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$\bullet$$\bullet$Which way and by whatangle does the accelero-meter in Figure 5.48 deflectunder the following condi-tions? (a) The cart is mov-ing toward the right withspeed increasing at 3.0 $\mathrm{m} / \mathrm{s}^{2}$ .b) The cart is moving to-ward the left with speed de-creasing at 4.5 $\mathrm{m} / \mathrm{s}^{2} .$ (c) Thecart is moving toward the left with a constant speed of 4.0 $\mathrm{m} / \mathrm{s}$

a) $17^{\circ}$b) $\left[25^{\circ}\right]$ towards left

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

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All right. So the hanging masses, the rest would respect the cart. So we usually the second law to solve to result forces and themselves to different parts of this problem. So you start off with free body diagrams here, so heart, I am. This is the free body diagram. Okay? Acceleration is to the right. So there's a So there's a deflection. Beta, obviously on exploration is to the right, as speed is increasing to the right. Okay, in the direction of motion. So very simply needing second law. So for part air, very simply, we use Newton's second law and let's was all forces in both directions in the white direction you have. So let's look into components of this tension here. Right? So this is t uh, in this in this scenario, and the data from the vertical is beta. So the so the horizon's sort of vertical component here is T co signed Maeda. That's this. Okay. And the horizontal components t signed beta. And so if you take 90 minus beta, this will be equivalent to thi co signed 90. My speed. It's the same thing. You just have to pick which angle you're choosing here anyway. And the Y direction that you have an upward force of tea co sign data, which is equal to M. She, of course, is the downward force. And there's no Net IX elevation of the wider action. And that's why, uh uh and that's why there's no term there. Yeah, because acceleration is purely in the extraction. So in the white direction that we have that tea equals mg over co signed beta. Okay, let's call that. Let's call that equation one. Ah, and that and so very similarly in the ex direction we resolved force is used in second law. No, any setbacks. And so here we do have a net acceleration next direction, but in the but that is equal to Teo. Sign beta. Because remember, that's the horizontal component tension. No horizontal torrent of weight s. So this is just equal to m A s. A acceleration to acting surely horizontally. And so we have the tea is m a over side beta and that this equation too. Okay. And so we have to equations for tea. So let's said them equal to each other. Um, what's that one quoted to What we get is em. She over co side beta equals m a over sign made up. Okay, EMS, cancel out all you can see and you pull out the sign. Beta to the other side of what you get is a tangent of data equals acceleration over acceleration due to gravity. Um, so data. It's just our 10 grand for inverse tangent off over tea. Hey, we know to be equal to B meters per second squared, and she is 9.8 meters per second squared. Therefore, we have beta of 17 degrees, so the angle in this case would be 17 degrees to the left of vertical. Okay. Ah, for part B, what's happening here is add velocity is in the direction opposite to the direction of motion. So it's in that direction. The blue era for Barbie. Um, so basically the same three body diagram replies, except velocity is in the opposite direction of Yorkshire motion. And very crucially, it is decreasing range. So decreasing velocity in the direction opposite motion means that acceleration will be in the direction of motion and so acceleration. It's still to the right, and therefore we have the same ah equations as part of A. Therefore, we still have this. We still have 10 beta equals a Richie. And so this gives us the beta equals inverse tangent of in this case, 4.5 meters per second squared over nine pointy. You know, spare second square. Those cancel and you get 25 degree deflection to the left of the vertical. So this would be 25 please. Okay. And finally in part, see if constant acceleration our sorry of constant velocity, which implies that acceleration is zero and saw beta zero meaning no deflections, and that's it.

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