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7. Friction, Inclines, Systems

# Inclined Planes with Friction

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concept

## Intro to Incline Plane with Friction 5m
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example

## Pushing A Box Up A Ramp 6m
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Hey, guys, let's get started here. So we have a 20 kg block or box that we're trying to push up a moving truck. We're applying it to 110 Newton Force up the incline. So I'm gonna draw a quick little schedule. What's going on here? So I got my block like this. I know this is 20. Basically, I'm just trying to get it up this ramp by pushing it with 110 Newtons. What we want to do is what we're given the coefficients of friction. I want to figure out the magnitude and direction of the acceleration of the box. So let's get started. We know we're going to have to draw a free body diagram and then tilt our coordinate system. So let's go ahead and do that. So basically, I've got my free body diagram that's like this. I've got my mg, which is gonna act downwards, and then I've got my normal force, which acts up like this or sorry, perpendicular to the surface. I've got my applied force that pushes this way, and so we know we're gonna have some friction. But first, before I draw the friction What I want to do is I want to break up this mg. So this is going to be M g y. And we know this is going to be MDX, so we can just get rid of this mg here. So we have this this friction force. But what happens is we have these two forces one of this pushing up the ramp. One of them is pushing down the ramp and we want to figure out what the friction force is going to be. So we're actually going to the second step here. We're gonna go to the second step. We have to determine what kind of friction we're going up against and in what direction is going to push. So here's the deal. If you have ever have the direction of that, friction is not known, right? If this is ever unknown, what you have to do first is you're gonna have to find the net of all the non friction forces along your axis of motion. Basically, what we're gonna do is we're gonna look at the non friction forces F and M G X and figure out which one of them is stronger. Remember that friction always wants to oppose the direction that the object would move if there were no friction, meaning if the if the block was going to go up the ramp, friction would go down. If the block was going to go down the ramp, then friction actually point up. So what we have to do is we have to figure out which one of these stronger which one of these forces are stronger, and then our f is going to be opposite to that. So let's get so basically we do is we look at our F force, which we know is 110 and we're gonna compare this to MG X, which we know is MG times the sine of theta. And so this is basically 15 times 9.8. I'm sorry about 15 is 20. So it's 20 times 98 times the sine of 50. Sorry, that's 15 degrees. So if you go ahead and work this out, you get an MG X that is equal to 50.7. So if you take a look here we have our force that points up the ramp. That's 1 10 mg X is equal to 50.7. So basically we're pushing up the ramp stronger than gravity is pulling it down. What that means is that f is going to be opposite to our that direction. So that means our friction forces actually going to point down the ramp. All right, so now that we know the direction of friction, we still have to figure out what type of friction that is F s or F k. And if we don't know from the problem text, we're going to have to look at the sum of all forces along the axis of motion and figure out whether it's enough to overcome the static maximum threshold. And if it does, then our friction turns to kinetic. So basically, we're going to do here is we have to look at the sum of all forces on the axis of motion. That's basically our F and M g X. So you're some of all forces that are non friction is really just gonna be f minus mg X, and we actually know what those are, right? So this is 1 10 minus 50.7 and you get 59.3. Basically, what witnesses is once you cancel out the 1 10 this way and the 50.7, It's as if you had one force that was going up. That was 59.3. How does this compare to Fs Max? Well, F s Max is going to be mu static times the normal. Which arm you static is 0.3 and then remember, our normal force on inclined planes is going to be M G y, which is mg times the cosine of data. So really, we're just going to do is 20 times 9. times the coastline of 15 degrees. So you end up getting 56 points. Uh, let's see, this is 56.8, I believe. Yeah, so you get 56.8. So what happens is the our forces are net of all the non friction forces is 59.3. So what happens is here are Sigma F is stronger than the F S. Max. So we've overcome the static Max, the static Max threshold. And so therefore, our friction is kinetic friction, which is mu k times the normal. So we have here is our friction are kinetic friction. Here is the force that is pushing down the ramp, Um, the actually getting the box to start moving. So to calculate this kinetic friction force, I'm just gonna use 0.2 times 9.8. Sorry. Uh, times the mass, which is 20 times 98 times the co sign times the coastline of 15 degrees and what you get is you get a kinetic friction force of 37 points. Uh, nine. So that's our kinetic friction. So now what we have to do in order to figure out the acceleration, that's all this whole problem is about is now we have to write our r f equals M A. So we're gonna write out f equals m A along the axis of motion. Now we just pick a direction of positive. So I'm just going to choose the direction up the ramp to be positive, because we know that basically our forces going to go up the ramp, and that's like the one that's stronger, right? So I'm just gonna choose that to be positive. So we're gonna expand our forces. We have f Then we have minus mg X minus. F K is equal to m a So this is what we're looking for here. Okay, so we actually basically know what all of these forces are. We just calculated what this f k is. This kinetic friction is 37.9. We already know what Mg X is because we calculated earlier and we also know what are applied forces. It's the 1 10, So we're just gonna plug everything in. So this is gonna be 1 10 minus 50.7 minus 37.9 is equal to the mass, which is 20 times a. So if you go ahead and plug all this and you're gonna get 21.4 is equal to 28 so a is equal to 1.1 meters per second square. Remember that when we saw for acceleration in these problems, this sign will give us the direction because we got a positive sign here. That just means it points a longer direction of motion. So we look at our answer choices and it has to be answered. Choice C. So that's it for this one. Guys, let me know if you have any questions.
3
example

## Block Launched Up Ramp 7m
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example

## Force Not Parallel To Ramp 6m
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concept

## Incline with Friction & Critical Angle 8m
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example

## Critical Angle 1m
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