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Problem 41 Medium Difficulty

A car is traveling up a hill that is inclined at an angle $\theta$ above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) $\theta=15^{\circ}$ and $(\mathbf{b}) \theta=35^{\circ}$.

Answer

0.97
0.82

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Video Transcript

In general, we have a car that is going up a hue with an Internation given by Tita. So what is the relation between the normal force on the weight force? Well, we have to decompose the wait for us to answer that question. One component off the weight force is pointing these direction and the other component is pointing in that direction. So there are two components. Let me call this W why? And these w x note that this is a directing a triangle because the season right angle so we can work with that rectangle. Try and go as we are used to Let me do a bigger drawing off these rectangle. Try and go right here. Nice Now knows that the normal force is countering the Y component off the weight force. So the normal force is the coast to the y component off that weight force. Now, how can you relate the y component off the weight force with the weight force itself? So it takes a little bit of German tree, so take a look. We can extend these down here and note the following the easiest directing will try and go because this is a right angle on this angle. Let me call it Alfa is such that Alfa pasta is a costume 90 degrees No note something when we some these angle offer. With this angle here we must get a right angle because this is a right angle. So Alfa plus these angle is equals to 90 So this angle must be equals to teach that they're a form. This angle right here is also teeter Then we have some information about our triangle the's angle s data and as a consequence, this other angle is outfit So how can we relate? W why we've w two? We can use the coastline to do that. Why can we use the coastline? Because they co sign is the coast to the address and side off the triangle w why divided by the hypotenuse which is he goes to w then the way you write is equal studio value times they co signed off teeter. Therefore the normal force is equals to the weight forced times the co sign off Tita And finally, the ratio between the magnitude off the normal force and the magnitude off the weight force is the coast to the co sign off Peter. Then on the first item. The ratio between the normal force and the way force is it goes to the co sign off 15 degrees, which is approximately 0.97 on the second item we have that Tita is equals to 35 degrees. The reform. The ratio is a Costa Rico sign off 35 degrees, which is approximately 0.82.