# Question Video: Determining the Forces Acting on an Object on an Inclined Plane and Its Acceleration

A block of mass 2.0 kg is on a perfectly smooth ramp that makes an angle of 30° below the horizontal. What is the magnitude of the block’s acceleration down the ramp? What is the magnitude of the force of the ramp on the block? What magnitude force applied upward along and parallel to the ramp would allow the block to move with constant velocity?

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### Video Transcript

A block of mass 2.0 kilograms is on a perfectly smooth ramp that makes an angle of 30 degrees below the horizontal. What is the magnitude of the block’s acceleration down the ramp? What is the magnitude of the force of the ramp on the block? What magnitude force applied upward along and parallel to the ramp would allow the block to move with constant velocity?

Let’s highlight some key information given in the statement. We’re told the mass of the block is 2.0 kilograms and that it rests on a ramp angled at 30 degrees below the horizontal. Now let’s draw a picture or a diagram of the situation.

Alright so here we have our block. We’ve labelled it 𝑚 for mass, and we show it sitting on this inclined plane at an angle of 30 degrees below the horizontal, and we know that the mass of our block is given as 2.0 kilograms. So the first question we wanna answer is what’s the magnitude of the block’s acceleration down the ramp.

To figure this out, let’s begin by drawing the forces that are acting on the block. Well, we know one force for sure is acting on the block; that’s gravity, and that points straight down. We’ll draw that in our diagram, and we know there’s one other force that acts on the block. Now the block isn’t moving into the plane. That means the plane is resisting the block in some way, and it’s resisting through the normal force, which acts as you might remember perpendicular to the surface between the block and the ramp.

So let’s draw that in, and we call that normal force 𝐹 sub 𝑁 to complement the gravity force 𝐹 sub 𝐺. Now you might wonder if there’s a frictional force. But remember in the problem statement, we’re told that the ramp is perfectly smooth, and that’s another way of saying that we can neglect friction in this case.

Now since there’s no friction, you may realize that the block wouldn’t sit still as we’ve drawn it. Rather it would start to slide down the ramp due to the forces acting on it, and indeed we wanna find out what is its acceleration as it does slide down the ramp. To figure that out, we need to find out the magnitude of the force that acts parallel downward in the direction of the ramp.

Let’s start that process by drawing in a coordinate plane to our scenario, but it’ll be a tilted coordinate plane in line with the normal force in the slope of the ramp rather than purely horizontal and vertical as usual, so let’s draw that in now. Alright so now we have our new coordinate plane; we have a 𝑦-axis, instead of moving vertically, it moves normal to our ramp and an 𝑥-axis that moves parallel to our ramp.

And if we can figure out the sum of the forces that act in the positive 𝑥-direction, we can use that information along with the mass of our block to solve for the block’s acceleration down the ramp. So what forces act in the positive 𝑥-direction? Well, as we look at the force of gravity, let’s break it into components and see that one of its components indeed acts that way.

Now as we look at the components of the gravity force, we see that one component, we can call it 𝐹 sub 𝑥, acts in the positive 𝑥-direction. That is what gives our block its acceleration down the ramp. If we can find the magnitude of 𝐹 sub 𝑥, then we’ll be well on our way. Now at this point, take another look at the triangle we’ve drawn to represent the ramp. We see that one corner is at 30 degrees, the other is a right angle, meaning that we have a 30-60-90 triangle.

Now this triangle is similar, believe it or not, to the triangle we’ve drawn that represents the force of gravity in the components of that force. So we can actually draw in in a corner of this triangle representing the components of the gravity force an angle here which is 30 degrees, the same as the angle that our slope is above the normal. The reason that’s so useful is because that angle and the magnitude of the gravitational force acting on the block will let us solve for 𝐹 sub 𝑥, the force pushing the block down the ramp.

So let’s do that now. Let’s figure out what is the magnitude of 𝐹 sub 𝑥. As we look at that triangle, we see that 𝐹 sub 𝑥 is equal to 𝐹 sub 𝐺, the force of gravity, multiplied by the sine of 30 degrees, just like that. Now 𝐹 sub 𝐺, the force of gravity, can itself be rewritten as the mass of the block times lowercase 𝑔, the acceleration due to Earth’s gravity. So let’s rewrite 𝐹 sub 𝑥 in those terms.

