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A $5.00-\mathrm{kg}$ block is placed on top of a $12.0-\mathrm{kg}$ block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.600 . What is the maximum horizontal force that can be applied before the $5.00-\mathrm{kg}$ block begins to slip relative to the 12.0 $\mathrm{kg}$ block, if the force is applied to $(\text { a the more massive block and }$ less massive block?

29.43 $\mathrm{N}$

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Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

McMaster University

we begin this question by applying Newton's second law in these reference frame, these will be the Y axis and these will be the X axis. Then we will apply Newton's second law to both blocks in both directions. Let us begin by applying to block number one to on the X direction we have that the net force that acts on that direction is equal to the mass off block number one times its acceleration the X direction. But there is only one force that is acting on block number one that direction. And is this defection all force number one, which happens because someone dispassion block number two so it tends to move in this direction. Therefore, what happens is that block number one tends to move in this direction. We've respect to block number. True, both blocks were moved to the right, but these block tends to move in a slower fashion. Therefore, it tends to fall behind in orderto avoid that from happening the fictional for seven years and try to bring the way for the block number choo. Then this is why we have to frictional forces. One frictional force is acting from block number two on two block number one and another fictional force is acting from block number one on block number two, trying to make it not move. So for a block number one, we have the following. Frictional force number one is equals to the mass off the block number one times its acceleration in the X direction, then doing the same for the vertical direction. We got the net force in the Y direction is the coast the mask off block number one times its acceleration in the direction. But it's not moving, and it's not going to move in the vertical direction. Therefore, its acceleration is it goes to zero. Then there are two forces acting that access and one which is the normal forced that block number two exerts done Block number one on the weight forced off block number one. Then we have n one minus the waste number one meaning close to zero. So the normal force number one is, of course, the weight off block number one. No letters applying Newton's second law to the second block on the X direction we have that the net force and that the reaction is because the mass off block number two times its acceleration. That direction. There are two forces acting own blocks number choo on the horizontal direction, and those forces are these. Apply it force and the frictional force that block number one exerts on block number two trying to make it not move. Then these results in the following F that points to the positive direction. Mine is the friction number. True, that points to the negative direction is because of the mass off like number two times its acceleration. Okay, now let us apply in Utah State with lot block number two on the vertical direction. Doing that, the net force in the wider action is given by the mass off the block times its acceleration in that direction. Again, it's not moving, and it's not going to move on the wider action. So the acceleration is the question. Zero then noticed that the Net Force the weather action is composed by three forces to weights that points downwards on the normal force that the table exerts on block number two, then into minus w one. Miners that were true is equals to zero, meaning that the normal number two is he close to this some off the weights. OK, but what we want? Better mind. We want to get your mind. What is the maximum possible value for these force f such that one book isn't sleeping on top of the order. And what is the condition for this to happen? If they are not sleeping with respect to one another, then they are moving with the same acceleration. So the key condition is this one. It must be true that a one X is there close to a two X. This must be true. And are you call these accelerations eight on Lee. Then we have something with these equation at these equation at least it means that using this condition equation number one can be written ized. Fictional force number one is he goes to the mass number one times acceleration and for equation number two F minus frictional force number chewed is the cost of the mass number two times its acceleration. Then notice that we want to discover what is the maximum possible value for the force. So these means that we should also use the maximum possible value for the frictions and the friction in this context is the static frictional force. So the fictional forced number one is equals to static friction off refuge in times the normal force that acts on block number one. Similarly, the frictional force number chewed is it goes to the static, which no confusion times. And these is important to tell you their normal force that these block exerts on this block. So this is multiply it by the weight off block number one because the weight off block number one is the force that this block is exerting on these older block. Then it's even by these equations. But we also know that the normal force number one is he close to the weight number one. Then then we conclude that both frictional forces are equal. Therefore, let me write it like this on both frictional forces. I will call f f. Therefore, we can write these equations one and true as follows F f