Video: Determining the Average Drag Force on an Object Falling through a Fluid

A ball bearing of mass 150 g is dropped into a tube filled with oil. The ball bearing hits the oil’s surface with a vertically downward velocity of 1.5 m/s and after falling through the oil for 1.5 s, the ball bearing’s velocity becomes 6.3 m/s. What vertically upward average drag force does the oil apply to the ball bearing?

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Video Transcript

A ball bearing of mass 150 grams is dropped into a tube filled with oil. The ball bearing hits the oil’s surface with a vertically downward velocity of 1.5 meters per second. And after falling through the oil for 1.5 seconds, the ball bearing’s velocity becomes 6.3 meters per second. What vertically upward average drag force does the oil apply to the ball bearing?

Okay, so there’s lots of information in this question. So let’s underline all the important bits so we don’t miss anything out. So firstly, we know that we’ve got a ball bearing which has a mass of 150 grams. And we know that it’s dropped into a tube filled with oil. We are also told that the ball bearing hits the oil surface with a vertically downward velocity of 1.5 meters per second. And after falling through the oil for 1.5 seconds, the ball bearing now has a velocity of 6.3 meters per second.

We’ve been asked to calculate the vertically upward average drag force applied by the oil to the ball bearing. Now let’s write down all this important information in a very succinct way. Firstly, we know the mass of the ball bearing, which we’ll call 𝑚. And it’s 150 grams. Secondly, we know the velocity with which the ball bearing hits the oil surface. We’ll call this 𝑢. And we know that it’s 1.5 meters per second. Thirdly, we know that the ball bearing continues to fall in the oil for a time, which we’ll call 𝑡, of 1.5 seconds.

We also know that after this ball has been falling for 1.5 seconds in the oil, its velocity becomes 6.3 meters per second. And we’ll call this 𝑣. We’ve been asked to work out 𝐹 sub drag, which is what we’ll call the vertically upward drag force. We could also draw a little diagram to help us out with this question. So here’s our tube. Here’s the oil. And here’s the ball bearing when it first hits the surface of the oil. We know that it has a mass 𝑚. And its initial velocity is what we’ll call 𝑢.

It then continues to travel for what we’ll call a time 𝑡. And at this point, it has a velocity which we’ve called 𝑣. Now at this point, we’ve basically sucked the question dry from any useful information. So we can get rid of it. Ahh! That’s so much better! Look how much space we have to work with now! Okay, so first things first. Let’s notice that the ball bearing as it falls into the tube of oil, it’s changing velocity. Initially, it had a velocity 𝑢, and later on it has a velocity 𝑣.

That means that the ball bearing is accelerating. And we can calculate what this acceleration is. An acceleration, which we’ll call 𝑎, is defined as Δ𝑣, the change in velocity, divided by the time taken for this change in velocity to occur, 𝑡. Now we can say that 𝑎, therefore, is equal to, well the change in velocity is simply the final velocity, which is 𝑣, minus the initial velocity, which is 𝑢, and then we divide this by the time taken for this velocity to change, which is 𝑡.

Now we won’t plug in the numbers just yet because we want to be working in symbols as long as possible so that our mathematics doesn’t get too messy. Anyway, so we’ve got a formula that gives us the acceleration of the ball bearing through the oil. Why is that useful? Well if we know the acceleration of the ball bearing, then we can work out the resultant force on the ball bearing using what’s known as Newton’s second law. Newton’s second law tells us that the resultant force on an object, 𝐹, is equal to the mass of the object, 𝑚, multiplied by the acceleration of the object, 𝑎.

Now we’ve just worked out what 𝑎 is right here. And we know what the mass of the object is, 𝑚. So we can work out what the resultant force on the object is. In this case, of course, our object is the ball bearing. So we say that the resultant force is equal to 𝑚𝑎. And we can substitute in what 𝑎 is, which leaves us with 𝐹 is equal to 𝑚 multiplied by 𝑣 minus 𝑢 over 𝑡. So we’re a step closer to finding out the value of 𝐹 sub drag.

However, what we found here on the right is simply the resultant force. This is the net force on the ball bearing. If we zoom in a little bit to the ball bearing, then we can consider the forces acting on this ball bearing as it accelerates through the oil. Well, one of the forces naturally must be the weight of the ball bearing, which we’ll call 𝑊. And we can recall that the weight of an object 𝑊 is given by multiplying the mass of the object by 𝑔, which is the gravitational field strength of the Earth. So one other force is acting on the ball bearing is the weight, 𝑊, which we’ll say is equal to 𝑚𝑔.

