Question Video: Motion on Rough Inclined Planes with Friction Mathematics • Third Year of Secondary School

Video Transcript

A body, starting at rest, slides down a rough plane inclined at an angle of 45 degrees to the horizontal. The coefficient of friction between the body and the plane is three-quarters. Let 𝑡 one be the time required to traverse a certain distance down the slope and 𝑡 two be the time required for the same body to travel the same distance down a smooth slope at the same angle of inclination. Express 𝑡 one in terms of 𝑡 two.

Okay, in this situation, we have a plane that’s inclined at 45 degrees to the horizontal. We then imagine a body on the plane that starts at rest. Maybe we set it on the plane with our hand, then starting to slide downward. This motion is resisted by the roughness of the plane, creating an opposing frictional force. Nonetheless, the body does slide downhill, and after some amount of time, 𝑡 one, it’s moved a distance we’ll call 𝑑. We’re then told to imagine an identical scenario with the same kind of body and the same kind of slope except that as the body slides down this time, it’s not resisted by friction. That is, the slope is smooth, so the coefficient of friction between the body and the plane is zero.

Just like before, this body too moves through a distance 𝑑, but now it takes a time 𝑡 two. This time will be different from 𝑡 one because slope one is rough, while slope two, we can call it, is smooth. Our goal is to develop a mathematical equation relating 𝑡 one to 𝑡 two. As we get started, let’s record the fact that the coefficient of friction, we’ll call it 𝜇, for our first plane is three-quarters. And then clearing some space on screen, we’ll begin by solving for this time 𝑡 one. To figure this out, we’ll need to know how our body moves down this rough slope. We know that, initially, it’s at rest and that over time it picks up speed, so this body must be accelerating as it slides downhill.

To solve for that acceleration, let’s consider the net force that acts on this body sliding down a rough slope. Newton’s second law tells us that this sum of forces is equal to the body’s mass times its acceleration. Say that this is an up-close view of our body as it slides down the rough slope. In terms of the forces acting on it, we know there’s a weight force, 𝑚 times 𝑔, a reaction or normal force, 𝑅. And then also because the slope is rough, there’s a frictional force. Since our body is moving down the incline and friction always opposes the direction of motion, that force, which we’ll label 𝐹, points up the incline.

In this sketch, let’s establish the downward direction along the slope as the positive 𝑥-direction. Likewise, the positive 𝑦-direction will be perpendicularly away from the incline. If we consider the forces acting in the 𝑥-direction then, we see that there is a component of the weight force acting in the positive 𝑥-direction and then a frictional force which acts purely in the negative. Considering the right triangle made by the components of our weight force, we know that this angle here in that triangle is equal to 45 degrees. It always matches the angle of inclination of our plane. Therefore, the 𝑥-direction component of the weight force is 𝑚 times 𝑔 times the sin of 45 degrees. And knowing that the sin of 45 degrees equals the square root of two over two, we can write this overall force as the square root of two over two times 𝑚 times 𝑔.

From this, we subtract the frictional force acting on our body. The reason the frictional force is negative is because it acts up the incline, whereas positive 𝑥 is down the incline. For a moving object experiencing friction, that force is equal to what’s called the coefficient of friction times the reaction force experienced by the object. In the case of our body sliding down this incline, the magnitude of the reaction force is equal to the magnitude of this component of the weight force. That component equals 𝑚 times 𝑔 times the cos of 45 degrees. So, overall, the frictional force is equal to the coefficient of friction 𝜇 times 𝑚 times 𝑔 times the cos of 45 degrees. But then cos of 45 degrees, like sin of 45 degrees, also equals the square root of two over two. And we now have, in simplified form, all of the forces acting on our body on the rough plane in the 𝑥-direction. By Newton’s second law, this sum is equal to the body’s mass times its acceleration in this dimension.

Notice, on the left-hand side, that we can factor out square root of two over two times 𝑚 times 𝑔 and that, on both sides of the equation, we have the mass 𝑚 of our body. Therefore, we can divide both sides by this mass and cancel this term out completely. And lastly, we can substitute in the value we know for 𝜇. It’s given as three-quarters. So the acceleration of our body sliding down the rough incline equals the square root of two over two times 𝑔 times one-quarter, or simply the square root of two over eight 𝑔. In order to distinguish this acceleration from the acceleration of our body as it slides down the smooth incline, let’s call it 𝑎 sub one and record it at the top of our screen.

Knowing this, we move on to our smooth incline, and now we want to calculate the acceleration of this body. We can start doing this by revisiting our free-body diagram. All of these forces still apply except one, the frictional force 𝐹. This force is now gone because our block is smooth. And so if we go back to our original Newton’s second law force balance equation, we can say that now this whole term here is zero because there’s no coefficient of friction between our smooth incline and the body. We simply have then that the square root of two over two times 𝑚 times 𝑔 is equal to 𝑚 times 𝑎 sub 𝑥, the acceleration of the body on the smooth incline in what we’ve called the 𝑥-direction.

Note that once more the mass of our body cancels from this equation. 𝑎 sub 𝑥, in this case, equals the square root of two over two times 𝑔. We’ll record this value as 𝑎 sub two, and note that it’s four times greater than 𝑎 sub one. However, it’s a relationship between 𝑡 one and 𝑡 two that we really want to solve for, so let’s keep going. It’s possible to relate the accelerations we’ve solved for with these times. We do it using what’s called an equation of motion. These are equations that describe object motion whenever they experience uniform or constant acceleration. That’s true for both of these bodies, and so we’re free to use any equation of motion that helps us.

A useful one for our purposes tells us that the displacement of a body is equal to the original velocity of that body times the time elapsed plus one-half the body’s acceleration times that time elapsed squared. If we apply this relationship to the situation with the rough plane, we can say that 𝑑, the distance that our body moves through, is equal to its initial velocity times 𝑡 one plus one-half 𝑎 one times 𝑡 one squared. Since our body started from rest, 𝑣 zero is zero. And so if we rearrange this equation to solve for 𝑡 one, we find it’s equal to the square root of two times 𝑑 over 𝑎 one. If we then substitute in the value for 𝑎 one, square root of two over eight times 𝑔, we find that 𝑡 one equals the square root of 16𝑑 over root two 𝑔 or, recognizing that 16 is four squared, four times the square root of 𝑑 over the square root of two times 𝑔. This then is the value we’ll use for 𝑡 one.

Now let’s apply the same equation of motion, but this time to solve for 𝑡 two. This is equal to the square root of two times 𝑑 over 𝑎 two. And if we then substitute in the square root of two over two times 𝑔 for 𝑎 two, 𝑡 two we find is equal to the square root of four times 𝑑 over root two times 𝑔 or two times the square root of 𝑑 over the square root of two times 𝑔. This is the value we’ll use for the time 𝑡 two. And notice how this time value compares with the time 𝑡 one. Both have this factor of the square root of 𝑑 over the square root of two times 𝑔. But in 𝑡 one, that factor is multiplied by four, while in 𝑡 two, it’s multiplied only by two.

We can say then that 𝑡 one is twice as big as 𝑡 two or 𝑡 one equals two times 𝑡 two. This is our answer. And now let’s check if it makes physical sense that 𝑡 one is greater than 𝑡 two. 𝑡 one is the time it takes our body to slide across a distance 𝑑 on a rough surface. 𝑡 two is the time required to slide across that same distance on a smooth surface. We would expect then that 𝑡 one is greater than 𝑡 two because the body in that case is opposed by friction. So our finding that 𝑡 two is less than 𝑡 one does make physical sense and gives us greater confidence in our result.