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A Mercedes-Benz 300 $\mathrm{SL}(m=1700 \mathrm{kg})$ is parked on a road that rises $15^{\circ}$ above the horizontal. What are the magnitudes of $\quad$ the normal force and (b) the static frictional force that the ground exerts on the tires?

$1.61 \times 10^{4} \mathrm{N}$ $4.316 \times 10^{3} \mathrm{N}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

University of Winnipeg

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in this question. There are three forces acting on the car, the weight force, the normal force on the frictional force. We'll begin this question by calculating the value off the normal force. So how can we do that? We do that by the composing, the weight forced into components. One component is these one. Let me call it W. Why, on the other component is that one? Let me call it the Value X. Now we can choose our reference frame as follows, so these accents will be the Y access. And these other axis will be the axe axis. So everything that is pointing the direction it's positive, positive or negative. Then we have flying Newton's second over that. Why access? So over that access we have that the net force is equals to the mass off the car times acceleration on that access look that the car is either going in this direction or in that direction so it doesn't move in this axis. There afford the acceleration on that axis. It's equals 20 and the net force is composed by two forces. The normal force and W. Why one of the components off the weight force and this is equal to zero. Therefore, the normal force is equal to the Y component off the weight force. Now, what is the Y component off the weight force? To figure it out, we have to do a little bit off geometry, so this angle is 15 degrees. If we extend the weight w until it reaches this point, we see erecting will try and go arising. These angle is 90 degrees on. This angle is, as a consequence, 75 degrees because this angle plus this angle must be costing 90 degrees at the same time. Note that this a new here is a right angle. So 75 degrees plus some now angled must be close to 90 degrees. And we know that 75 plus 15 is a close to 90 race. So these angle is an angle off 15 degrees. Now we can calculate the coastline ar 15 degrees using the definition of the co sign. So the co signed off 15 degrees. He's equals to the I just stay inside though you why divider? But they had better news, which is? He goes to W. Therefore w why is equal to W times that co sign off 15 degrees. Finally, the normal force is given by the weight force times the co sign off 15 degrees. But the weight force is it close to the following the mass off the car, times acceleration off gravity and then we have that co sign our 15 degrees. In this question, the mass off the car is a close to 1000 and 700 kilograms on the acceleration of gravity is approximately 9.8 meters per second squared. Therefore, the normal force is equals to 101,000 and 700 times 9.8 times their co sign off 15 degrees. And this gives us approximately 16,000 Newtons. So this is the normal force. Now we have to calculate what is the frictional force to do that we apply Newton's second law on the X axis. So on that axis, then that force is close to the mass times acceleration On that access note that the car is park it Therefore it isn't moving and it's not going to move on. Acceleration on that access is equal to zero true, but the net force is composed by two forces. The weight force component X on a different channel force. These is equal to zero. Therefore, the frictional force is equals to the acts component off the weight force by using a similar reasoning that we had already use it. We can capitalise won his decks component off the weight force we have to do. But to do that you have to calculate the sign off 15 degrees. The sign is he goes to opposite side off the triangle, which is W X, divided by the high party news. Therefore, W X is equals to W's times the sign off 15 degrees. Therefore, the frictional force that is acting is equals to the weight force times this sign off 15 degrees Using the information we already have, we have that the fictional forest is you close to 1000 and 700 times 9.8 times this sign 15 degrees and these gives us a frictional force off approximately 4000 and 300 new tones. So this is deflection off course, and this is the normal force that is acting on our car

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