Lesson Video: Motion of a Body on a Rough Plane Mathematics

In this video, we will learn how to study the motion of a particle on horizontal and inclined rough planes against friction force.


Video Transcript

In this video, we’re talking about the motion of a body on a rough plane. We’re speaking then about objects moving across surfaces, both inclined and level where friction is involved. To analyze and understand this motion, we’ll be using coefficients of friction, free-body diagrams, and equations of motion, including Newton’s second law.

We can start off by imagining an object at rest on a rough surface, and we’ll see that this object has a mass 𝑚. If we draw the forces currently acting on this body, we see that there are two. First, there’s the weight force acting straight downward with a magnitude of 𝑚 times 𝑔, where 𝑔 is the acceleration due to gravity. And there’s also an equal and opposite normal or reaction force. Now, at this point, we can recall Newton’s second law of motion, which tells us that the net force acting on an object is equal to that object’s mass times its acceleration.

For our object here, the forces acting on it cancel one another out, and so the net force is zero. And therefore, according to this law, it has no acceleration. But now imagine that we apply a force on this box pushing it from left to right. If the surface that our object was at rest on was smooth, that is, if there was no friction between the surface and the body, then 𝐹 sub 𝐴 would give a net force in the horizontal direction and our object would start to accelerate. But in this case, because there is friction between the two surfaces, an opposing frictional force is generated.

The interesting thing about this frictional force is that it exists purely in response to the applied force. If there were no applied force, there would be no force of friction. And even more interesting, as our applied force increases in magnitude, say, for example, we were to push that much harder on the box, then, up to a certain point, our friction force will increase to match it. We say up to a certain point because there’s a certain maximum value the force of friction can attain. That maximum value is given by this equation. The Greek letter 𝜇 represents what is called the coefficient of friction. And as we saw, 𝐹 sub 𝑁 is the normal or reaction force. When it comes to the coefficient of friction, this is a unit list number that is typically but not always between zero and one.

There are two factors that determine the value of 𝜇 in a given scenario. The first is what specific materials are involved in the interaction, in other words, what material is our body made of and what material is our surface made of. The second factor affecting 𝜇 is whether our object is stationary or in motion. If our object is not moving, then we use what’s called the coefficient of static friction, a 𝜇 with an 𝑠 subscript.

On the other hand, say that we increase the applied force enough that our box does start moving despite the opposition due to friction. In that case, there’s still a frictional force opposing the box’s motion. But it’s of a different magnitude because now we’re using what’s called the coefficient of kinetic friction, 𝜇 sub 𝑘. In general, these two coefficients are not the same. Typically, the coefficient of kinetic friction is less than that of static friction. We use one or the other in our equation for the frictional force, depending on how our object is moving. One other point about our equation for the friction force is that it’s important that we use the normal force here rather than the weight force. This is especially true when our body is on an inclined rough surface, as we’ll see later on in an example.

To get a sense for how the friction force can change over time, let’s go back to our original condition where we only had vertically acting forces on our body. If we were to make a plot of the force of friction acting on our body against the applied force acting on it, at this moment, we know that the applied force is zero and so is the force of friction. Then, say that we gradually increase the applied force from zero up to one unit where we’ll leave the units unspecified. And say further that as we do this, we don’t exceed the maximum static frictional force that can act on our box. This tells us that in response to this applied force, our frictional force kicks in, so to speak, and matches it.

All throughout this time then, our object is still not in motion. It still has a zero net force acting on it. Say then that we continue to increase our applied force from one to two units. And that as we do this, our frictional force is still able to keep up, so our box remains stationary. In that case, our graph would look like this where the magnitude of frictional force is always equal to the magnitude of the applied force. So far, the coefficient of friction we would use in our friction equation is 𝜇 sub 𝑠, the coefficient of static friction because our box is not in motion. But as we know from practical experience, if we continue to push harder on the box, eventually it will start to move. When that happens, we’ve overcome the maximum force of static friction.

We might imagine that our graph then would look like this where the force of friction remains at this level while our applied force increases. Actually, though, it typically has more of this shape. The maximum frictional force achievable is 𝜇 sub 𝑠, the coefficient of static friction, times the normal force, whereas the flat part of this curve represents 𝜇 sub 𝑘 times 𝐹 sub 𝑁. This is confirmation that indeed 𝜇 sub 𝑘 is typically less than 𝜇 sub 𝑠. Considering this vertical dashed line on our graph, we can say that, to the left of this line, our box is stationary, whereas to the right our box is in motion.

For each condition of motion, we’ll want to be careful to use the correct coefficient of friction. These values, by the way, are typically either given to us in a problem statement or we can look them up in a table. Whether or not an object is moving, Newton’s second law of motion can help us analyze the forces acting on it. When a body is in motion, we then may be able to apply other equations of motion, sometimes called the kinematic equations and sometimes called SUVAT equations. These names both refer to the same thing, a set of four equations that help us describe the motion of objects in motion. These equations involve initial and final velocities, accelerations and displacements, and times.

