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mmh The drawing shows a large cube (mass $=25 \mathrm{kg}$ ) being accelerated across a horizontal frictionless surface by a horizontal force $\overrightarrow{\mathbf{P}}$ . A small cube (mass $=4.0 \mathrm{kg} )$ is in contact with the front surface of the large cube and will slide downward unless $\overrightarrow{\mathbf{P}}$ is sufficiently large. The coefficient of static friction between the cubes is $0.71 .$ What is the smallest magnitude that $\overrightarrow{\mathbf{P}}$ can have in order to keep the small cube from sliding downward?

400 $\mathrm{N}$

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Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

Hope College

in this question have the following set up There is one big cube and there is one is more cube. A Force P is exerted on the big cube, which tends to move the right because of the action off that force off course. This big cube also suffers the action off the weight force. And because of the weight force, the ground exerts a force and one on these big cute. Now, for the small cube, there is another set of forces. There's more. Cube is touching the big Cube. And as the big cube is moving to the right, there appears a normal force in between the small cube and the big cube. And that normal force is exerted by the small cube on the Big Cube on by the bean cube. On the small cube outside, the weight force tries to push the small cube towards the ground, while the frictional force tends to hold it in place. It is because the fictional force always approves two movements or tendencies off movement on. The attendance we have here is that this more cube, we was lied down to be cute and to avoid that from happening there appears a frictional force. And then we have to discover what he's the smallest magnitude off this force. Be such that the small cube was not slight on the Big Cube. For that, let's choose a reference freeing as these one. These is the Y access, and these will be my X axis. Now, to get some equations, we apply Newton's second law to both this small cube and the B cube. So for the Big Cube, we got the following on the X axis. The net force that acts on the Big Cube is because the mass off the big cube times its acceleration but that net forces composed by two forces beat on and true, so well, you got be minus and true is equals to m one times a one. Let's keep this equation now. Do the same for the y axis. The net force new alliances that acts on the Big Cube Is there close to the mass off the big Cube times, its acceleration in the Y direction. I noticed that it's not moving in the right direction. Therefore, its acceleration is it was 20 then the net force in the Y direction is it close to zero. And the net force is composed by the normal and one and the weight that were you want then and one miners. W one is the question zero, meaning that the normal force that acts on the key on the Big Cube is it goes to its weight. These wasn't love your results. So this equation, I believe, is not really helping us to solve this problem. Now let us do the same thing for this small cube for the small cube on the exact Since we have the following the net force in the X direction is in close to the mass off. There's more cube times its acceleration. But the net force that acts on the small cube on the X axis is only one force and true then and chew is equals to em to times a true And now So that is Do it for the Y axis. Then on the Y axis, the net force is there close to the mass off the cube number true times. It's vertical acceleration. We wanted not to move so exactly aeration should be close to zero. Then the net force and the Y axis should be zero. That means that the fictional force minors the weight number two should be close to zero. And finally, the frictional force must be equals to the weight number two. And now how can we use those equations that we acquired to solve the problem? Let me raise the board and summarized these equations. We end up with this. Four equations now noted the following There is no relation between equation number two or on the four speed. So we are not using this equation at all. And there also seems to be no relation between the question number four under four speak. But these is not the case, because when he is the smallest possible, the force must be the biggest possible. The reform. We must be working with the biggest possible frictional force, which, in the static situation, has a relation with the normal force. So question number four actually becomes muse. Started times normal number. Choo is there close to doing number two so it has a relation before speak because it has a relation. We've the normal and chew. Then we will be using equations 13 and four to solve this problem at me here. Right Them. Okay, Now we begin by solving equation number one for Pete. These results in P is he goes to M one times a one waas and truth. Now we have to discover what is the value off a well. And Andrew, we could discover the value off entry using equation four from equation forgets that entry is the coast. Ow. Two divided by mu static then b is equals to m one. Thanks a one. Plus well, the truth divided by mu static. Now we have to discover what is the value off a one. And for that we can use the equation. Number three equation number three tells us that and true is equals two and two times a truth. Therefore, each year is and two divided by m two. And at the same time into we already know is w two divided by me static. So we get this for extradition number two it happens that both blocks moves at the same time to reform a one is there close to a true then B is he goes to m one times the more you chew. Divided by muse Static times m true plus W. Chu Divided by mu static, we can write it as the way you true, divided by muse static times and one divided by M. Chu plus one. Now remember that the weight force is given by the mass times that the acceleration of gravity, which is approximately 9.8 meters per second, squared near the surface of the Earth show. Plugging the values that the problem gives us we got beat is equals True m 2 25 times G 9.8 divided by new static 0.7 to 1 times and 14 divided by m true, 25 plus one, then B is approximately 400 Newtons, or four times 10 to the second new terms.

Brazilian Center for Research in Physics