Guys, let's check this one out. We have a 15-kilogram block that's at rest. We're given the coefficient of static friction, and we want to figure out how hard you have to push down on this block to keep it from moving. So basically, let's sketch this out. We have a 15-kilogram block that's on a flat surface. Right? And basically, what we're going to be doing is pushing down. I'm going to call this force Fdown. That's what we're trying to find. And we want to push down hard enough so that we can generate enough friction so that a horizontal force, which we know is 300, cannot get it moving. Alright? So, what we want to do is first start off with a free body diagram. So let's go ahead and do that. So, we've got our box, we've got our mg that's downwards, and then we also have any applied forces. We know there are 2. Our Fdown is what we're trying to solve for. And we also have an applied force that acts to the right. This is F = 300. Now there is also going to be some normal force because you're on the floor. And then finally, what happens is without friction, the box would start moving to the right. But if we're going to keep this thing from moving, that's because there has to be some static friction that is opposing this. So there's some friction here that is opposing that. Right? So that actually goes so that actually brings us to the second step, which we've already just talked about here. We have to determine whether this friction is static or connected based on the problem text or by looking at all the forces involved. And what we've seen here is that we're pushing down on the block to keep this box from moving, basically. So what happens is we know that this friction is going to be static. Alright? So let's go ahead and now write our F = ma. So I'm going to write out my F = ma here. I'm going to just pick a direction of positive up and to the right. Now, usually, we would start with the x-axis, but because we're looking at a force that's in the y-axis, we're going to go and start with our y-axis first. So we've got our sum of all forces equals may. Now this box isn't going up or down in the vertical axis, so we know the acceleration is going to be equal to 0. So, therefore, when we expand our forces remember our normal is positive minus mg minus Fdown and this is equal to 0. Those forces have to cancel. So here's Fdown. And if I go ahead and just solve for this variable over here by bringing it to the other side and flipping the equation around Fdown is really just equal to the normal minus mg. Okay? So this Fdown, the force that I need to get this object to prevent this object from moving is going to be equal to the normal minus mg. But the problem is I actually don't know what this normal force is. So to figure this out, whenever I get stuck in one axis, I usually just go to the other axis to solve. So I'm going to go to the x-axis forces to now solve for this. So I've got F = ma in the x-axis. So what are our forces? We have our F, which is 300 minus your Fs. And what's the acceleration? Well, here's the kicker, guys. If we're trying to figure out how much we need to push on the block to prevent it from moving, we're basically trying to figure out what is the minimum force that we need so that the static friction exactly balances out the 300. And so what happens is this is going to be Fsmax. So the minimum force is going to be where the Fsmax is going to balance out the 300. If 300 were just a little bit more, then it would actually get the object to start moving. So this Fs here is actually maximum static friction. And so because of that, because the object doesn't move, then that means that the acceleration has to be equal to 0 and these forces have to balance. So that means that your F is equal to your Fsmax which is equal to, remember, μstatic times the normal. So remember we came to the x-axis because we were stuck, and we wanted to figure out what that normal force is, now we can figure it out. Our normal force is really just going to be equal to your F divided by μstatic. So your F is 300 Your μstatic is 0.7 You'll get a normal force of 429. So your normal force is 429. Basically, if the normal is 429 then your Fsmax is going to be 300 to balance out the 300 that you're pushing it with. So now we have our normal force here, which means that Fdown is just equal to 429 minus your mg, which is 15 times 9.8. So if you guys go ahead and plug this into your calculators, you're going to get 282 newtons. So if we look at our answer choices, that's answer choice C. Alright, guys. Let's hit this one.
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7. Friction, Inclines, Systems
Static Friction
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