Video Transcript
A cyclist supplies a force of 250 newtons to her bicycle. She and the bicycle together have a mass of 130 kilograms. The bicycle accelerates at 1.5 meters per second squared as it travels into a headwind that applies a 15-newton force in the opposite direction to the bicycle’s velocity. And friction acts on the bicycle in the same direction as the wind. How much force, in newtons, is supplied by friction?
Okay, so let’s start by underlining all of the important parts of the question, so we don’t miss anything out. Firstly, we know that we’ve got a cycle and a cyclist. And the cyclist applies a force of 250 newtons to her bicycle. She and the bicycle together have a mass of 130 kilograms. We’re told that the bicycle accelerates at 1.5 meters per second squared. Unfortunately for the cyclist, she’s travelling into a headwind that applies a 15-newton force in the opposite direction to the bicycle’s velocity. And there’s also the force of friction which acts on the bicycle in the same direction as the wind. We’re asked to work out how much force, in newtons, is supplied by friction.
So now that we’ve underlined all the important parts, let’s label some quantities. Let’s start with the force that the cyclist applies to her bicycle. We’ll call this 𝐹 sub cyc. This force is 250 newtons. And of course, this force is going to act in the direction of the bicycle’s velocity because on a bicycle, when you push with your legs, you want to move forward, right? Now, when we come to drawing a diagram later, we’ll set that the cyclist is travelling to the right. This is an arbitrary choice, but it’s important to be consistent. So let’s just say the cyclist is trying to travel to the right. In that case, the force 𝐹 sub cyc is also acting towards the right. This will become important later. For now, let’s move on to the mass of the bicycle and the cyclist together.
Let’s call this mass 𝑚, and this happens to be 130 kilograms. We could then also look at the acceleration of the bicycle. The acceleration is 1.5 meters per second squared. And from contacts, we can realize that this is towards the right as well, in the direction of travel. Because remember, the cyclist is pushing with a force of 250 newtons. Unless she’s going up against a massive headwind or trying to climb up a hill, she’s not going to be accelerating in the opposite direction to her travel. In other words, she is not going to be slowing down because it’s really rare on a bike to be pedaling along hard and slowing down. You can probably tell I’m not a cyclist, right?
Anyway, she would only be slowing down if there was a massive force of gravity because she was trying to go up hill or something. Or, if there was a huge headwind, which there isn’t. We know that the headwind force is 15 newtons. The other option for her to be slowing down as if friction was massive. But generally, in everyday life, we can pedal hard enough to overcome friction. Otherwise, cycling wouldn’t be possible. So anyway, she’s accelerating to the right as well.
Now, the next thing that we know is that the force applied by the headwind, we’ll call this 𝐹 sub wind, is 15 newtons. Now, we also know that this force is in the opposite direction to the bicycle’s velocity. So this force is trying to slow her down. And hence, it acts towards the left. But it’s a relatively small force. It’s only 15 newtons, compared to the 250 that she’s putting into the bike so that she can move forward. And finally, the force that we’re trying to actually find out, we’ll call this 𝐹 sub fric, for the frictional force. And we don’t know what this is. But we do know that the frictional force acts in the same direction as the wind. So it acts towards the left.
Now, it’s all well and good discussing all of these directions of travel. But, we don’t actually have a diagram to show us what’s going on properly. So let’s draw one. So here it is, our slightly surrealist interpretation of a cyclist and a bicycle. But the important thing is that she is wearing a helmet for safety reasons. So anyway, let’s get to labelling all the forces on the bike and the cyclist. We’ve said the 𝐹 sub cyc acts to the right and 𝐹 sub wind and 𝐹 sub fric are acting towards the left. As well as this, we know the mass of the bicycle and the cyclist combined. And we know the acceleration as well. And actually, these last two quantities are quite useful. Knowing the mass and the acceleration is going to allow us to use Newton’s second law of motion.
This 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. And since we already know the mass and the acceleration, we can therefore work out the resultant force on the bike. So let’s say that the resultant force on the bike — 𝐹 sub res, that’s what we’ll call it — is equal to the mass of the cyclist and the bicycle together multiplied by the acceleration of that whole object. Now, this is a really useful expression. But we can find another one using the values 𝐹 sub wind, 𝐹 sub fric, and 𝐹 sub cyc. Because remember, the resultant force on any object is simply the sum of all those forces, when you take into account the directions in which they’re acting.
In other words, 𝐹 sub res, the resultant force, is equal to 𝐹 sub cyc, the force exerted by the cyclist towards the right, minus 𝐹 sub wind — the force trying to hold back the cyclist due to the headwind, and it’s negative because it’s acting in the opposite direction — and also minus 𝐹 sub fric, the other force trying to hold back the cyclist. And at that point, we’ve accounted for all of the forces and all of the directions. But, oh look! We’ve got two expressions for 𝐹 sub res. Why don’t we equate the two. 𝐹 sub res is equal to 𝑚𝑎, as we’ve seen here. But also, this is equal to, 𝐹 sub res is equal to 𝐹 sub cyc minus 𝐹 sub wind minus 𝐹 sub fric. And at this point, we don’t need to worry about the resultant force anymore. So let’s get rid of it, bye bye.
And now we’re left with a really useful expression because we already know what 𝑚 is. We already know what 𝑎 is, same with 𝐹 sub cyc, same with 𝐹 sub wind. And we’re trying to find out what 𝐹 sub fric is. So we know all of the quantities in this equation apart from the one we’re trying to find out. Brilliant news! Let’s substitute some stuff in then. 𝑚 times 𝑎 becomes 130 times 1.5. And the right-hand side becomes 250 minus 15 minus 𝐹 sub fric. And 250 minus 15 becomes 235. And the left-hand side becomes 195. At this point, we can rearrange. Add 𝐹 sub fric to both sides. So it cancels on the right-hand side, which leaves us with 𝐹 sub fric plus 195 is equal to 235. And then subtract 195 from both sides as well. It cancels on the left-hand side. So we’re left with 𝐹 sub fric is equal to 40. 40 is the value of the right-hand side. And remember, we need to put the units of newtons.
The question wanted us to give an answer in newtons. And luckily, all of the values that we used in our working out — newtons, kilograms, meters per second squared, and newtons — happened to be in the standard units. So 𝐹 sub fric is also going to be in the standard units of newtons. And so, this becomes our final answer. The force supplied by friction is 40 newtons.
