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Lesson Video: The Equilibrium of a Body on a Rough Horizontal Plane Mathematics

In this video, we will learn how to solve problems involving the equilibrium of a body on a rough horizontal plane.

15:43

Video Transcript

In this video, we will learn how to solve problems involving the equilibrium of a body on a rough horizontal plane. We will begin by looking at some key definitions and then consider how we can resolve vertically and horizontally to calculate unknowns. These will include the normal reaction force, the frictional force between the body and the plane, and the coefficient of friction πœ‡. We will also consider the angle of friction.

Let’s begin by considering a body resting on a rough horizontal surface as shown. If the body is in equilibrium, it has zero resultant force acting on it. Two forces act on the body: firstly, its weight, which is equal to the mass of the body multiplied by the acceleration due to gravity. Secondly, as a result of Newton’s third law, we have the normal reaction force 𝐍. For a body on a horizontal surface, this acts vertically upwards. As there is zero resultant force, these two forces must be equal such that 𝐍 is equal to π‘Š, which we can also write as 𝐍 is equal to π‘šπ‘”.

If the body was on a smooth surface, any net force applied horizontally on the body would accelerate the body horizontally. However, on a rough surface, the body will only accelerate if the horizontal force applied to it has magnitude greater than the frictional force between the body and the surface. In this video, we will label this frictional force π…π‘Ÿ. This frictional force acts in the opposite direction to the applied force 𝐅.

In this video, we will consider problems where the body is on the point of moving and, as such, the frictional force will be at its maximum. At this stage, the body will still be in equilibrium. And when we resolve horizontally, the applied force 𝐅 will be equal to the frictional force π…π‘Ÿ. The maximum frictional force is also called the limiting friction. And this limiting friction π…π‘Ÿ between a body at rest and the surface it rests on satisfies the formula π…π‘Ÿ is equal to πœ‡π. And since 𝐍, the normal reaction force, is equal to π‘šπ‘”, the frictional force π…π‘Ÿ is equal to πœ‡π‘šπ‘”.

We will now look at an example where the maximum applied force on a body that remains in equilibrium is to be determined.

A body weighing 25.5 newtons rests on a rough horizontal plane. A horizontal force acts on the body, causing it to be on the point of moving. Given that the coefficient of friction between the body and the plane is three seventeenths, determine the magnitude of the force.

We can begin by sketching a diagram to model the situation. We know that any body resting on a horizontal plane will have a downward force equal to its weight. And this weight π‘Š will be equal to the mass of the body multiplied by the acceleration due to gravity. In this question, we are told the body weighs 25.5 newtons. So there will be a force acting vertically downwards equal to 25.5 newtons. Newton’s third law tells us that there will be a normal reaction force acting in the opposite direction to this. Since the body is in equilibrium, we know that these two forces are equal. The normal reaction force 𝐍 is equal to 25.5 newtons.

We are told that a horizontal force 𝐅 acts on the body. And as the plane is rough, there will be a frictional force acting between the body and the plane. The body is on the point of moving, which means that the frictional force will be at its maximum. This is known as the limiting friction such that π…π‘Ÿ is equal to πœ‡ multiplied by 𝐍, where πœ‡ is the coefficient of friction between the body and the plane. In this question, we are told this is equal to three seventeenths. The frictional force π…π‘Ÿ is therefore equal to three seventeenths multiplied by 25.5, which is equal to 4.5. The maximum frictional force is equal to 4.5 newtons.

Since the body is on the point of moving and is still in equilibrium, the horizontal forces must also be equal to one another. The applied force 𝐅 must be equal to the frictional force π…π‘Ÿ. We can therefore conclude that the magnitude of the horizontal force acting on the body is 4.5 newtons.

In our next example, we need to calculate a resultant force of which the limiting friction is a component.

A body is resting on a rough horizontal plane. The coefficient of friction between the body and the plane is 0.2, and the limiting friction force that is acting on the body is 80 newtons. Given that 𝑅 is the resultant of the force of friction and the normal reaction force, find the magnitude of 𝑅.

We will begin by sketching a diagram to model the situation. If the body is in limiting friction, it will have four forces acting on it. Firstly, acting vertically downwards, we have the weight force, which is equal to the mass of the body multiplied by the acceleration due to gravity. Acting vertically upwards, we have the normal reaction force 𝐍. We will have an applied force 𝐅 that is causing the body to be on the point of moving, and finally a frictional force π…π‘Ÿ between the body and the plane acting in the opposite direction to this.

Whilst the weight and applied force are not mentioned in this question, we know that for a body at rest on a surface, the normal reaction force 𝐍 must be equal to the weight of the body. The limiting friction is the maximum frictional force that the surface exerts on the body before the body starts to move. And we therefore know that the applied force 𝐅 must be equal to the frictional force π…π‘Ÿ. This limiting friction is equal to πœ‡ multiplied by 𝐍, where πœ‡ is the coefficient of static friction between the surface and the body.