Knowing all this about how objects move when friction is involved, let’s get a bit of practice through an example.

Using the information in the figure, calculate the coefficient of kinetic friction, rounding the result to the nearest two decimal places. Given that the mass of the body is 28 kilograms and that the acceleration due to gravity, 𝑔, equals 9.8 meters per second squared.

Taking a look at our figure, we see our body which is on some surface, moving along with an acceleration to the right of 2.2 meters per second squared. This acceleration is due to a force of 155 newtons that’s being applied at an angle of 45 degrees to the horizontal. Knowing this, along with the fact that the mass of our body — we’ll call it 𝑚 — is 28 kilograms, we want to solve for the coefficient of kinetic friction between the body and the surface.

We can begin solving for this coefficient by first sketching in all the forces that are acting on our body. This is called a free-body diagram. For one thing, we know that our body is subject to the force of gravity. We also know that there’s an applied force, we’ll call it 𝐹 sub 𝐴, which is given as 155 newtons, acting at a 45-degree angle on the body. Furthermore, there’s a normal or reaction force acting straight up from the surface. And lastly, there’s a frictional force; we’ll call it 𝐹 sub 𝑓. And this opposes the motion of our object. That’s always true of friction, which is how we knew that it acted to the left.

Now that we know all the forces acting on our object, we can recall Newton’s second law of motion. This tells us that the net force that acts on somebody is equal to that body’s mass times its acceleration. Now because motion in the vertical direction is independent of motion in the horizontal direction, we can apply Newton’s second law to either one independently. In other words, we could say that the net force in the horizontal plane acting on our body is equal to the object’s mass multiplied by its horizontal acceleration. And indeed, this is the dimension we’ll focus on because we want to solve for the coefficient of kinetic friction, which is bound up in the force of friction.

When we say that, we just mean that the force of friction acting on an object that’s in motion is equal to the coefficient of kinetic friction multiplied by the normal force acting on the object. So, here’s what we’ll do. Focusing first on this horizontal direction, let’s clear a bit of space on screen and apply Newton’s second law in this horizontal plane. The first thing we’ll do is establish sign conventions, positive and negative.

Note that the acceleration given to us in our problem statement is a positive value and that acceleration is to the right. We’ll say then that any force or motion in that direction is positive, and any force or motion in the other direction is negative. As we study the forces acting horizontally in our free-body diagram, we see that there’s a component of the applied force that acts horizontally. And then there’s the friction force which acts entirely in this direction.

Given a 45-degree right triangle where the hypotenuse is our applied force magnitude, 155 newtons, we can solve for this horizontal leg of that triangle by multiplying 𝐹 sub 𝐴 by the cos of 45 degrees. That then accounts for the positive horizontal force acting on our body. We then subtract from that the frictional force. That force recall is equal to the coefficient of kinetic friction, what we want to solve for, times the normal force. This accounts for all of the horizontal forces acting on our mass. And therefore, by Newton’s second law, this sum is equal to mass times acceleration in the horizontal direction.

We’re given 𝐹 sub 𝐴 as well as the mass and acceleration of our body, but we don’t know 𝐹 sub 𝑁 to let us solve for 𝜇 sub 𝑘. To solve for it, we’re actually going to need to apply Newton’s second law again, but this time in the vertical dimension. As we do this, we can set up the conventions that motion up is in the positive direction and motion down is in what we’ll call the negative direction. So, let’s consider the vertically acting forces on our body. First, there’s the normal force, 𝐹 sub 𝑁. Added to that is the vertical component of our applied force, 𝐹 sub 𝐴.

Returning to our triangle, that side length is equal to 𝐹 sub 𝐴 times the sin of 45 degrees. And then lastly in the vertical direction, we have the weight force, 𝑚 times 𝑔, acting downward. By our convention, this force is negative. And by Newton’s second law, all of this is equal to the mass of our body times its acceleration in the vertical direction, what we’ll call 𝑎 sub 𝑣. We then realize that our object isn’t accelerating vertically, so this value is zero. That means these forces add up to zero. So, if we subtract this term from both sides of the equation and add this term to both sides, then we find an expression for the normal force, which is in terms of the weight force minus the vertical component of the applied force.

We can then substitute this whole right-hand side in for 𝐹 sub 𝑁 here. And now, we have an expression we can use to solve for 𝜇 sub 𝑘, the coefficient of kinetic friction. This is because we know the numerical values of 𝐹 sub 𝐴, 𝑚, 𝑔, and 𝑎, as well as the sin and cos of 45 degrees. Before we plug in those values though, let’s rearrange this expression so that 𝜇 sub 𝑘 is the subject. First, we’ll subtract 𝐹 sub 𝐴, cos of 45 degrees, from both sides. Then, we’ll divide both sides of the equation by this expression in parentheses. And lastly, we’ll multiply both sides by negative one. We end up with this expression. And now, we’re ready to substitute in our values.

