Video: Effects of Friction on the Motion of Objects across Surfaces

The stopping distance of a car is the sum of the thinking distance and the braking distance. Figure 1 shows the variation of the thinking distance and the braking distance with the speed of the car. What is meant by the thinking distance for a vehicle? The data in Figure 1 is for a car moving on a wet road surface. Explain why the stopping distance of the car would be shorter on a dry road surface. A student looking at the data in Figure 1 hypothesises the following: Thinking distance ∝ speed². Braking distance ∝ speed². Explain why the first hypothesis that the student makes is incorrect but the second hypothesis is correct. Applying the brakes with too much force can cause a car to skid. The distance that the car skids before stopping depends on the friction between the road surface and the car’s tyres as well as the speed of the car. Friction can be investigated by pulling a device called a “sled” across the surface at a constant speed. Figure 2 shows an apparatus that can be used to pull a sled with rubber contacts over a length of a frictional surface at a constant speed. The constant of friction of rubber on the frictional surface can be found by measuring the value of the force pulling the sled and the weight of the sled. What must the relationship between the gravitational force on the weight and the friction force on the sled before the sled to move at a constant speed over the frictional surface? Assume that the pulley produces negligible friction and that there is negligible air drag. The apparatus used to measure the constant of friction is set up incorrectly so that the string is not parallel to the frictional surface. The change made to the apparatus setup is shown in Figure 3. When the sled reaches the frictional surface, it comes to rest after moving a short distance. Draw a free-body force diagram showing the forces acting on the sled when the apparatus is set up this way. A car is travelling at 25 m/s and its brakes are used to decelerate at a rate of 6.25 m/s². Calculate the braking distance of the car.

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Video Transcript

The stopping distance of a car is the sum of the thinking distance and the braking distance. Figure one shows the variation of the thinking distance and the braking distance with the speed of the car. What is meant by the thinking distance for a vehicle?

Okay, so to answer this question, let’s first imagine that we’re driving along in a car and then suddenly a boulder just rolls into our path somehow. Now, it’s at this point that we as the driver realize that we need to brake.

However, there is a slight time delay between when we realize we need to brake and when we actually press the brakes because it takes some time for the neurological signal to reach our legs and tell the muscles to press the brake. Now, in this fairly small period of time where we’re reacting to the big boulder, the car continues to travel at a constant speed because we haven’t pressed the brakes yet. And in this time, the car travels a certain distance. Now, this distance is known as the thinking distance.

So we can say that the thinking distance is the distance moved by a vehicle in the time interval between when the driver decides to stop and when the brakes are actually applied.

Moving on then, the data in figure one is for a car moving on a wet road surface. Explain why the stopping distance of the car would be shorter on a dry road surface. In other words then, why does the car have to travel a larger distance before stopping on the wet road than it does on the dry road?

Well, before we discuss this, let’s understand that regardless of what road surface we’re on, wet or dry, if the car is initially moving, then in order to stop it, we have to do some work on it. This is because before braking, the car has some kinetic energy. And in order to stop the car, we have to get rid of that kinetic energy. So we do work on the car.

And then, we can recall that the work done on an object, which we’ll call 𝑊, is given by multiplying the force applied to that object by the distance that the object moves. So now in both situations, when the road is wet and it is dry, we have to do work on the car in order to stop it. And so the distance that we mean when talking about the work done on the car to stop it is the braking distance of the car because this is the distance travelled once we’ve pressed the brakes until the car stops.

So that’s the distance we’re talking about. But then what’s the force? Well, the force that we’re talking about is the frictional force on the tyres due to them contacting the road. Now, this frictional force acts in the opposite direction to the car’s motion — in other words, the way we’ve drawn it, it acts towards the left. And so we act to stop the car.

Now, interestingly, on a wet road surface, the tyres of a car don’t grip the road strongly. Therefore, the friction force is fairly small. However, on a dry road surface, the friction force is much larger. And to represent that, we’ll draw this massive arrow.

Now, we’ve already seen that in order to stop the cars in both cases, we need to do a certain amount of work on the car. And assuming the car in both cases is travelling initially at the same speed, then the work done to stop the car in both cases is the same, but then the values of the forces applied to the car in both cases is different.

We’ve already talked about this: the frictional force on the wet surface is lower than the frictional force on the dry surface. So to counteract this, the distance that the car moves on the wet surface must be much larger than the distance moved on the dry surface. This way the work done in both cases can be the same.

