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