Now at this point, let’s recall Newton’s second law, and remember that the second law states that the force on an object equals its mass times its acceleration. Now that is helpful to us because it means we can extend this equation for 𝐹 sub 𝑥 and write in that not only does it equal 𝑚𝑔 sine of 30 degrees, but it’s also equal to the mass of the block times its acceleration in the 𝑥-direction.

Now that is exactly what we’re trying to solve for, that acceleration. Looking at this equation, we see that the masses cancel out, so this whole result is independent of the mass of the block, and that acceleration the block experiences down the plane is equal to 𝑔, 9.8 meters per second squared, times the sine of 30 degrees. When we calculate that, we find that it is equal to 4.9 meters per second squared. That’s the acceleration of the block as it slides down the plane.

Now our next question is what is the magnitude of the force of the ramp on the block. In other words, what is the magnitude of the normal force? Let’s clean up the right side of our screen and look again at this diagram we’ve drawn. Okay, as we look at the diagram, we see that, yes, there is a normal force of the ramp on the block and that this force balances out and opposes the gravitational force on the block that moves in our negative 𝑦-direction. So the block is not moving in the 𝑦-dimension as we’ve drawn it, which means that the forces there balance out.

We can write that as 𝐹 sub 𝑁, the normal force, is equal to 𝐹 sub 𝑦, the 𝑦-component of the gravity force. Now the 𝑦-component of the gravity force, just like the 𝑥-component, can be rewritten in terms of the magnitude of the force of gravity on the block times a trigonometric function of 30 degrees. Now in this case, we won’t use the sine of 30, but we’ll use the cosine of 30 because we’re seeking the magnitude of the adjacent leg of this triangle. So all that to say, we can rewrite 𝐹 sub 𝑦 as 𝐹 sub 𝐺 times the cosine of 30 degrees, and again this can be rewritten as the mass of the block times gravity times the cosine of 30 degrees.

Now looking at this most recent expression, we realize we know all these values. We know 𝑚; we know 𝑔; and we know or can calculate the cosine of 30 degrees. When we plug in for those numbers, we find that 2.0 kilograms multiplied by 9.8 meters per second squared times the cosine of 30 degrees is equal to 17 newtons. That is the magnitude of the normal force of the ramp as it acts on the block. Now we’re almost finished, but we just have one more thing we want to find out. We want to solve for the magnitude of force which if it was applied upward along and parallel to the ramp would create a constant velocity motion for the block.

Right now we know the block is accelerating down the ramp, so the question is how hard would we have to push back on the block in order for it to move with a constant velocity. Now to figure that out, let’s return once more to our diagram. Now we can see that we do have a net force acting on the block down the ramp, and that’s represented by 𝐹 sub 𝑥, so the question is asking how much of a force would we need to apply in the opposite direction of 𝐹 sub 𝑥 in order to balance out those two forces. In other words, what hypothetical force — we could call it 𝐹 sub 𝑜 for the opposition force — what hypothetical force would we need to apply to balance out 𝐹 sub 𝑥 so that there is no acceleration of the block in the 𝑥-direction?

Well our diagram is helpful here. It shows us that that opposition force, 𝐹 sub 𝑜, would need to be equal in magnitude and opposite in direction to 𝐹 sub 𝑥. Now our question asks for the magnitude. So if we can solve for the magnitude of 𝐹 sub 𝑥, then we’ll know the magnitude of 𝐹 sub 𝑜. And indeed, we’ve done that earlier! We found that the magnitude of 𝐹 sub 𝑥 is equal to the block’s mass times gravity times the sine of 30 degrees. And what we’re saying is that this expression is equal to 𝐹 sub 𝑜, the opposition force we’re looking for.

So just like before, if we plug-in 2.0 kilograms for 𝑚, 9.8 meters per second squared for 𝑔, and sine of 30 on our calculator, we can solve for the opposition force. And we find that that is equal to 9.8 newtons. That’s the amount of force we would need to apply to the block upward and parallel to the ramp so that it would move at a constant velocity.