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The basic concepts in this problem are presented in Multiple-Concept Example 9. A $225-\mathrm{kg}$ crate rests on a surface that is inclined above the horizontal at an angle of $20.0^{\circ} .$ A horizontal force (magnitude $=$ 535 $\mathrm{N}$ and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?

$\mu_s = 0.666$

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in this question to have to use neutral second low. And for that I'm choosing the falling reference frame, these accents that I will call the Why access on these accidents that I will call the X access. Then, using this reference free, we can apply Newton's second law, the box, then my doing that we get the following for the y axis. There are three forces acting the normal force, a component off the weight force and hours every component off the force f. It's easier to see if we draw the components. The vertical component off the weight force is these one. But were you white? By the way, it also has a horizontal component. W X The vertical component off the force f is this one Afghan foreign twice and these is the component X off the force f. Okay, X. I was so already notice that this angle is opposite to this angle reform. Both angles are 20 degrees. Then now going back to neutral second law. We see that in that force is equal to the mass off the great times, its acceleration since its resting the acceleration is close to zero. So the net Force Zero, but the net force in the Y direction is composed by in normal force. That points to the positive y direction plus the force after why component that points in the positive direction minus the weight forced. Why component that points the negative y direction on these easy cost zero. With that, we see that the the normal force is given by the weight forced by component minus the force asked component Why now? We have to relate the white components with the magnitudes off the factors so front of actor f We have the following disease magnitude and then we have these component on these component together. The form directing will try and go where one side is the Y component and another side the X component. This unloads and angle off 20 degrees. Then we can capitalise the y component by using the sign off 20. So this sign off 20 degrees he's given by the opposite side. F component. Why divided by the hype continues F then f component. Why is equals two f times This sign off 20 degrees for the X component we have that the co sign off 20 degrees is equals two f component X divided by f But before the X component off F is given by f times the co sign off 20 degrees No, For the wait for us, we can do with something similar. But before we have to discover what is this angle for that we can do the following treek so we can extend the weight until it meets this point. Then this forms a rectangle triangle. Then these angles 20. This angle must be something that when we add up with 20 composed 90 before it must be an angle off 70 degrees. So this is 70 degrees at the same time. This angle is the right under 70 degrees plus these intern uncle, most of the 90 meaning that this imprint under is also 20 degrees. Then, using these information, we can draw the try and go for the wait, which is as follows we have you side on the side. On that site is the high part two news. This is the white components. Do these Dax component here we have a 20 degree angle. Then using the sign off 20 we calculate the X component. So the sign off plenty is given by the ex components divided by the hypotenuse. Perform the weight component X is given by the wait times. This sign off 20 degrees. By using the pro sign, we can calculate the white component. The code sign is the address and site divided by the hypotenuse. Therefore, wait. Why component is given by W times. They co sign off 20 degrees. Now we can go back to the normal force using this information. The normal force is weight. Climb school sign off 20 degrees minus f times the sign off 20 degrees. And then we can remember that the weight he's given by mass times acceleration of gravity. So the normal force is given by 225 times 9.8 times Niko sign off 20 degrees. Andi f is 535 times this sign off point degrees. This results in a normal farmers off approximately 1889.0 for one new terms. Now we keep this value and proceed to apply Newton's second Law on the X Direction. For that, let me organize my board. Okay? Now, applying to the X direction, we get that the net force is equals to mass times acceleration off the great in the X direction. The crate is at rest. It's acceleration is again zero. So I noticed that the Net Force and X direction is composed by 123 forces. Again, the frictional Force and the ex components off W and F. The fictional forced points in the negative direction, but the orders points to the positive. So nobody acts plus F acts. My new special force is equal to zero then the frictional force is equals to the weight X component plus f x component. We can complete this components using the equations we have so little you acts is the value times the sign off 20 degrees plus F x, which is f times the go sign off 20 degrees. Now we can plug in the values That problem gives what you got. 225 times 9.8 times this sign off. 20 plus 535 times the co sign off 20. These results in the frictional force off approximately 1256 point 890 new terms approximately no new in the frictional force and the normal force and can calculate the static frictional coefficient because you know that the friction is given by the static friction, local official times, the normal force. So the static frictional coefficient is the frictional forwards divided by the normal force, which is 1256 0.890 divided by 1889.0 for one. These results in a static vision, or fusion, off approximately 0.666 and these is the answer to this problem.

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