7. Friction, Inclines, Systems

Inclined Planes with Friction

# Intro to Incline Plane with Friction

Patrick Ford

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Hey, guys. So in previous videos, we've seen how to solve inclined plane problems, and we've seen friction problems. So we're really just gonna combine those two concepts together in this video because sometimes you're gonna run to problems where you have objects on rough inclined planes meaning inclined planes with friction. So let's go ahead and just jump right into the problem here. There's really nothing new. So we've got this 10 kg block that's on a ramp at 37 degrees. We have our coefficient of static and kinetic friction, and in part of what we want to do is calculate the friction force that's acting on the block when it's released. So in part of what we want to do is figure out. What is this friction Force F. It could be static or kinetic, so to do that, to start things off, we're gonna need a free body diagrams. Let's get started here. We've got this block, there's going to be an MG that's downwards, right? And then we have no applying forces or tensions, but we are going to have a normal forest that that points perpendicular to the ramps surface, and then finally we have some friction now whether the block actually starts to move down the ramp or is trying to move down the ramp. That friction force wants to oppose that motion, and so it's going to act up the ramp regardless of whether it's static or kinetic. So that's what we're trying to find here. Now remember that when we draw a free body, diagrams for inclined planes were gonna tilt our coordinate system so that the incline so that the X axis points down the ramp like this. So basically, this is a plus X plus y. So when we do that, remember, RMG basically just get split up into its components. So we have M g y. And then we have MG X. All right, so those are MG components here. So now what we have to do is we have to determine what kind of friction we're using, or we do what we have in this problem, whether it's static or kinetic friction. So that brings us Step Number two. We're going to determine whether this is static or kinetic friction, and there's two ways to do this. Since some problems you'll actually be able to tell from the text. It will say that the box isn't yet sliding or it is sliding. You'll know whether it's static or kinetic, but in this problem, it's kind of vague. We're supposed to calculate this friction force, but we don't know whether it's still stationary or starts moving. And so what happens is we're gonna have to compare all the non friction forces. Basically, in this problem here, that's just gonna be rmg X. So we're gonna have to compare those non friction forces along the axis of motion to the maximum static friction. We're basically trying to figure out whether our non friction forces like MG X are strong enough to get this object moving right. So we're basically trying to figure out whether we're dealing with F S or F K. And to do that, we're gonna we're just basically going to calculate MG X and F s Max and see which one is bigger. All right, so when we're doing MG X, remember, that's just mg times the sine of data, and we can calculate that we have all those numbers. This is massive 10 9.8, and then we're using the sign of 37 If you go ahead and work this out, you're going to get 59 Newtons for mg X So we know that this is equal to 59. Now what we do is we just compare this to the maximum static friction. Remember, that's that threshold. So remember that the equation for that is Mu s times the normal. We know what the US is. It's just 0.6 that was given to us in the problem here. So what about this? Normal force will remember on inclined planes if you're if you have friction and mg ex and the X axis and then only end in N g y and the Y axis, those forces have to cancel so and is equal to mg y, which is equal to mg times the coastline of data. So we plug in the normal force. We're really just gonna do 10 times 9.8 times not sign of but co sign of 37. So if you go and work this number out where you're gonna get is Newtons. So what happens here is that our Fs Max is 47. That's the maximum static friction but your MG X is stronger than that. So what that means is that if your MG X overcomes the maximum static friction, then therefore the actual friction basically just becomes kinetic friction. So this F actually is f k. And to calculate this, we're just going to use mu k times the normal and we can figure that out as well. This f k here is just going to be, um UK, which is 0.4 and then times the normal force, which again is 10 times 98 times the co sign of 37. So if you go ahead and work this out, you're gonna get 31.3 Newtons. Alright, so basically we know this is going to be 31.3 mg X is strong enough to get the object moving. That friction becomes kinetic and we have that kinetic friction force. All right, so that's the answer to part A. Now we're doing in Part, B is trying to figure out the blocks acceleration. So now in part B, we want to figure out a X. So now if we're trying to figure out a X now, we're going to get into our F equals m A. So we write all the forces in the X axis equals m a X. And once we have our direction of positive, we really just have our mg X. That's positive, minus our friction. Kinetic is equal to m a X, And this is actually really straightforward because we've already calculated all of these values. So we can do here is we can say the MG X is equal to 59. That's what we calculated over here and then going to subtract the friction force, which we just calculated is 31.3 and this is going to be 10 times a X. So we go ahead and solve this. You're gonna get to 100.77 m per second squared. We get a positive number for our acceleration. And that makes sense. It just means that accelerates down the ramp along the direction of positive. All right, so that's it for this one. Guys, let me know if you have any questions. I'd like to get some practice

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