𝐹 sub 𝐴 is 155 newtons. Both the cos and sin of 45 degrees is the square root of two over two. Our mass, 𝑚, is 28 kilograms. The acceleration, 𝑎, is 2.2 meters per second squared. And lastly, the acceleration due to gravity, 𝑔, is 9.8 meters per second squared. When we enter this expression on our calculator, to two decimal places, we find the result of 0.29. This is the coefficient of kinetic friction between our body and the surface.

Now, let’s look at an exercise where our body is on an inclined rough surface.

A body of weight 24 newtons is placed on a rough plane inclined at 30 degrees to the horizontal. The coefficient of static friction between the body and the plane is one over six root three. And the coefficient of kinetic friction is one over 12 root three. A force is pushing the body upward in the direction of the line of greatest slope. Determine the magnitude of the force that would make the body about to move.

Sketching in this scenario, it could look like this where our body is at rest on an inclined plane of 30 degrees. Situated like this, our body is subject to a static frictional force, which depends on the coefficient of static friction given as one over six root three. We’re told that there’s a force, we’ll call it 𝐹, that acts on the body in the direction of the line of greatest slope. And we want to figure out how big that force could be, such that the body would be about to move.

Let’s first give ourselves some space on screen to work. And then let’s include in our sketch all of the forces that are acting on our body. We’re told that there’s a weight force of magnitude 24 newtons. And we also know that there’s a normal force perpendicular to the surface of our slope. Along with all this, there is a static frictional force. Now, if we had no applied force, what we’ve called 𝐹, pushing our body up the incline, then our static frictional force would point in that same direction. This would be so it would resist the block sliding down the plane. But in our scenario, we want to solve for the maximum possible value of 𝐹 such that the block is just barely not moving up the incline. That means that, here, the static frictional force is actually acting down the incline.

As we begin analyzing these forces acting on our body, we can orient ourselves with a pair of orthogonal axes. We’ll say that forces in motion acting up the incline are positive, and those acting the opposite way are negative. Likewise, forces in motion perpendicularly away from our surface are positive, and those in the other direction are negative. Knowing this, we can apply Newton’s second law of motion to both of what we’ll call our 𝑦-direction and our 𝑥-direction.

Looking first in the 𝑥-direction, we see that our force 𝐹 acts in the positive 𝑥, while our frictional force acts opposite this. These, however, aren’t the only 𝑥-direction forces. That’s because our weight force has a component right here in the negative 𝑥-direction. Now, this angle right here in this triangle is equal to 30 degrees. And since this intersection is a right angle, we can see that this 𝑥-component of our weight force is 24 newtons times the sin of 30 degrees. The sin of 30 degrees we can recall is one-half, so we’ll substitute that in right away.

So these then are indeed all the forces acting on what we’ve called our 𝑥-direction. By Newton’s second law, these add up to our object’s mass times acceleration in this dimension. But remember that we’re solving for 𝐹 such that our object is not yet in motion. In other words, 𝑎 sub 𝑥 is zero. That means our whole right side is zero, which means we can rearrange and see that the maximum force we can apply to our body without it yet being put in motion is equal to the static frictional force plus 12 newtons. In general, the force of static friction is equal to the coefficient of static friction times the normal force acting on the relevant body.

In our scenario, we know 𝜇 sub 𝑠. But what about 𝐹 sub 𝑁? To figure that out, we’ll want to look at Newton’s second law of motion applied in the 𝑦-direction. Here, we have a positive, normal force but then a negative component of our weight force here on our sketch. This component is 24 newtons times the cos of 30 degrees where the cos of that angle equals the square root of three over two. Just like in the 𝑥-direction, because there’s no acceleration in the 𝑦-direction, these forces add up to zero. This tells us that the normal force equals 12 times the square root of three newtons.

Now that we know 𝐹 sub 𝑁, we can substitute it as well as 𝜇 sub 𝑠 into our equation for the frictional force. And multiplying these quantities together, we see that a square root of three cancels out and that the maximum frictional force, therefore, is two newtons. 𝐹 then is 14 newtons. This is the magnitude of the largest force that could act directly up the incline without our body being put in motion.

Let’s review now some key points from what we’ve learned. In this lesson, we saw that a body resting on a rough surface can experience a static friction force, while a body in motion across a rough surface experiences a kinetic friction force. Along with this, we’ve learned that, mathematically, the friction force is equal to the coefficient of friction times the normal force acting on the relevant body. The coefficient of friction may be represented 𝜇 sub 𝑠 if the body is stationary or 𝜇 sub 𝑘 if it’s in motion. Lastly, we saw that Newton’s second law of motion and free-body diagrams are useful tools for understanding motion involving friction.

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