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A book of mass 2.7 kg is pressed against a wall where the coefficient of static friction is _s-0.35 and the coefficient of kinetic friction is |_k-0.25 between the ...

Question

A book of mass 2.7 kg is pressed against a wall where the coefficient of static friction is _s-0.35 and the coefficient of kinetic friction is |_k-0.25 between the wall and the book: What is the minimum pressing force required to keep the book at rest(in Ni?

A book of mass 2.7 kg is pressed against a wall where the coefficient of static friction is _s-0.35 and the coefficient of kinetic friction is |_k-0.25 between the wall and the book: What is the minimum pressing force required to keep the book at rest(in Ni?



Answers

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs 31 N. The coefficient of static friction between his hands and the book is 0.40. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

In this problem. The first step is to figure out which way the box will accelerate. Yeah, and the answer is, it'll axillary up since the 96 news of force applied overpowers gravity and overpowers gravity because sometimes ji is the force of gravity. And since masses five n. G s, 9.8, I get a value 49 Nunes. And so even though the forces being applied and an angle, it is so large that it overpowers this and you can check that the vertical force of the Applied Force is greater than this mountain. But once you realize that the box is going to move up, then you realize that the friction force is going to be in the dower direction. And so when we construct our force diagram, we're two forces downward mg and the friction force. We're not a normal force to the right because it's resting on the wall. And so there needs to be in normal force to the rights at the book doesn't go through the wall. And then there's an applied force up this way at an angle of 60 degrees. And so some of the forces in the extraction is equal zero. Since the acceleration, the extraction is equal to zero. By the way, I should set up a recording system positive. Actual peaches, right? Positive life will be upwards like that. And we also have an acceleration in the white direction upward. And so back here, some of the forces equal zero. This implies that in minus f co sign, uh, 60 is equal zero, which tells us that and is equal to 48 noons. Once you plug in what f is. It's a given. This immediately tells us that the force of friction which is equal to the coefficient of kinetic friction times the normal force, is equal to four point for Nunes after we plug in point through here, which is also given now that we know what the force of friction as we can apply Newton's second law in the wind direction. So some of the forces in the white direction in this case it's not equal to zero since it's accelerating in the white direction is just equal to this. This implies that he applied force times sign of 60 minus version Force Linus lg Is it gonna last time's exploration? Why these have minus signs on them because they are directed in the minus y direction. Now, if you solve this for the acceleration in the white direction because masses known, we just figured out what this is and F is known, and so we just solve it, using a calculator to find extortion. Ally, I get 3.95 meters per second squared now because the acceleration is constant. We can use a cinematic equation in order to find out what we actually want to find out, which is the velocity after his travel 0.4 meters. And so just writing down our knowns, why my swine on which is the distance traveled is your plate for initial speed, zero movies are unknown and the celebrations 3.95 And so to solve this or I'm going to use this kind of magic equation right there. Now this term goes okay since being on zero and you just pulling in these values here into here and then take the square root and you get that the velocity is 1.17 years for second, and that's the final answer to the problem

In this question. By pressing the sides off the book, a frictional force is produced pointing upwards. We have to discover what is the minimum magnitude off the present force that is needed to keep the book in place. In order to answer that question, have to use Newton's second law are you? Choose the following reference frame a vertical axis pointing upwards and the horizontal axis pointing to the right, then applying. Newton's second law, the vertical axis, results in the following the Net Force. The Y direction is given by the mass off the book times its acceleration in the direction it should not be moving say its acceleration is constrained to be close to zero and then that force is a question zero. But the net force in the Y direction is composed by two forces the frictional force which points to the positive Y direction the weight force which points to the negative y direction. So fiction off warrants miners. The weight forced is it goes to zero before different No force is he goes to the weight force. Okay, this is a nice result. No, that is moved to the X direction in order to get something with the force f. So in the horizontal direction we have the following the net force on that direction is there close to the mass off the book times. It's horizontal acceleration, which is a question zero because the book should stay in place. Then there are two forces acting on the horizontal and they are equal forces acting on opposite directions before we got to zero equals 202 These applications off Newton's second law on the horizontal direction have not helped us. Then what we can do? You can remember that if the force f is the minimum force for the book not to fall, then the frictional force should be the biggest possible frictional force, which in the case that the books not moving is a static frictional force given by the static snow coefficients times the normal force. But then, what is the normal force that is acting on the book? In the situation, the force F is behaving as the normal force on. There are two forces that are pressing, so we have the normal force that is equals two times F because the total pressing force is the close to true times f. So that's it, then these give us the following equation. The static original coefficients times half times true is it goes with the weight force. Then the necessary force is given by the weight forced, divided by two times less static friction or proficient. Then we have to use the data that the problem gives us, which is that the weight force is 31 new terms, and the static frictional coefficient is 0.4. Then the pressing force is given by 31. Divided by two times is Europe 14 which is 31 divided by 0.8. And these results in a Princeton force off approximately 39 new tones. And this is the answer to these questions.

