### Video Transcript

A half-full recycling bin has mass 3.0 kilograms and is pushed up a 40.0-degree incline with constant speed under the action of a 26-newton force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at a constant velocity?

Let’s start by highlighting some of the critical information this problem statement gives us. First, we’re told the bin has a mass of 3.0 kilograms and that the incline it’s on is at 40.0 degrees. We’re also told that the force acting on the bin has a magnitude of 26 newtons. We’re told further that the incline has friction. In other words, it’s not a frictionless surface.

And we’re asked what magnitude force must be applied parallel and up the incline so that the bin moves down rather than up the incline at a constant velocity? So let’s call that force that were searching for capital 𝐹. Now as we approach this problem, we make a few assumptions. The first assumption is that the friction between the bin and the incline is constant, so we’ll say that the coefficient of friction that exists between these two objects doesn’t change. We’ll also assume that 𝑔, the acceleration due to gravity, is exactly 9.8 meters per second squared.

Now let’s draw a diagram of the bin on the incline to see how we can start to solve this problem. Here’s a picture of the bin on the incline. The incline is at an angle of 40.0 degrees above the horizontal, and our 3.0-kilogram bin is being pushed up the incline by a force we’ll call 𝐹 sub 𝑝 for 𝐹 push. And we’re told the magnitude that force is 26 newtons.

Now are there any other forces acting on the bin? The answer is yes. First, gravity acts on the bin, and the direction of that force is straight downward. So we’ll draw that in and label that force 𝐹 sub 𝐺.

From previous problems involving objects on inclines, you may recall another force acting on the bin; it’s the normal force. And true to its name, it points perpendicularly away from the surface of the incline. We’ll label that Force 𝐹 sub 𝑁

And in this case, remember that we’re told that the incline and the bin have friction, so there’s a frictional force involved here too. We’re told the bin is moving up the incline which means the frictional force, always in opposition to forward motion must, be pointing down the incline. We’ll call that force 𝐹 sub 𝑓.

So these are the four forces that act on the bin while it’s in constant velocity motion up the incline We want to take a closer look at the components of these forces that are parallel to the surface of the incline. So let’s draw a zoomed-in diagram of the bin on the incline to look more closely at those components.

Now in this expanded-in view of our bin, we look again to see what forces are parallel to the surface of the incline, and we see there are three forces that have components in this direction: first, the pushing force 𝐹 sub 𝑝; second, the frictional force 𝐹 sub 𝑓; and finally, gravity has a component in this direction too.

If we draw dotted lines in to represent the components of the gravitational force that are perpendicular and parallel to the inclined plane, we see that there is this bottom portion which is parallel to the incline. That is a component that we’re interested in to figure out how the forces balance parallel to the face of the incline.

Now before we go further, let’s recall two relevant facts. One is we’re interested in solving for what we’ve called capital 𝐹, which is the magnitude of the force that we would need to use to push on the block so that it can descend down the slope with a constant speed, and the direction of our force, the force 𝐹, is the same as the direction of 𝐹 sub 𝑝, up and parallel to the slope; that’s item number one.

Item number two is a law you may be familiar with, Newton’s second law. The second law tells us that the net force acting on an object is equal to the mass of that object multiplied by its acceleration, and this is true for each direction of motion as long as it’s independent. So in this case, our independent directions of motion are up the incline, like this arrow shows, and normal to the incline, as shown by the second arrow. Now we’ll define both of these directions in our problem to be positive, meaning motion up the incline is positive and motion down the incline is negative.

Now let’s write out Newton’s second law as it applies to motion parallel to the face of the incline. Our forces there are 𝐹 sub 𝑝, which is positive because it acts in the direction we’ve defined as positive, minus 𝐹 sub 𝑓, the frictional force, minus the component of the gravity force that acts down the ramp.

Now before we can write that in, let’s figure out what that is. We’ve been told that our ramp is at an incline of 40.0 degrees. Now it turns out that the triangle created by our incline and the triangle created by 𝐹 sub 𝐺 and its components are similar, meaning that 𝐹 sub 𝐺 is also a right triangle where the angle between the 𝐹 sub 𝐺 vector and the component of 𝐹 sub 𝐺 perpendicular to the incline is also 40.0 degrees.

Now what that means is we can now solve for the component of the gravity force that’s acting down the incline. That component is expressed in the terms of the magnitude of the gravity force 𝐹 sub 𝐺 multiplied by the sine of 40 degrees.

So we now write that term in to our Newton’s second law equation. So all of these three forces combined this way by the second law is equal to the mass of our bin multiplied by its acceleration. And we’re told in the problem statement that acceleration of the block is zero, that it moves at a constant speed, which means that these three forces combined this way equal zero.

