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On vacation; your 1250-kg car pulls _ 590-kg traller away from stoplight with an acceleration of 1.60 m/s2 You may want to review (Pages 130 - 133)What is the net f...

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On vacation; your 1250-kg car pulls _ 590-kg traller away from stoplight with an acceleration of 1.60 m/s2 You may want to review (Pages 130 - 133)What is the net force exerted by the car on the trailer? Assume that the positive axis is directed toward the direction of motion:Azd880SubmitPrevlous Answers Request AnswerIncorrect; Try Again; 5 attempts remainingPant BWhat force does the trailer exert on the car?Assume that the positive x axis is directed toward the direction of motion_AZd

On vacation; your 1250-kg car pulls _ 590-kg traller away from stoplight with an acceleration of 1.60 m/s2 You may want to review (Pages 130 - 133) What is the net force exerted by the car on the trailer? Assume that the positive axis is directed toward the direction of motion: Azd 880 Submit Prevlous Answers Request Answer Incorrect; Try Again; 5 attempts remaining Pant B What force does the trailer exert on the car? Assume that the positive x axis is directed toward the direction of motion_ AZd



Answers

On vacation, your 1400 -kg car pulls a 560 -kg trailer away from a stoplight with an acceleration of 1.85 $\mathrm{m} / \mathrm{s}^{2}$ . (a) What is the net
force exerted on the trailer? (b) What force does the trailer exert on
the car? (c) What is the net force acting on the car?

This question. We have this situation. A car and trailer is, uh, accelerating it. 1.85 m per second square, Okay. And yeah, actually, parts in the question. You want to calculate a few things? Okay. Such as, uh, that falls on the trailer. The force that the Children exists on the car and that falls out of the car. Okay, so in part, A. Okay, So in this case, we are working with Newton's second law, which says the Nets force is mass times acceleration. Okay, so if you want to find the net force on the trailer Yeah, you think f Net It goes to m A. D. This is the equation for Newton's second law. Internet calls on trailer is equal to mass times acceleration. Mt. So the mass of the trailer is five and 60. The acceleration is 1.5. Okay. Did you get 1050 weapons? Okay. Hey. And then maybe you want to find a force that the charter X on the car he so, uh, force. Uh huh. Sheila, Thanks. On the car is equal to the force that the car X on the trailer. Okay, so this statement is Newton's, uh it is the law. Mhm. In my thing equal here is to say that many tools are equal. Okay. And so, um so the force that the car X on the trailer is the next force on trailer, uh, false that the car X on the shoulder is equal to the next force. Right shoulder. Okay, so this is the answer in part A, which is 10 for zero mutants because that's the only force that the car at the church, Uh, X on the Children. Okay. Okay. Next in proxy, you want to find a net force on the car next for us on the car? Is he going to the mass of the car? Hence the acceleration. Okay, so you have 1400 times, 1.85 and you get 2590 methods. Okay, so this is the answer for Patsy, and that's all for this question.

For this problem. On the topic of Newton's laws of motion, we are told that a 1000 kg car is pulling it 300 kg trailer. We have the acceleration of the car and trailer combination to be 2.15 m per square second, and this acceleration is need positive X direction. If we neglect frictional forces in the trailer, we want to determine firstly the net force in the car. The net force on the trailer, the magnitude and direction of the force that the trailer exerts on the car as well as the resultant force is exerted by the car on the road. Now, if we draw a force diagram for the trailer and the car, we can see that the same tension acts the same tension in the rope that connects the car and trailer acts to the right for the trailer and then to the left for the car. There's a normal force for the trailer and car acting from the road onto the car and into the trailer, as well as the weight of the trailer and car acting vertically downward. Finally, we have a driving force F, which pulls the car and trailer forward, so the result in force acting on the car, we can see acts at an angle theater above the horizontal. Now we choose the X direction to be horizontal positive, X direction to be horizontal to the right and the positive Y direction to be vertically in vertical and upward. So we know the common acceleration is 2.15 m per square second and that's purely horizontal. Now the net force on the car is horizontal and we can find this net force by taking the sum of all the forces along the horizontal direction acting on the car and this is equal to the driving force F minus Attention T and my Newton's second Law. This must equal to the mass of the car times the acceleration in the X direction and so this is 1000 kg times the acceleration of two 0.15 m per square second, which gives us to 0.15 times 10 to the power three Newton's and that me forward direction. So that's the net force acting on the car in this problem. Next in part B, we want to find the net force and the trailer, the net force and the trailer is also horizontal and has given similarly by the some of all the forces in the horizontal direction that act on the trailer. And this is purely just detention, which is positivity in this case and positive T by Newton's second Law must equal to the mass of the trailer times the acceleration a X massive, the trailer we know to be 300 kg and acceleration. Again. It's 2.15 m per square second, and so the resultant force acting on the trailer is 645 Newtons forward for part C. We want to find the magnitude and direction of the force exerted by the trailer from the car. Now, if we consider the free body diagrams of the car and trailer, the only horizontal force acting on the trailer is the tension, which we calculated to be 645 Newtons, and this is exerted on the trailer by the car by Newton's third Law. Then this states that the force the trailer exerts on the car must be 645 Newton's toward the end of the car. So that's the force that the trailer will exert on the rear of the car. Now, lastly, for party, we want to find the resultant force that the car exits on the road. Now the road exerts two forces in the car F and N. C has shown in the free body diagram now from Part A. We can find F to be t last two 0.15 times to enter the power three Newtons and we've calculated the tension t. So this is 645 Newtons. Less too 0.15 times 10 to the Power three Newtons, which gives us the driving force F to be too 0.8 times 10 to the power three Newton's at the same time, the vertical forces acting in the car. If we take the some of these, this is equal to the normal force of the car minus its weight. And by Newton's second law, this must equal to the mass of the car times. It's hard, it's vertical acceleration, which we know is zero. So this means that the normal force acting on the car must equal the weight of the car, which is it's mass times acceleration due to gravity. And so that's 1000 kg times 9.8 m per square second, which gives us 9.8 times 10 to the power three Newton's. So now, using the vector triangle we drew above, we can find the resultant force exerted by the car on the on the car by the road. So we'll call this force are and our car in magnitude is equal to the square root of F squared plus in C squared, the driving force squared, plus the normal four squared. And so this is the square root of two 0.8 times 10 to the power three Newtons squared plus 9.8 times 10 to the power three Newtons squared. If we calculate this, we get the magnitude of the result in force. Acting on the road or on the car by the road is 1.2 James 10 to the power for Newton's. Now we need to calculate the angle at which this forces being applied. So the Angle Theater is the octagon of the normal force divided by the driving force. So this becomes the Oct 10 of three 0.5. This gives us an angle of 74.1 degrees above the horizontal and forward. So they're Newton's third law than states that the results in force exerted on the road by the car is one 0.2 times 10 to the power for Newton's, and this is an angle of 74.1 degrees below the horizontal and real wood.

