in the first page and so we can just plug in. And so this would be 96 Newtons force and then the sum of all the masses would be 30 kilograms. Let's I need six points. Are Nunes uh, divided by three 0.0 kilograms. And so this is giving us 3.20 meters per second squared now. The net force on each mass force net would be the same for all of them and so this would be m A. So this would be 10 kilograms multiplied by 3.20 meters per second squared, giving us 32 Nunes. So that would be the net force on each mess. And to find the magnitude of a B equaling f sub b A, this would be equal to the acceleration times, massive B plus massive sea, and this would be equal to 20 kilograms multiplied by 3.20 meters per second squared. And this is giving us ah 64 new needs now at this point for four sub BC. This is equaling for some C B. And at this point it's pretty apparent that this illegal again 32 Nunes so this would be our final answer for a Part E and our final answer for the second final answer for a party that is the end of the solution. Thank you for washing

In this example, I'm going to be looking at forces and force the hunting equations to get some information about our system here, Let me tell you a little bit more about the system. We have two blocks. One's resting on a table and it's attached by a string passes over a pulley and attach is to block to here. And what we want to do here is find a force that we need to apply to give block one an acceleration of 0.75 m per second squared to the left. All right. Considering that we have it attached to this pulley that is free to fall. Let me write down the information that the problem gives us. So we're looking for force. I'll call this life force left. We don't know that, but we do know the mass of the blocks, massive block one equals 25 kg. A massive block two equals 15 kg. And we have a coefficient of friction between the table and block one and that is U. K. Equals serial point. Okay, so, how do we solve this equation or problem? We don't have an equation yet. Um what we're going to do is look at our force diagrams for each block. Okay, the first thing I want to do is to find a positive and negative direction. I'm gonna call to let the positive direction and this will be the negative direction. Okay, so let's what we have my force that I'm looking for that's in the positive direction. We have the force of attention from the string and that will be pulling it in the negative direction and then we have the frictional force which opposes the motion. Center motion is in the positive direction, friction is also going to be negative. So that gives me for my forces. I have course let's minus force of tension minus force of friction and I equals mass times acceleration for that truck. Right now we want to look at block to this one only has to forces acting on it. We have The 4th to the gravity going in the negative direction. And now here my tension is acting in the positive direction. So I have some of force and blocked two equals mm to acceleration legal of course attention minus forest gravity. Okay now we're all set up to solve the equation. What we want to do is equate the variables that we know are equal for each system. So what do we know? We know acceleration is going to be the same for each box and we are getting that value of We want our acceleration to be zero sorry zero point 75 m/s squared you know that's equal. We also know that the tension force is going to be equal. So I'm going to use our attention for us because we don't have a number for that. So we can we can manipulate these equations to eliminate it completely. Alright, so from our first block I have force of tension equals Course left -118 minus force of friction. And from this one I get yeah he looked F. T. Equals. And to a plus force gravity. All right now I can set these to equal the right hand side equal to each other. And when I do that gives me F. L minus M. One a minus bush friction equals. And to a plus force gravity I want to sell for fl. So that just becomes FL equals M. one plus M. two A. Plus for friction Plus Force of gravity. Such force friction acting on Black one. That's the force of gravity acting on. But you all right so now we have all of these numbers we can go ahead and solve. Um I won't two out the math explicitly. I'll just give you my answer because we're running out of time here. So we have asked 1 1525 acceleration of points. 75 m/s. Force of gravity is just I mean sorry, force of friction is musa K times MG of black juan. Force gravity is just massive block, two times acceleration due to gravity. And when I do that I get an F. Oh equals 226 unions. It's positive. So it is in the left direction which is exactly what we hoped to find. So that's just looking at signs like that can be a good check to make sure that you have amazed, horrible, horrible earth. I think that's how you solve visible course balancing problems using force, uh huh various forces.


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