Now let’s write this force balance equation out in as much detail as we can. We know that 𝐹 sub 𝑝, the pushing force, is given as 26 newtons. 𝐹 sub 𝑓, the frictional force, you may recall is equal to the coefficient of friction multiplied by the normal force. Because forces balance out in the direction perpendicular to our incline, we know that the normal force is equal in magnitude to the gravitational force multiplied by the cosine of 40.0 degrees.

So we can substitute this expression for the normal force into our equation for the frictional force, which then gives us a frictional force expression of 𝜇, the coefficient of friction, times 𝑚 times 𝑔 times the cosine of 40.0 degrees. When we plug that into our 𝐹 sub 𝑓 in our force balance equation and we replace 𝐹 sub 𝐺 the force of gravity with 𝑚𝑔, we get a fully expanded force balance equation that reads 26 newtons minus 𝜇𝑚𝑔 cosine of 40.0 degrees minus 𝑚 times 𝑔 times the sine of 40.0 degrees equals zero.

Now as we look across this equation, we realize that we know all of the values written down here except for one; we don’t know 𝜇 the coefficient of friction. And if we can solve for 𝜇, that will help us to work back to figure out what is the magnitude of the force of friction that the bin experiences when it moves along this incline.

So let’s rearrange this equation to solve for 𝜇. First, let’s add 𝜇𝑚𝑔 times the cosine of 40.0 degrees to both sides. When we do that, that entire term cancels on the left side of our equation. Now if we divide both sides of equation by 𝑚 times 𝑔 times the cosine of 40.0 degrees, then we see on the right-hand side that the 𝑚, 𝑔, and cosine of 40 degrees terms all cancel out and we’re left with 𝜇, the coefficient of friction, by itself.

Let’s write a cleaned up version of the expression for that coefficient of friction. So now we have an expression that reads 𝜇, the coefficient of friction, is equal to 26 newtons minus 𝑚 times 𝑔 times the sign of 40.0 degrees all divided by 𝑚 times 𝑔 times the cosine of 40.0 degrees. When we plug in 3.0 kilograms for 𝑚 and 9.8 metres per second squared for 𝑔, we find an overall expression for 𝜇 which when we enter into our calculator yields a value of the coefficient of friction of 0.32.

Now that we figured out the coefficient of friction value, let’s work back to solve for the force that is needed when the block instead of sliding up the plane at a constant speed is sliding down the plane. Now let’s revisit for a moment our diagram in the middle of the screen that shows the forces acting on the bin when the bin is moving up the incline. And we’ll even give it a label, we’ll call it “UP” to remind ourselves of the motion of the bin.

Now this designation is important because if we look at the forces acting on the bin in this case, those force directions depend on the fact that the bin is moving up the incline rather than down. Specifically the frictional force always opposes the direction of motion. So in the case when the bin is moving up, our frictional force as we’ve drawn it points down the incline.

But in the opposite scenario, when the bin is moving down the incline, our frictional force, as always opposes, that motion by acting now up the incline. So let’s draw a free body diagram for the case when the bin is moving down rather than up the incline at a constant speed.

Now take a look at this comparative free body diagram for when the bin is moving down the incline rather than up. All the force directions are the same except for the frictional force which now points up the incline.

Also our blue labeled force rather than being 𝐹 sub 𝑝 in the case when the bin is moving up the incline is now simply 𝐹; that’s the force we want to solve for. To do that, we’ll once again refer to Newton’s second law, and we’ll create a force balance equation again in the direction parallel to the incline.

Let’s write the parallel components of the forces involved on a line at the top of our screen. And as we do, that let’s recall that we’ve defined motion up the incline to be in the positive direction. So that means in our force balance equation, we have our force, 𝐹, that we’re trying to solve for plus the frictional force, which is 𝜇 times 𝑚𝑔 times the cosine of 40.0 degrees, minus the component of the gravity force acting down incline, which is 𝑚𝑔 times the sine of 40.0 degrees.

These three forces written this way all add up to zero because again our bin is moving with a constant velocity. In other words, acceleration is zero. Now as we look at this equation, we’re in good shape. We have 𝐹, the force we wanna to solve for, and every other value in this equation is known. We’ve solved for 𝜇. We’ve been given 𝑚 and 𝑔 as a constant, 9.8 meters per second squared. So we now rearrange this equation to solve for 𝐹.

We subtract the cosine term from each side and add the sine term to each side, to isolate 𝐹. We can now plug in for 𝑚, 𝑔, and 𝜇 the values we know; 𝑚 is 3.0 kilograms; 𝑔 is defined as 9.8 meters per second squared; and 𝜇 we solved for is 0.32.

When we plug all these numbers in and calculate these two terms out, we find that 𝐹, the force needed to be applied up the ramp in order for the bin to slide down the ramp at a constant speed, is 12 newtons. That number has two significant figures to match the number of significant figures of the values we were given.

Notice how this result is different from the pushing force of 26 newtons, and that has to do with the fact that the frictional force changes direction depending on whether the bin is moving up or down the incline.