And so we can say that the work done to stop the car is equal to the friction force on the wet surface, which is small, multiplied by the distance travelled by the car on the wet surface, which is large. And this is the only way that the work done in both cases can be equal. In this situation, the friction force is large and so the distance travelled is small. Hence, the braking distance on a dry road is smaller than the braking distance on a wet road. But then stopping distance is equal to thinking distance plus braking distance. And on a dry road, the braking distance is smaller. So the stopping distance is smaller.

So to answer the question, we can say that more friction is present on a dry road surface. Therefore, the decelerating force on the car is larger and the braking distance is smaller in order to do the work needed to stop the car because, of course, work done is equal to the decelerating force in this case multiplied by the braking distance of the car.

Okay, let’s now look in more detail at figure one.

A student looking at the data in figure one hypothesises the following: thinking distance is directly proportional to speed squared; braking distance is directly proportional to speed squared. Explain why the first hypothesis that the student makes is incorrect, but the second hypothesis is correct.

Okay, so first things first, the speed in both hypotheses that the student is referring to is the initial speed of the car before we think about braking. Secondly, in figure one, we’ve got the distance travelled by the car against this initial speed of the car. And we’ve got two different lines: one for the thinking distance and one for the braking distance.

So looking at the line for thinking distance, we can immediately see that this hypothesis — the first one — is incorrect because the student has suggested that thinking distance is directly proportional to speed squared. However, the line for thinking distance against the initial speed is a straight line. And this implies that thinking distance, which we’ll call d𝑡, is directly proportional to speed, not to speed squared.

Now, we might say, “Hang on! The line doesn’t go through the origin. So how can it be directly proportional?” Well, actually, if we look carefully at the graph, it very much might go through the origin because this is not the origin. Because look on the horizontal axis, we start at 10 meters per second not zero. Therefore, the true origin is actually somewhere off the graph.

So, anyway, why does thinking distance have to be proportional to speed? Well, remember earlier when we had to brake for a big boulder in our way, when this happened, we said that before we pressed the brakes, the car continued to travel at a constant speed. At least, it did until the driver finally pressed the brakes. And the distance that the car travelled between when the driver realized they need to brake and when they actually pressed the brakes was known as the thinking distance.

Now, in this period of time, the car continued to travel at a constant speed. And we can recall that speed is defined as the distance travelled — in this case, the thinking distance — divided by the time taken to travel that distance — in this case, the reaction time of the driver.

Now, we’re assuming we’re talking about the same driver in every case, regardless of what the initial speed is. So the driver’s reaction time stays constant, in which case the thinking distance of the car is directly proportional to the initial speed of the car and that’s why the first hypothesis is incorrect.

Looking at the second hypothesis then, the student has said that braking distance is proportional to speed squared. Now, looking at the graph, this is entirely plausible because we don’t have a straight line and so it could be proportional to speed squared.

Thinking back to the previous part of the question, we can recall that we said that before we stopped the car, it has some amount of kinetic energy. Now, kinetic energy is given by multiplying half by the mass of the car by the velocity of the car squared.

Now, in this case, the velocity of the car is the same as the speed of the car in the particular direction that the car was travelling. And more importantly, this is the initial speed before we brake because this is the kinetic energy that the car had before we brake.

Now, in order to stop the car, we need to get rid of all of this kinetic energy. And so the amount of work that we need to do in order to stop the car is equal to half 𝑚𝑠 squared. But then, we can recall that the work done in order to stop the car is equal to the decelerating force on the car multiplied by d𝑏, which is the braking distance of the car.

Now, at this point, we can compare these two expressions. We can see that during braking, the mass of the car is not randomly going to change. It’s not like a front bumper is gonna fall off every time we decelerate. And in this case, we’re assuming that the friction force is constant because we’re no longer talking about the difference between say a wet road and a dry road.

So the only things we’re trying to vary in this equation are the initial speed of the car that’s going along this axis on the graph and the braking distance; that’s this axis. And we can see that the braking distance is proportional to the speed of the car squared. Therefore, the second hypothesis is actually correct.

So to summarize this, we can say that the thinking distance is equal to the initial speed of the car multiplied by the reaction time of the driver. And because the car moves at a constant speed whilst travelling the thinking distance, this means that thinking distance is directly proportional to speed. Therefore, the first hypothesis is wrong.