Let’s now substitute the values we are given in this question. We are told that πœ‡ is equal to 0.2. We are also told that the limiting frictional force is equal to 80 newtons. Substituting these into π…π‘Ÿ is equal to πœ‡π, we have 80 is equal to 0.2 multiplied by 𝐍. We can then divide through by 0.2. 𝐍 is equal to 80 divided by 0.2, which equals 400. The normal reaction force acting on the body is 400 newtons. Whilst it is not required in this question, we could then calculate the weight force and the applied force 𝐅. These are equal to 400 newtons and 80 newtons, respectively.

We need to calculate the magnitude of 𝑅, where 𝑅 is the resultant of the force of friction and the normal reaction force. This will act in a direction shown on the diagram. We can calculate the magnitude by creating a right triangle. And using the Pythagorean theorem, we have 𝑅 squared is equal to 400 squared plus 80 squared. The right-hand side is equal to 166,400. We can then take the square root of both sides of this equation. And since the resultant force must be positive, we have 𝑅 is equal to 80 root 26. The magnitude of the resultant force is equal to 80 root 26 newtons. As a decimal answer, this is equal to 407.92 newtons to two decimal places.

This question leads us onto a key fact about any body on a rough surface. A body on a rough surface has an angle of friction. This is the angle between the normal reaction force and the resultant of the normal reaction force and the limiting frictional force. If we consider the following body on a rough horizontal surface, then if the body is in limiting friction and on the point of moving, there will be four forces acting on it in the horizontal and vertical directions. We have the normal reaction force 𝐍, the weight force π‘Š which is equal to π‘šπ‘”, the applied force 𝐅, and the limiting frictional force π…π‘Ÿ.

We recall that this limiting frictional force π…π‘Ÿ is equal to πœ‡ multiplied by 𝐍. And whilst it is not required here, it is worth recalling that when resolving vertically and horizontally, our forces are equal. Drawing on the force 𝑅, which is the resultant of the normal reaction force and the frictional force and not an additional force, we see that the angle of friction πœƒ is as shown. We can sketch a right triangle with this resultant force 𝑅 together with the normal reaction force and the limiting frictional force. Replacing π…π‘Ÿ with πœ‡π, we can use the tangent ratio to calculate πœƒ.

We know that in any right triangle, the tan of angle πœƒ is equal to the opposite over the adjacent. In our diagram, tan πœƒ is equal to πœ‡π over 𝐍. And dividing through by 𝐍, we have tan πœƒ is equal to πœ‡. The angle of friction for a body on a rough horizontal surface can therefore be determined by tan πœƒ equals πœ‡.

Let’s now look at a specific example where we need to calculate the angle of friction.

Given that the coefficient of friction between a body and a plane is root three over four, what is the angle of friction? Round your answer to the nearest minute if necessary.

We will begin by recalling what we mean by the angle of friction. The angle of friction is the angle between the normal reaction force on a body and the resultant of the normal reaction force and the limiting frictional force on the body. Since the limiting frictional force π…π‘Ÿ is equal to πœ‡π, and using our knowledge of right angle trigonometry, we know that the angle of friction πœƒ satisfies the equation tan πœƒ is equal to πœ‡π over 𝐍, which is equal to πœ‡, where πœ‡ is the coefficient of friction and in this question is equal to root three over four. The tan of angle πœƒ is therefore equal to root three over four.

Taking the inverse tangent of both sides of this equation gives us πœƒ is equal to the inverse tan or arctan of root three over four. Ensuring that our calculator is in degree mode, typing in the right-hand side gives us 23.4132 and so on. We are asked to give our answer to the nearest minute. We can do this by using the degrees, minutes, and seconds button on the calculator or by multiplying the decimal part of our answer by 60. This is equal to 24.79 and so on minutes, giving us an answer to the nearest minute of 23 degrees and 25 minutes. This is the angle of friction when the coefficient of friction between a body and a plane is root three over four.

We will now summarize the key points from this video. We saw in this video that the normal reaction force for a body on a horizontal surface has a magnitude equal to the weight of the body. The frictional force between a body and a rough surface acts in the opposite direction to the net force on the body and parallel to the plane. The maximum frictional force between a body and a rough surface is given by π…π‘Ÿ is equal to πœ‡π, where 𝐍 is the normal reaction force on the body and πœ‡ is the coefficient of static friction between the body and the surface. Finally, we saw that the angle of friction for a body on a rough horizontal surface is given by πœƒ, which is equal to the inverse tan of πœ‡, where once again πœ‡ is the coefficient of static friction between the body and the surface.

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