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Force Not Parallel To Ramp

Patrick Ford
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Hey, guys, Welcome back. Let's check this one out. So we're pushing a mini fridge up this incline and go ahead and sketch this out like this and we have a force. All right, 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 inclines. We know this F here is equal to 1 20. What we want to do is we want to 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 RMG into M g y N X. So that's our M G force like this. All right. And then we have an applied force as well. We know this applied force is actually gonna point not up the ramp like this. It's actually gonna point at some angle. This is f equals 1 20. 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 into its components. We have to do the same thing with the F. So we separate this into its f x, X and Y components like this. All right, The last thing we have to do is look for friction. We're told that we have some coefficient of static and kinetic friction. Unfortunately, based on the problem, we actually don't know whether we're pushing this thing up or whether basically, like, we don't know the direction of that friction force. So our friction force here is unknown. And just like we did in a previous problem, whenever we have the direction of friction is unknown. We first have to do is we have to look at all the non friction forces like FX and MGM X and figure out which is stronger. Basically, we have to do is figure out where would this block move If there weren't any friction. So let's go ahead and do that right. So you got our FX, which is really just f times the co sign of 30 degrees. So this is 1 20 times the co sign of 30 and you're going to get 104 So how does this FX compared to mg X So rmg X's mg times the sine of the angle Except we're not gonna use 30. We're going to use 20 because that's the angle of the ramp. So there's two angles here. Don't get confused there. And so basically, this is just gonna be 30 times 9.8 times the sine of 20 degrees. If you go ahead and work this out, you're gonna get 100.5. So what happens is our f X. The X component is one of four. That's the component that's going up the ramp. But our MG X 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 f. So that's the first step. We have that friction now. At least we have the direction. Now we move onto the second step of any problem. Basically, 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 some of all forces, all the non friction forces is really just gonna be FX minus mg X. And we already calculated those right? We just have 100 100 and four minus 100.5 and you're gonna get 3.5 here. So that is the sum of all forces that are non friction. So how does this compare to F. S? Max will remember, Um, we have to use the equation mu static times, the normal we have metastatic. Now we just have to figure out what this normal forces normally pun intended. Your normal force is going to be m g y, which is mg cosign theta here. But remember that only works if all of your forces are along the incline. But that actually doesn't happen here because this force that we're applying actually isn't perfectly along the incline. It's inclined at some angle. So we have this f y to consider. So we have to do first is 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 zero because it doesn't go flying off the ramp or anything like that. So we still have normal plus f y minus mg Y is equal to zero. So basically, what happens is we just have another force to consider. Your normal is equal to mg cosign theta minus your F y. All right. So if we go ahead and expand this what this really means is that we have, uh this is 30 times 9.8 times the co sign of 20 degrees minus. And then remember this f y here is really just gonna be times the sine of the angles. This is going to be a sign of 30 degrees. So basically what happens if this becomes to 76.3. 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 F. S Max is equal to use static, which is 0.3, I believe. Yeah. So it's 0.3 times to 16.3. You get an F S. Max, that is 60. Let's see, 64 0.9 Newtons. All right, 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 are non friction. Forces only really add up to 3.5 Maximum static friction is almost 65 Nunes. So what happens is because your non friction forces are less than F. S. Max. Then that means that your F 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. Newtons down the ramp. And because this is static friction, that means that the acceleration of the fridge is equal to zero. Basically, we haven't overcome the maximum static friction. And so therefore, that's your answer. The fridge actually doesn't move the acceleration zero.