All right, we've got a car pulling a trailer, so, draw a little car here, there's my car and it's gonna pull a trailer. So, the mass of the car Is 1000 kg. The mess of the trailer Is 300 kg. Its acceleration is that neglecting friction? What's the net force on the car? Well, there's a pushing force on the car and then there's tension pulling backwards. So F minus T. Which would be the net force is going to equal the mass of the car, times acceleration 1000 times 2.15 is 2150 newtons. B. What's in that force on the trailer? Well, that is just gonna be T. So t. is going to equal M. two times A. So, um substituting for a into the first equation, I need salt right first A. Is T over Em too. So, I am one T over. Em too is 21 50. So T. Is going to be 21 50 times M two over M. One. Yeah. Mhm. M two is 300 M one is 1000. So, that's going to give me 645. Yes. Okay, This is question 37. So, I'm gonna make sure that I'm still on the right track and I am okay, see force exerted by the trailer On the car. Well, that's just t. So that's just 645 newtons again. D. Force exerted by the car on the road. So, I think that would have to be F. So f it's gonna be T plus M one A. Which is just t. plus the 21:50. So we just add those two numbers together. 27 B. 28:00 and Okay, let's check the answers. Question 37 21 50. Okay. And it's saying an angle so D. Is incorrect. I'm not sure what's going on there. The force exerted by the car on the road. Well, let's think about this again. We got a car. I ignored MG. And this is M. one. So I had ignored M. One G. But then there's also a force in this direction which when it's applied to the road, it's going to be M. One G. Here and M. Well, F. Here car on the road. Yeah. Because if we draw the road could be a force And there'd be M one G. All right. Not quite sure how I'm going to get the answer in the book, but let's let's just go with it. Um The square root of F. Squared plus M. One G. Squared would be the magnitude of the force on the road. So F. Is T plus 21 50. So I need the square root of F squared plus M. one G. The quantity squared. It looks like I didn't declare M. One yet. M one is 1000. Yeah. Okay. And that does give me the right answer. Um Which is three significant digits one 02 times 10 to the fourth Power news. Okay. Now I move them one gee over here and I draw fi here then Fi is going to be the inverse tangent of M one G, which is in the Y direction over F. So let's get the angle. Okay? You get into calculator 74°. Yeah, but which is correct. Okay. Thank you for watching.

Hi, everyone. Here it is given no a car having the mass. 1000 kg is moving over the origin tal smooth surface with acceleration off 2.15 m per second squared. And it is pulling to the probably off much 300 cages in the first part. We have to calculate force acting on the car force acting on the trolley. See, we apply force applied by that probably on the car, on deeper force applied by the car on the ground. Let us see it probably and courage connected by the rope, which is ideal. So here is the tension teeth. So if we see the free body diagram off the cut, this is the car. Suppose there is the force f. This is attention, T. Then it is moving with the acceleration off 2.15 meter per second square. Let me correct the unit. Just a moment, please. So from here it can. Britain is f minus T net force will be equal to mass into acceleration. So most of the car and to acceleration the F minus T is called to you will get 1000 and tau 2.15 That is 2.5 00 Neutral. Safe question one. No free body diagram off the trolley. This is the early on which the forced ease acting since both are connected together. So they're moving like a single body. So it's acceleration is a attention. T must be equal toe mass off, probably into its acceleration. Massive naturally. 302 2.15 Yeah. So it will be 6. 45. Find 45 nuclear from equation one and two, if you will get Have you will get 27 15 Newton First part force acting on the car Net force acting on the car 2715 minus 6. 45. So it is to be 2150 Newton Yeah. Beat force. Acting on the trolley It is 6. 45 Newton She well force applied by trolley on the car minus 6. 45 Newton, deep forced, played with the car on the ground So two forces will act. One is F that is networks on the cut 2150 Newton and the weight of the car M G 1000 in tow 9.8 98 Jiro Jiro nuclear. So total force applied by the car on the ground will be resultant off it. So this will be the result in foods you may take the symbol f sorry by using parallelogram law. So it is Toby. 191 Newton at angle al fais culture. 10 in worse off 9800 upon 2150 which is 74.8 That's all. Thanks for watching it.


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