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(I) The coefficient of static friction between hard rubberand normal street pavement is about $0.90 .$ On how steep ahill (maximum angle) can you leave a car parked?

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$42 \degree $

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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the system. We see that the angle, let's say the angle that, um the pain the hill is making is data. Now, if we draw the free body diagram on the car, let's let's actually choose the access first. Now, in this kind of problem, we can choose our axes occurring to our convenience. So let's actually choose this way or, uh, the along the Steve off the hill as X axis on perpendicular to that as y axis. And if we do so, we see that first off, all md is vertically downwards. Let's call it mg. Then if we if we draw a line parallel to y axis, we see that the angle that engine makes with that parallel line toe. Why access is angle, Seda. Also, the normal force should be parallel to why access on Let's call it F. So since the car is parked so it will have a tendency to go downwards. So that means the frictional force is acting this way on. Let's call it forced you do friction. Also, the components off mg will be one along the X direction one along. Sorry, One of the wind Direction one along the x direction. So in the extraction, it's mg sign Leda. And in the wide direction it's empty call sign data. So now the free body diagram is come complete So two. So if if we want to stabilize the car if the car wants to be parked here, then ah, this force, the component of MG should be equal. Do deep friction force. It should not be more than that. If this force is dominant over this than, uh, the car will start a slight. So first, let's solve the question in wide relation. So in why directions? As we can see, we have only two forces. And since the car has no movement in the wind direction, we have f n which is equal to M g co signed data then in the extraction we have again. If we want the car tow be stable, will have the total force in white extraction is zero. So for that case MG science data minus if off, if our should be zero or in other words, mg signed data should be equal to efforts are now from the definition of frictional force. We know that f off f R must be equal to mu s or the coefficient of static friction times F n and ah, we can use this relation right here and blood that over here to find out fr so if we do so, we see that MG science data is equal. Do mu s times mg cause I ended up from here, We can see that MG cancels out and we have new s which is equal to sign fade over causing data which is tan theta From there it was all for theater. It's 10 in verse. I'm us and mule. Ah. So, um, the coefficient off static friction us is given, which is 0.9. And if we use that number over here, you see that 10 inverse 0.9 is equal to 42 degrees. Thank you.

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