Guys, welcome back. Let's check this one out. So we're pushing a mini fridge up this incline. I'm going to go ahead and sketch this out like this, and we have a force. Alright. So we've got our mini fridge right here. We have a force that is 120 degrees, but it's not actually perfectly up the incline. We have a force that's basically angled at some angle which we know is 30 degrees above the axis of the incline. So we know this force here is equal to 120. What we want to do is figure out the acceleration of the box. So let's go ahead and get started. We know we're going to have to draw a free body diagram. So let's do that over here. So I've got my box like this, and I've got my mg like that. Now I've got the normal force that points perpendicular to the surface, which means that we split up our mg into mgy and x. So that's our mg force like this. Alright. And then we have an applied force as well. We know this applied force is going to point not up the ramp like this. It's actually going to point at some angle. This is f equals 120, and we know that this angle here is actually equal to 30 degrees. So what that means is that just like we did with mg, we had to separate it into its components. We have to do the same thing with the f. So we separate this into its x, x, and y components like this. Alright. The last thing we have to do is look for friction. We're told that we have some coefficients of static and kinetic friction. Unfortunately, based on the problem, we don't know whether we're pushing this thing up or, you know, which direction that friction force is. So our friction force here is unknown. And just like we did in a previous problem, whenever we have the direction of friction as unknown, we first have to look at all the non-friction forces like fx and mgx and figure out which is stronger. Basically, we have to figure out where this block would move if there weren't any friction. So let's go ahead and do that. Right? So you've got our fx which is really just f times the cosine of 30 degrees. So this is 120 times the cosine of 30 and you're going to get 104. So how does this fx compare to mgx? So our mgx is mg times the sine of the angle. Except we're not going to use 30; we're going to use 20 because that's the angle of the ramp. So there are 2 angles here. Don't get confused there. And so, basically, this is just going to be 30 times 9.8 times the sine of 20 degrees. If you go ahead and work this out you're going to get 100.5. So what happens is our fx, the x component, is 104 that's the component that's going up the ramp, but our mgx is only 100 if we just kind of round it. So what happens is our fx is actually slightly winning. Without friction, this fridge would actually go up the ramp. So what does that mean? That means our friction force actually points down the ramp like this. So this is our fr. So that's the first step. We have that friction now. We have at least we have the direction. Now we move on to the second step of any problem: We're going to determine the type of friction whether we're going up against static or kinetic, and to do that we have to look at all the non-friction forces and see how they compare to the maximum static friction. So what happens is our sum of all forces, all the non-friction forces, is really just going to be fx minus mgx, and we already calculated those right. We just have a 100. That's where a 104 minus a 100.5, and you're going to get 3.5 here. So that is the sum of all forces that are non-friction. So how does this compare to fs max? Well, remember, we have to use the equation times the normal. We have mu static now; we just have to figure out what this normal force is. Normally, pun intended, your normal force is going to be mgy, which is mg cosine theta here. But remember that only works if all of your forces are along the incline. But that doesn't happen here because this force that we're applying isn't perfectly along the incline. It's inclined at some angle. So we have this fy to consider. So we have to go here and look at the sum of all forces in the y-axis which equals mass times acceleration. Now in the y-axis along the incline, that acceleration is 0 because it doesn't go flying off the ramp or anything like that. So we still have normal plus fy minus mgy is equal to 0. So, basically, what happens is we just have another force to consider. Your normal is equal to mgcosine theta minus your fy. Alright. So if we go ahead and expand this, what this really means is that we have, this is 30 times 9.8 times the cosine of 20 degrees minus and then remember this fy here is really just going to be 120 times the sine of the angles, this is going to be a sine of 30 degrees. So, basically, what happens is this becomes 276.3. This becomes 60. And so your normal is really just equal to 216.3. So that's the number that we plug into here. So now our fs max is equal to mu static, which is 0.3, I believe. Yeah. So it's 0.3 times 216.3. We get an fs max that is 64.9 Newtons. Alright. So this is our fs max. Remember the whole reason we did this is because we're trying to figure out whether the non-friction forces are enough to overcome the maximum static. Our non-friction forces only really add up to 3.5. Maximum static friction is almost 65 Newtons. So what happens is because your non-friction forces are less than fs max, then that means that your friction force is equal to static friction, and so therefore, the static friction that's acting on the fridge is really just equal to 3.5 Newtons down the ramp. And because this is static friction, that means that the acceleration of the fridge is equal to 0. Basically, we haven't overcome the maximum static friction. So therefore, that's your answer. The fridge actually doesn't move; the acceleration is 0.
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7. Friction, Inclines, Systems
Inclined Planes with Friction
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