Secondly, the braking distance is governed by the kinetic energy of the car before braking. And this kinetic energy depends on the speed squared. Therefore, the braking distance is directly proportional to the speed squared. And the second hypothesis is correct.

Okay, let’s now move on to looking at how we can investigate friction in the laboratory.

Applying the brakes with too much force can cause a car to skid. The distance that the car skids before stopping depends on the friction between the road surface and the car’s tyres as well as the speed of the car. Friction can be investigated by pulling a device called a “sled” across the surface at a constant speed. Figure two shows an apparatus that can be used to pull a sled with rubber contacts over a length of a frictional surface at a constant speed. The constant of friction of rubber on the frictional surface can be found by measuring the value of the force pulling the sled and the weight of the sled.

So this is figure two then. And we can see first of all that we’ve got a sled, which is two little feet made of tyre rubber. And then, we’ve got this string that connects to the sled, which pulls the sled along in this direction because the other end of the string is connected to a weight. Now, this weight gets pulled downwards due to a gravitational force on the weight or in other words the weight of the weight. So let’s call this weight 𝑊.

And as well as this, there is a little frictional surface here. This is what’s going to be providing friction on the sled in the opposite direction to the sled’s motion. So let’s call the frictional force on the sled 𝐹. And of course, it’s worth noting that 𝐹 is only that value once the sled has moved onto the frictional surface.

Now, the rest of the question reads “what must the relationship between the gravitational force on the weight and the friction force on the sled be for the sled to move at a constant speed over the frictional surface? Assume that the pulley produces negligible friction and that there is negligible air drag.”

In other words, we’ve been asked to find the relationship between this — the gravitational force on the weight — and this — the friction force on the sled. Specifically, we need to find the relationship between these two when the sled is moving at a constant speed. Now, as soon as we mentioned forces and a constant speed, we need to think of Newton’s first law of motion.

Newton’s first law of motion tells us that an object will continue to be at rest or to move with a constant speed unless a net force acts on the object. Now, in our situation, we want the sled to move at a constant speed. This is only possible according to Newton’s first law if there is no net force acting on the object because if they were a net force, then the sled would no longer move at a constant speed; it would either accelerate or decelerate. So the net force on the sled is zero.

Now, we’ve already discussed one of the forces acting on the sled: the friction force. But then, this can’t be the only force on the sled; otherwise, there will be a net force on the sled. The other force acting on the sled in the opposite direction is the tension in the string. Let’s call this tension 𝑇.

Now, because the friction force and the tension act in opposite direction on the sled, they can cancel each other out to give a net force of zero on the sled. But this is only true if the size or magnitude of the friction force is equal to the size or magnitude of the tension. This way, they cancel each other out.

Now, to make some more progress with this problem, we actually need to consider the weight. Note this weight is actually moving downwards. But because this weight is connected to this string and the string is taut and the string is connected to the sled, this means that the whole system — the sled, the string, the string, and the weight — all move as one unit together. In other words, this weight will be moving downwards at the same speed that this sled is moving to the right. And so the weight also has a net force of zero acting on it.

So we’ve already seen one of the forces acting on the weight. But the other force on the weight you guessed it must be the tension in the string. This tension is called 𝑇 once again. Now in order for the weight to have a net force of zero on it, we’ve got two forces — the weight and the tension — acting in opposite directions. And so the magnitude of the tension must be equal to the magnitude of the weight.

But then at this point, we’ve seen that the friction force is equal to the tension and the tension is equal to the weight. So this means that the friction force is equal to the weight. And hence, we found the relationship between the gravitational force on the weight and the friction force on the sled.

We’ve also accounted for the last sentence in the question, which tells us that the pulley produces negligible friction because we haven’t considered any forces due to the pulley and there is negligible air drag because we haven’t even thought about air drag on the sled at all.

So to answer the question then, we can say that the two forces — the gravitational force on the weight and the friction force on the sled — are equal in magnitude if the sled is to be moving at a constant speed.

So this is all well and good. But let’s see what happens if we accidentally set up this whole apparatus incorrectly. The apparatus used to measure the constant of friction is set up incorrectly so that the string is not parallel to the frictional surface. The change made to the apparatus is shown in figure three. When the sled reaches the frictional surface, it comes to rest after moving a short distance. Draw a free-body diagram showing the forces acting on the sled when the apparatus is set up this way.

Okay, so first things first, let’s have a look at figure three and see that the string which is meant to be set up like this has actually been set up so it’s pointing downward at some angle. In other words, the tension in the string will be pulling this way rather than this way.

