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NOTE: This IS 3 mutti-part question Once an- Block Ahas mass of 65 kg; and answer IS submitted You will be unable to return to this part block Bhas are Us mass 0f 1...

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NOTE: This IS 3 mutti-part question Once an- Block Ahas mass of 65 kg; and answer IS submitted You will be unable to return to this part block Bhas are Us mass 0f 13kg 0.20 and Uk= 015. The coefficients of friction between all sunaces of contactdetermlne the acceleration ol block(You must provlde an answer before moving on to the next part )The acceleratlon cf block B Isms2 + 25

NOTE: This IS 3 mutti-part question Once an- Block Ahas mass of 65 kg; and answer IS submitted You will be unable to return to this part block Bhas are Us mass 0f 13kg 0.20 and Uk= 015. The coefficients of friction between all sunaces of contact determlne the acceleration ol block (You must provlde an answer before moving on to the next part ) The acceleratlon cf block B Is ms2 + 25



Answers

In Fig. 5-34, $A$ is a $4.4-\mathrm{kg}$ block and $B$ is a $2.6$ -kg block. The coefficients of static and kinetic friction between $A$ and the table are $0.18$ and 0.15. (a) Determine the minimum mass of the block $C$ that must be placed on $A$ to keep it from sliding. (b) Block $C$ is suddenly lifted off $A$. What is the acceleration of block $A$ ?

The execution. It is given that a block of 35 Kg is resting on the floor, and another block of Mars seven Kg is resting on it. And we have given the coefficient of friction between the block and the surfaces. And it is told that a horizontal force of 100 newton is applied to the block B. We have to find the acceleration of block. So this is our blockade which is resting on the surface and this is our blog which is placed on this block. So this is the block B. This is block and we apply the force of 100 newton on this block. And we have given the value of coefficient of static friction and kinetic friction. So, as you noted, the static friction, fourth between the two blocks is calculated by that is FS should be less than equal to the limiting for that is no into And so from here we can say that the normal forces equals two. That is N equals two MG. And this is given by seven into 10. So we can say this is equal to 70 newton. So we can figure it. Our Static friction forces that is FS should less than equal to this is 0.5 into 70. So this will consult to be 35. So the first should be less than equal to 35 Newtown. As we can say that our applied fourth is greater than static friction force, which means our body is in motion. So there is kinetic friction acting between the blocks. So we can see that the value of kinetic friction is given by that S. F. K. Is equal to this is new cane to and and this is equal to 0.4 into 70. So this will comes out to be 28 newton. This is the value of kinetic friction. Now we can say that for mosque 35 kg. If we draw the free body diagram. So this is the March 35 kg and there is kinetic frictional force acting on this and this is equal to F. K. And the expiration of this block is equal to that is So we can say that this force should be close to that if if they should equals to disease. I'm into circulation of this block so we can see that This is 35. He goes to this is 20, sorry. From here we can say that this is 28 base because this is 35 into a. So from here we get the acceleration and this will come out to be 0.8m/s square. So this is the acceleration of this guy block. This block is moved because of the kinetic friction. There is no other driving force on this block. So this is our answer. And for that we can say option is the correct choice. Thank you

So here we can apply Newton's second law to the X in my direction. So the horizontal force minus the magnitude of the frictional force equaling the mass times acceleration in the extraction. Here, the vertical force plus the normal force minus the gravitational force equals zero. We're here F this equaling 6.0 Newton's and the Mass is gonna be equaling 22.5 kilograms. So we can say that four part A, Given that the vertical forces equaling 8.0 Newton's, we can say that the normal force is I'm gonna be equaling to 2.5 kilograms multiplied by 9.8 meters per second squared and there's gonna be minus 8.0 Newton's and this is gonna give us 16 0.5 Newton's. Now we can say that the maximum the maximum static frictional force is going to be equaling to the coefficient of static friction multiplied by the normal force. This would be equaling 2.40 multiplied by 16.5 Newton's. This is equaling 26.6 Newtons, which is greater than F of 6.0 Newton's. So here block doesn't move acceleration of zero and we can say that. Then the force, the frictional forces equaling the magnitude of the the magnitude of the the vertical force. My apologies. And this is equaling 8.0 mutants. So we can then say that he rather my college is six point. And so here we can then say that for part B P is equaling 10 Newtons and we can say that then the normal force is gonna be equaling two 14.5 Newtons. The Frictional Force, the maximum static frictional force will be equaling two 0.40 multiplied by 14.5 Newtons. And this is giving us 5.8 Newtons, which is actually less than F so here, given that it's at less than half of 6.0 Newton's, we can say that here the block does move And here the Canada, the friction will be going to the kinetic friction equaling 0.25 multiplied by the normal force of 14.5 Newton's. And this is giving us 3.6 Newton's for part C, then P equaling 12. Newton's then leads to the normal force equaling 12.5 Newtons, which is giving a maximum static frictional force equaling 0.40 multiplied by 12.5 Newton's. This is giving us five brother, 5.0 Newton, which is, of course, less than half of 6.0 Newton's. And so the forced the frictional forces equaling the kinetic frictional force of point to five multiplied by 12.5 new tins. And this is giving us approximately 3.1 Newton's. This would be our answer for part C. That is the end of the solution. Thank you for watching.

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.


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