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Problem 109 Hard Difficulty

The central ideas in this problem are reviewed in Multiple-Concept Example 9. One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and the magnitude of the force is slowly increased. When the force reaches 47.0 $\mathrm{N}$, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block so that it just begins to slide out from under the upper block?

Answer

141 $\mathrm{N}$

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Bf

Basma F.

June 30, 2021

Planet X is very large, having a mass and radius that are, respectively, 336 and 10.9 times that of earth. Suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. Ea

Bf

Basma F.

June 30, 2021

Video Transcript

several forces are acting on the system. Block number two. We have its weight force acting on it. Then it's late Force pressed down the ground and produces the normal force Number two, which is exerted on block number two. Then on top of block number two, we have Block number one, which exerts a force w weren't on block number chewed, which further increases the normal force off block number two. And then block number two exerts a force on block number one, which is normal, and one on top of that. We have a force that is trying to push block number one to the right under block number one. We'll try to move, we just back to the block number two. So a frictional force will appear between blocks one and true that will try to keep block number one in place at the same time. From the point of view off block number one, Block number two, we'll try to move to the left. So what happens is that Block number one produces a frictional force F F one that acts on block number troop, trying to push it ahead a top of that. We also have a frictional force on the ground that tries to keep block number two from moving. So there is even a turd frictional force that acts on block number two. And then we have to consider a very similar situation with not an identical situation. It's similar because it has almost all the same forces. But the force F isn't acting on the top block where it's acting on this lower block and do this F brain. Then we have to calculate what is the minimum value off F prime such that, uh, block number one. We'll move with respects to block number two. We will have to use Newton's second law a lot of times, so I don't choose the falling reference frame vertical access that I am calling. Why that point upwards on a horizontal axis that I'm calling X, which points to the right. Then let this be situation number one on these be situation of the truth. We are interested in situation over one, so that's beginning by analysing such situation. Then, using Newton's second law under situation number one own block number one on the X direction results. In the following the Net force acting on the X direction. Block number one is equal to its mass, which I'm calling em because both blocks are identical times. Next generation off block number one. Then notice that the net force on the X axis that is acting on Block number one is composed by a single force, which is the frictional force number two. Then we have f F coupe equals to the mass off block number one times its acceleration. Now we applying Newton's second law to block number two still on the X axis. So it tells us that the net force acting block number two the ex directions is give equals to the mass of block number two times its acceleration. Notice that the net Force that acts and block number two you on the X axis is composed by three forces the frictional force that the upper block exerts on the lower block, trying to keep it in place. The frictional force exerted by the ground on block number two on the force after prime there is apply it to the lower block. So it tells us that f prime minus frictional force one minus fictional forced three is equals to mass off block number two times its acceleration. Then we have to notice that on the limiting situation that the problem asks us where the block just starts to light. All the frictional forces are the maximum possible. So it's true that we can use the frictional forces as the kinetic frictional coefficient times the normal force. So back toe the first situation these well, we can write it as the kinetic coefficients times the normal force and number one which is a normal force acting on block number one. And these is equal to the mass off this block times its acceleration. Now notice that on the vertical axis there are two forces acting on block number one, the normal force on the weight force. So as a block, number one is not living in this access. The only possibility is that the normal force is equal to the weight forced so we can write it as they can at the crescendo coefficient times the weight off block number one and these these m times a one. But we know that the weight force is given by the mass times acceleration of gravity. So this is UK times, M times D and the easiest him times a one so we can simplify the masses and get a value for the acceleration. Number one A one is Mu Kate. Times Jeep. Then keep this value now going toe the other equation and doing a similar thing We got that f prime minus mu k times the contact force that the block number one is exerting on block number two and such contact force is equal to the weight off block number one so times that we want minus mu Kate times diffraction all force number three The frictional Force number three is given by the kinetic frictional coefficient times the normal force that the ground exerts on block number two and such force Is he close to the weight off the block Number two. Then we have this and it is equal to the mass times acceleration off the block. Then we can again right the weights as mass s times gravity to get UK times m times V minus mu K times, M times G on the Zico's two AM times a true and now notice that we have lots off these mu K thanks deem UK times Jean right here we can write it as F prime miners. 81 times M minus a one m one ending is substituting UK times G by a one, as we have discovered from our last equation. And these is m times a true now in the limiting situation. The problem. Ask this. A one is equals to intrude. Let me call it a So we have f prime minus troop times him Times eight equals two m times A. So finally f prime is equals 23 times, M times eight. Now we know what is their value off f brain in his three times the mass times the acceleration off the blocks. And then how can I discover the acceleration off the blocks? We have to return to the first situation because in the first situation, all the frictions are out. Sue the maximum possible frictions on the first situation Analyzing the block number one we have that true forces are acting on the X axis so that the net force in the X direction he's given by its mass times its acceleration then Demet forces R f minus half true of these is M times eight but the situation the blocks are not moving yet, so the acceleration is zero. Then f is Equus defection all force number two. But you know that F is 47 neutrons because the problem totals. So with that, we discover that the frictional force number two is equals to 47 neutrons. And why is the is important? Because on the first situation in the situation, we discovered that the mess times the acceleration off the block number one is it causing the magnitude off the maximum frictional force F F number two and the maximum fictional force F F number two is equals to 47 new terms, as we had just discovered from the second situation. Therefore, F half number tree is he goes to m times D from these equation, but F F two is equals to 47 from these equations up here, no f prime is three times 47 which is the cost to 141 new terms. So disease, the neccessary force that one should apply in the lower block in order to make the block number one starts was light on top off block number two