Now, we need to draw a free-body diagram. So to do this, let’s clear some space. Okay, so when we’re drawing a free-body diagram, there are lots of simplifications that we can make. Now, in our case, we’re gonna go as simple as possible. Forget the fact that the sled has two feet. Forget the fact that the sled even is a box shape. Let’s just draw the sled as a big blob.

Now, what are the forces acting on the sled once it reaches the frictional surface? Well, naturally, it’s going to have the friction force acting towards the left and is going to have the tension in the string acting towards, well not quite, the right, but at some angle, to the horizontal when normally it will be acting in this direction.

Also, we’re gonna assume that this angle here, the angle between the horizontal and the direction in which the string is actually pointing, is quite a small angle. We’ll see why this is important in a second.

But it’s also important to note that this length here represents the size or magnitude of the friction force and this length here represents the magnitude of the tension force. The longer the arrow, the larger the magnitude of the force. That means that this length and this length need to be the same because even though the tension is acting in a different direction to before, it’s still dependent on the weight of the weight.

And so the size of the tension force is still going to be the same as before and so is the friction force because when the sled is moving over the frictional surface, it’s going to have a certain friction force; that hasn’t changed compared to before.

What this ends up meaning is that this length here, which is the component of the tension force in the horizontal direction, is slightly smaller than this length here, which is the magnitude of the friction force. And this is why the sled decelerates and stops once it gets onto the frictional surface because the force acting towards the left is slightly larger than the force acting towards the right which causes the sled to decelerate.

However, because we’ve said that this angle is very small, we can basically ignore the component of this tension force that’s acting in the downward direction. And this actually leads us very nicely onto discussing forces acting in the vertical direction because of course the sled itself has a certain weight. So we need to draw that on the free-body Force diagram as well.

And because the sled is in contact with the surface, regardless of whether it’s the frictional surface or the table, there is going to be an upward contact force on the sled as well. Now, it’s worth noting again that because we’ve decided that our sled is going to be a blob in our free-body force diagram, we’ve completely ignored the fact that there are two feet on the sled.

Instead, we’re just considering the overall forces acting on the sled. And because we decided that the downward acting component of the tension force in the string is negligible, then that means that this length here, the length representing the weight of the sled, must be equal to this length here, the contact force. In other words, those two must be basically equal because otherwise the sled would accelerate either upwards or downwards.

And at this point, we’ve considered all of the forces acting on the sled as well as their magnitudes. And so this is our free-body force diagram.

So now that we’ve discussed this experiment, let’s quickly go back and look at cars again.

A car is travelling at 25 meters per second and its brakes are used to decelerate at a rate of 6.25 meters per second squared. Calculate the braking distance of the car.

Okay, now in this part of the question, we’ve been given a few pieces of information. Firstly, the initial velocity of the car, which we’ll call 𝑢, is 25 meters per second. Secondly, the acceleration of the car is negative 6.25 meters per second squared because remember we’re using the brakes to decelerate and therefore we’re gonna have a negative acceleration. And thirdly, we’ve also implicitly been told the final velocity of the car, which we’ll call 𝑣. This happens to be zero meters per second because what we’ve been asked to calculate is the braking distance; this means the distance travelled by the car before it completely stops.

Now, when it stops, the final velocity of the car will be zero meters per second. And of course, because we’ve been asked to calculate the braking distance, we’ve definitely been asked to calculate the distance travelled by the car in a straight line, which we’ll call 𝑠.

The reason we call it 𝑠 is because then we can use one of the SUVAT equations. Specifically, the equation that we’re looking for tells us that the final velocity of the car squared is equal to the initial velocity squared plus two times the acceleration of the car times the distance travelled in a straight line.

Now, in order to find the distance travelled in a straight line by the car or in other words the braking distance, we need to rearrange this equation. When we do, we find that the distance travelled in a straight line is equal to the final velocity squared minus the initial velocity squared divided by two times the acceleration.

Now, a quick look at the units in all of the quantities that we’ve been given shows us that they’re all in standard units. Therefore, when we calculate the distance, it’s going to be in its standard unit as well, which is the meter. And this is perfect because if we look ahead to the answer blank, we’re expected to give our answer in meters.

So now, let’s plug our values into this equation. This answer is being zero squared minus 25 squared over two times negative 6.25 meters or in other words 50 meters. And so our final answer is that the braking distance of the car is 50 meters.

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