Discussion cover the concept of Newton's 2nd law. So let's show the fretboard diagram of the book. The forces acting on the book is the weight of the book, which is acting weird degree down world, and a normal reaction force which act along the horizontal end, and our friction force which acts downward. There is no times. And And the force which is acting on the book at an angle 60° above the horizontal. Okay, let's show the court taxes along or dental is X. And along vertical is why So from Newton's second law, the book is in equilibrium along or gentle. So we can write sigma Fx. Is it close to zero or and minus of course 60 degree is it goes to zero. Are an easy question F off Cause 16 weeks. This is the question one And along why direction sigma F. Y is equal to M. Into A. Or we can write F sign 60 degrees minus M g minus no times N. Is equivalent to em into a Or from the question, when we can write F sign 60 degrees minus M g minus no times, Of course succeeding is that is equal to M into eight. Are the acceleration is mhm F sign 60 degrees mhm mm hmm Minour mealtimes, cost 60° minus mg upon mm. So should the values the acceleration is the forces 96.0 Newton's. And to sign off six Jalili's minus The new 0.3 and cause of 60 degrees minus the masses. Five kg into 9.80 m for second square upon 5.0 Khe or the acceleration is 3.95 m four seconds square. Another question asked is, what is the speed? So from can dramatic motion we Squire as equals to two A S R V. S equals to square root off to a y E r S S. Y. Or we can also ride to A. Y. So the street v square root off to into a 3.95 meters per second square. And to the virus 0.40 m are the speed is 1.78 m for 2nd. Mhm.

Mhm. This question covered the concept of Newton's 2nd law and the coefficient of friction. So, to solve this question, this first draw the free world a grandma. The way we get along downward that is MG. And vertical force offered by the ground. That is a friend. And let's say ours until forces pulling the block with a magnitude of F. And the friction force which as opposed to the fourth F. Okay, so let's say our dental X. And the vertical direction is why uh since the block is in equilibrium in the right direction. So we can write so my son of all forces around the wind direction in syria. And from that we can write the normal reaction forth by Miss M. G. Is a Cuban and 20. Other normal reaction forces secure into empty. We can call with the normal reaction process. The masses to cage Into the gravitational expressionist 9.8. They just all 2nd square and normal reaction force is almost equal to continue this. Mhm. This is back Now in Part three. Uh The friction force very the friction force will apps must be less than the maximum static friction force. Okay. And in this case when the forces less than the maximum static friction falls, let's say the maximum static friction forces in cuba into Milton's. The effort that is continue them are the maximum satisfaction forces 0.4 and to 20 students. Or that is a Cuban into eight students. So if the fools have is less than the static friction force, then the friction force will will we always be close to the magnitude of the force? So in part B the magnitude of the force is zero of them. In that case the friction force will also be zero. So no affection. 4th will be active now in part C. The maximum static friction is eight newton. So in order to move the love the fourth must be greater than or equal to the maximum static friction. And from this the f minimum required to pull the block is egg yolks. Yeah. And party when the block moves. In that case let's say the block is moving with a speed V. Uh In that case the friction force which act access the kinetic friction force F. K. And it hasn't magnitude of milky times the normal reaction force effort. So in party the section is kinetic friction. Yeah. Yeah. No one party. The kinetic friction Fks Milk A Times The Normal Action for seven. All the kinetic friction affairs milk 0.2. And the normal reaction forces to continue dance. And that scares the kinetic friction is four notice


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