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The drawing shows box 1 resting on a table, with box 2 resting on top of box 1. A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2, and the other end is connected to box 3. The weights of the three boxes are $W_{1}=55 \mathrm{N}, W_{2}=35 \mathrm{N}$ and $W_{3}=28 \mathrm{N}$ . Determine the magnitude of the normal force that the table exerts on box 1 .

62 $\mathrm{N}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

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Hope College

McMaster University

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in this question. The following forces are in action. We have the weight force off the box number one which act on the box number one itself on on the ground. Then the ground acts in box number one by a normal force, which I had called. And one The second box has also its weight force number you to acting on it. But it also acts on box number one with its weight force. And because of that, the box number two receives another force called the Normal Force and chew, which is the force that is exerted by the box number one on the box number two as a response to do weight off most number truth. Then the box number two is out, sir. Pulling down his rope with some force that I had called F. And as an answer to that forced half the rope pulls the box number two up. Uh, we have we have a force that I had call it t that forced it is also present here, and it is reacting to another force f which is It goes to the force w t three, which is no wait. Forced off the turned box, Then in blue are the forces that are exerted by the boxes and red. There are the forces that are exerted only boxes we want to calculate. What is the value off the normal force that the table exerts on box one. So we want to complete, and one How can you do that? Well, by applying Newton's second law in these direction, which is the vertical direction, let me call it white. Then we have to apply Newton's second law, probably to several objects in the situation. So the force we are interested the force and one is acting on box number one. So letters apply Newton's second law to box number one, so box number one than that force on the Y direction. But as it is the only direction off interest in this problem, I'm not be writing F and component why or anything like that. So the Net force in the interesting direction is equal to the mass off box number one times its acceleration in that direction. But note that the box everyone is standing still, so its acceleration is it close to zero than the net force that acts on box number one is the cost of zero. But this net force is composed by several forces. So acting in box number one, we have the normal and one which points in the positive direction. We have the weight off the books number one which points in the negative direction. And we also have a few more things here. So we have the weight off the box number two off course. But it's not the world link that acts on box number one, because box number true is being pulled up by a four steep. So these forced E makes the force that the box number shoe exerts on box number one different from the weight forced off box number two. In order to take that in account, we have to include here in this equation the tension force. So plus the tension force, this is because this is the force that box number two I exerts on box number one. Then these wolfing is a cause for zero. So the normal number one and his equals two w one plus it over your true minus the sanction force. And these is equals True 55 Close 35. Mina's steep. Then and one is equals to 90 minus detention force. Then how can we calculate what is attention force? So we have to ask ourselves, what is tension in the rope? And in this situation, what distinction in the rope is the box number three? Then let's apply Newton's second law, the box number three, to apply to box another tree. We have the following in that force. Acting in box number three isn't close to the mask off the Marks Times its acceleration, which is close to zero because the box number three is a standing still, Then the net force acting on box number three is equal to zero. What are the forces acting in? Box number three? There are only two forces acting owned at the Box Detention Force and the Weight Force. The older Force F He is exerted by box number three, and it's not being exerted by something else Own Box number three. This is why it's not included in this equation. Then detention force is equals to do it off the box number three, which is the Costa 28 new terms, then going back to the equation for the normal and one we got and one is in question 90 minus 28. And these gives is a normal forced that the table exerts on the box number one off 62 neutrons.

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