Guys, let's get started here. So we have a 20 kilogram block or box that we're trying to push up a moving truck. We're applying a 110 Newton force up the incline. So, I'm going to draw a quick little sketch of what's going on here. I've got my block like this. Basically, I'm just trying to get it up this ramp by pushing it with 110 newtons. What we want to do is when 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. I've got my free body diagram that's like this. I've got my mg which is going to 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 going to have some friction. But first, before I draw the friction, what I want to do is I want to break up this mg. m g y . And we know this is going to be m g x . So we can just get rid of this mg here.
So, we have this friction force. But what happens is we have these two forces. One of this is pushing up the ramp. One of them is pushing down the ramp. We want to figure out what the friction force is going to be. So we're actually going to stick to the second step here. We have to determine what kind of friction we're going up against and in what direction that is going to push. If you ever have the direction of that friction is not known, what you have to do first is you're going to have to find the net of all the non-friction forces along your axis of motion. Basically, what we're going to do is we're going to 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 block was going to go up the ramp, friction would go down. If the block was going to go down the ramp, then friction would actually point up. So what we have to do is we have to figure out which one of these forces is stronger and then our f is going to be opposite to that.
So basically, what we do is we look at our f force which we know is 110 and we're going to compare this to m g x which we know is sin ( θ ) . And so this is basically 20 times 9.8 times the sine of 15 degrees. So if you go ahead and work this out, you get an m g 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 110. m g 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 direction. That means our friction force is actually going to point down the ramp.
Alright, so now that we know the direction of friction, we still have to figure out what type of friction that is, fs or fk. 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. So basically, what we have 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, your sum of all forces that are non-friction is really just going to be f minus m g x . And we actually know what those are. Right? So this is 110 minus 50.7, and you get 59.3. Basically, what this is, is once you cancel out the 110 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 f s max ? Well, f s max is gonna be μ s static times the normal which our μ s static is 0.3. And then remember our normal force in inclined planes is going to be m g y which is mg times the cosine of theta. So really you just do is 20 times 9.8 times the cosine of 15 degrees. So you end up getting 56.8, I believe. Yes. So you get 56.8. What happens is the forces are net of all the non-friction forces is 59.3. So what happens here our Σ f is stronger than the f s max , so we've overcome the static maximum threshold. And so, therefore, our friction is kinetic friction which is μ k times the normal. So what we have here is our friction, our kinetic force, I'm just going to use 0.2 times 9.8. Oops. Sorry. Times the mass, which is 20 times 9.8 times the cosine times the cosine of 15 degrees, and what you get is you get a kinetic friction force of 37.9. So that's our kinetic friction.
So now what we have to do in order to figure out the acceleration that's what this whole problem is about is now we have to write our f equals ma. So we're going to write out f equals ma 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 force is going to go up the ramp and that's, like, the one that's stronger. Right? So I'm just going to choose that to be positive. So we're going to expand out our forces. We have F then we have minus m g x minus fk is equal to ma. So this is what we're looking for here. So we actually basically know what all of these forces are. We just calculated what this fk is, what this kinetic friction is. It's 37.9. We already know what m g x is because we calculated it earlier, and we also know what our applied force is it's the 110. So we're just going to plug everything in. So this is going to be 110 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 in, you're going to get 21.4 is equal to 20a, and so a is equal to 1.1 meters per second squared.
Now, remember that when we solve for acceleration in these problems, this sign will give us the direction. It's because we got a positive sign here, that just means it points along the direction of motion. So we look at our answer choices, and it has to be answer choice C. So that's it for this one guys. Let me know if you have any questions.