Now the other force that must be acting on the ball bearing is the drag force. And this force opposes the motion of the ball bearing. So we know that the ball bearing is travelling in this direction downwards. And the drag force will resist this motion. So the drag force will be acting in the vertically upward direction. We’ll call this 𝐹 sub drag. And this is what we’re trying to find out. Now it’s important to note that the question asked us just to find the average drag force, which means that we can technically assume here in this question that the drag force on the ball bearing stays constant, 𝐹 sub drag.

And we can do this because we’ve also assumed that the acceleration 𝑎 of the ball bearing is constant. Now this is not technically true. The acceleration changes because the drag force on the ball bearing changes as well as the velocity of the ball bearing increases. But to simplify our scenario, we’ve just assumed that the acceleration is constant and 𝐹 sub drag is constant. And this is the average drag force. We also know that this is the average drag force because we’ve got values of the velocity of the ball bearing here and here and, like we said earlier, we just assumed a constant acceleration between these two points.

Anyway this just serves to simplify our problem. Now let’s go back to looking at the two forces on the ball bearing, the weight and the drag force. Well these two forces are basically competing against each other. One is acting downwards, which is the weight, and one is acting upwards, which is the drag force. And so if we combine these forces, then we can find out what the resultant force is. We can say that the resultant force, which we know is what we’ve called 𝐹, because that’s what we calculated using Newton’s second law. That’s 𝐹 here and 𝐹 here.

And we can say that this 𝐹 is equal to the resultant of the two forces 𝑊 and 𝐹 sub drag. However, 𝑊 acts in the downward direction. So we’ll call this positive 𝑊 and 𝐹 sub drag acts in an upward direction. So we subtract it. In other words, we’ve assumed that any force acting downwards is positive and any force acting upwards is negative. Also worth noting therefore that we’ve assumed that 𝑊 and 𝐹 sub drag are simply the magnitudes or the sizes of the forces. In other words, we haven’t accounted for the fact that they’re working in a certain direction.

But it’s fine because we have accounted for that with the arrows that we’ve drawn in the diagram. Anyway, so coming back to this expression here for 𝐹, we can rearrange to find out what 𝐹 sub drag is. To do this, we can first add 𝐹 sub drag to both sides of the equation, which means that 𝐹 sub drag cancels on the right-hand side and leaves us with 𝐹 sub drag plus 𝐹 is equal to 𝑊, the weight. Then we can subtract 𝐹 from both sides, which means that 𝐹 cancels from the left-hand side of the equation. And hence, we’re left with 𝐹 sub drag, the average drag force on the ball bearing, is equal to 𝑊, the weight of the ball bearing, minus 𝐹, the total resultant force on the ball bearing.

At this point, we can substitute in all the expressions that we’ve already got. 𝐹 sub drag is equal to 𝑚𝑔, which is the weight, minus 𝑚 multiplied by 𝑣 minus 𝑢 over 𝑡, which is the resultant force. At this point we know all of the values on the right-hand side of the equation. We know 𝑚. We know 𝑔. We know 𝑚 again. We know what 𝑣 is. We know what 𝑢 is. And we know what 𝑡 is. So we could happily plug in all the values and find out what 𝐹 sub drag is. But before we do that, let’s notice something about the value of 𝑚.

This value is given in grams. Now this is not ideal for us. We want to convert this to the standard unit of mass, which is the kilogram. So in order to do that, we must recall that one kilogram is equal to 1000 grams. So if we divide both sides of that equation by 1000, we get that one thousandth of a kilogram is equal to one gram. But we want to know what 𝑚 is in kilograms. In other words, we want to know what 150 grams is in kilograms. Luckily, we can find this out by multiplying both sides of the equation by 150. This way we find that 150 grams is equal to 150 over 1000 kilograms. And this ends up being 0.15 kilograms.

And at this point, we’ve got all the other values in their standard units. 𝑢 and 𝑣 are in meters per second, and 𝑡 is in seconds. So we can plug all the values in to this equation on the right. Doing this gives us the following: we’ve got 𝑚 multiplied by 𝑔 which is 9.8 meters per second squared, that’s the gravitational field strength of the Earth, minus 𝑚 multiplied by 𝑣 minus 𝑢 over 𝑡.

Those are all our values. so we can evaluate this to give us a value of 𝐹 sub drag as 0.99 newtons. And we know that it’s in newtons because we used standard units in all of the other quantities. So we don’t need to worry about the units. We just put newtons, which is the standard unit of force at the end of our calculation. And hence, our final answer is that the average upward drag force on the ball bearing is 0.99 newtons.

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