is equals to m one times eight for the second equation F minus. F f is equals to M Chu times a. Then we solve both equations wanting to for the acceleration so eight is equals. Two F f divided by m one and also a Is it close to F f? It's already f minus f f divided by the mass number Truth. Then we can equate both sides to get f f divided by m one is equals two f minus f f divided by m True. Now we can solve this equation for the force f and these gives the following send this term to the other side to get and true. Divided by m one times f f is equals two f minus f f and I will say this term to the other side to get at the force f Is it close to m two? Divided by m one times f f plus f f finally factor divisional force to God fictional forced times and choose divided by m one close one is equals two f So this is expression for the force f. Now let me clear the board to calculate what is its maximum value. Okay, so the maximum value off force F happens when the frictional force is also a maximum. So f is given by the static frictional coefficient times the normal force number one times am true, divided by M one plus one But remember that and one is because of the way to number one. Then f is it close to the static friction aquisitions times and one times geek times entry divided by M one was one. Remember, In that G is approximately 9.8 meters per second squared. We get that f is it close to 0.6 times. Five times 9.8 times 12 divided by five close one And these gifts and maximum force of approximately 100 new terms. So this is the answer for the first item? No, To solve the second item, let me do a little change in the drawing and keep the result of the first item here. So 100 new times. Okay. Now, instead of acting here, the Force F is acting dear directly on block number one. Then we have to calculate the maximum force again. For that, we have to use new toe Second law, let us begin by a plane. You don't second law on this block on the horizontal axis because these real involved value off the force f. So for block number one direction, the net force next direction is equals. True, the mass off block number one times its acceleration again. Both blocks need to have the same acceleration. If they are not sleeping on top off on an order, then let you write it like that, I would just put eight in here. The net force that acts on block number one is composed by true forces now the frictional force that tries to push this block ahead. Then, because of the action off this force block number two, we'll try to keep this block in place. So instead, off what we had before, we now have that the frictional force that acts on block number one is pointing to the left. And on the other hand, when Block number one tries to move to the right, it tries to bring block number two with it. So now we have a frictional force acting block number two to the right. Then, according to Newton's second Law for Block number one on the horizontal direction, we have plus F minus frictional force number one, So F minus frictional force number one is because the mask off the first block times its acceleration them remember that the maximum possible frictional force is given by static friction coefficient times the contact force, which in this case is the normal number one. It's very easy to see that the normal number one equals with the weight number one we have already don't that calculation actually, Then we can use the result to write F minus mu static times the weight number one Z goes to the mass number one times acceleration, then the acceleration according toe, the Newton's second law. Apply it on block number one. Is he close to F minus mu static times and one times G, divided by the mass number one now doing the same for the order block. We get that in that force. Acting eat on the X direction is because the mass off that block number two times its acceleration, which is also ate. Then I noticed that the Net force that acts on block number two is only the frictional force number two. So we have the frictional force number two being equal to the mass number two times acceleration. Then I noticed that the frictional force number two must be the maximum, so it's given by the static frictional coefficient times the contact force between blocks one. And truth of that contact force is the wait for us off block number one and these results in mews static times that mass number one past G so mu static times the mass number one times g Zico's true. The mass number True times acceleration. The reform we have the acceleration being given by muse Static times The mass number one divided by the mass number True pines acceleration off gravity. Then we can equate these two equations to get an equation for the force. F By doing that to get the following f divided by the mass number one minus mu static times G. I have simplify it. These masses on the second term is equals to the static friction coefficients times G Times am one divided by M. Chu Then f is the question You static times Jeep times m one divided by mt plus muse static times G and this is F By AM one they only can factor out new static times G to get f divided by M one is the question You static times G times am one divided by m two plus one and finally sent this term to the other side. So the force is mu static m one g and one divided by m. True was one then we just have to plug in their values that were given by the problem. And these we will result in the following. F is equals to 0.6 times, five times 9.8 times and 15 divided by 12 close one. This gives us a force off approximately 41.6 neutrons so that his dance or for the second item, then this is the final answer for the complete problem.

Brazilian Center for Research in Physics