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$\bullet$$\bullet$ Two boxes are connected bya light string that passes over alight, frictionless pulley. Onebox rests on a frictionless rampthat rises at $30.0^{\circ}$ above the hori-zontal (see Figure 5.50 ), and thesystem is released from rest.(a) Make free-body diagrams ofeach box. (b) Which way will the 50.0 $\mathrm{kg}$ box move, up theplane or down the plane? Or will it even move at all? Showwhy or why not. (c) Find the acceleration of each box.

a) $w_{x}=m_{1} g \sin 24^{\circ}$ $w_{y}=m_{1} g \cos 24^{\circ}$b) The 50.0 kg box moves up the ramp, and the 30.0 kg box falls to the ground.c) 0.6125 $\mathrm{m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

McMaster University

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question 29 say that two boxes are connected by light string that passes over light. Friction was poorly. One box rests on a frictionless ramp that rises at 30 degrees above the horizontal showing. The figure that I drew here and the system is released from rest A make free body diagrams of each box and be which way with a 50 kilogramme box, move up the plane or down the plane, or will it move at all? So why or why not? And c find the acceleration of each box. Okay, so this is our our set up here we are two boxes. Um, first and shoot before you do anything else is labeled my access. Angle it along the direction of my inclined plane, the ex at least along the inclined plane. The first question when you make free body diagrams So you mass one is the object that's on the ramp, of course, that the tension in the string will pull it towards the police that the box has a weight which pull street down the y component of this weight. If Alfa is the angle of the ramp, this angle here is also Alfa. This is the weight in the Y component and by our triangle that this component here also this component is our rate. The extraction and, of course, isn't normal for us pushing away from the surface of the box. So you know how easy this is for M one, the one on the inclined plane, which I call them one. It wasn't in one of the question, but we don't do it with friction in this problem. That's a little bit easier. Uh, mass to is a little bit easier. This is vast, too. Attention pulling the box up and the weight of box, too. So first I'll say, Wait one for mass one. It's over. This one. It's weight of two. A little bit simpler than the object on the ramp purpose and get fined. If there's acceleration of the blocks. Well, im makes me well, the blocks move which But of course, if it starts from rest with cause on examination, So I'm just going to assume I mean, I did assume if it does travel in a given direction, it's that helps me formulate my force equations so I can use the fact that the forces equal and a. At least we'll say the X direction because of the wind direction, it should be zero. Um, so I would assume that M one will move in this direction. So in my direction to make in this equation will assume that scenario. And if I were to get a negative value, such as in a negative acceleration, then that would tell me that I chose the wrong direction and the object would simply move in the opposite direction. So I'm gonna seem and one move up the ramp and therefore m two will move downward most articles third from that premise and go from there. So for blocks, 1 may this new FX equals and a and one day, I'll say, is moving up, Gramp. So looking at our directions from a free body diagram, that means the attention to be positive. My hasert w X wish. Well, based on again, if this is angle theater Alfa, then this component are X component would be a sign of the angles of w sine omega. Sorry, Alfa, which again this is W X. So this is have you sign Alfa and W is the weight Coast Alfa and then it s equal acceleration of box able solution of the system is the same for both objects. But that corresponds to mass one. This isn't Tell us a lot of information, Mrs for box one. We'll get box too. We know that again. In the scenario F fly force in the white direction has to equal m a m to a this time because we're seeing that looking at our box our direction here or attention is punky pointing upward and are free by diagram here minus the weight which is M two g. That equals M two A. So what I'll do now That was self for attention in box too. So be and to a my plus Sorry, m two g when I should should state that my m two a here would be negative because I'm moving in the downward direction. So this term here is negative. Excuse me. So native sign here doing substitute this term into our equation from part for the box of a house we have I'll still be done here. So negative m 2 a.m. And because of the negative direction and to a plus mtg minus the wait times sign Alfa equals m one A. And so I can click my like terms for acceleration. So who moved my and to turn to the right hand side of M one plus m two times a with the equivalent Teoh mtg minus the way time. Sign over. And so you obtain an expression for acceleration to be m two g. Also this W Scialfa. I should label that says his w one sign Alfa. So all these is W one the weight of box one which matters, of course. So when do we pull a G term out in front? Because they both have king, of course, minus m one sign Alfa G chairman for my wheat term comes up to the front Close that so that simplifies this turn here, right there. And this whole term that divided by the sum of the masses and one plus two. So this is also kind of solution to part see what is excellent in the box. But again, the direction of this acceleration will tell me if I assumed correctly that the box box system will move to the right and downward and so looking are givens were given massive one given massive tumor given the angle so we can solve for this in 9.8 meters per second. Coming out the front, I asked you is the smaller box that's 30 minus 50 times sign of 30 degrees divided by to some of the masses, which is 80 kilograms. You find that acceleration therefore 0.61 three meters per second squared. And so since this is positive, we can take that mass one. The one on the ramp moves up. Ramp was up the red. What is your solution? Are given an answer for part B and part subjugated. We already calculated this, but we do find the acceleration is 0.613 meters per second squared.

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