# Lesson Explainer: The Equilibrium of a Body on a Rough Inclined Plane Mathematics

In this explainer, we will learn how to solve problems on the equilibrium of a body on a rough inclined plane.

The normal reaction force, , on a body on an inclined surface is shown in the following figure:

The reaction force is the result of the weight of the body acting on the surface supporting it. The weight of the body, , is given by where is the mass of the body and is acceleration due to gravity.

The direction of is always perpendicular to the surface. The magnitude of is equal to the component of the weight of the body acting perpendicularly to the surface. is therefore given by where is the angle from the horizontal at which the surface slopes.

The resultant force on a body on an inclined plane, , is the sum of and . We can express this as

The direction of the line of action of is shown in the following figure,

where

The direction of is downward, parallel to the plane.

We see that and are perpendicular components of ; hence,

For a body in contact with a rough surface, a frictional force acts on the body in the opposite direction to the net force on the body. For a body in equilibrium, the frictional force has a magnitude equal to the magnitude of . Increasing the angle of inclination of the plane supporting a body increases net force acting on the body. The frictional force on the body also increases as the net force on the body increases. The upper limit of the magnitude of the frictional force is called the limiting friction. The limiting friction, , on a body at rest on a surface is given by where is the coefficient of static friction between the body and the surface and is the magnitude of the reaction force.

The forces acting on a body in equilibrium on a plane inclined at the maximum angle of inclination for which the body can be in equilibrium are parallel to the plane, as shown in the following figure.

For the body to be in equilibrium, these forces must have equal magnitudes, so we can see that and hence that

The greatest angle of inclination of a rough plane that can maintain a body in equilibrium is given by

Let us look at an example where the maximum frictional force on a body in equilibrium on a rough inclined surface is determined.

### Example 1: Calculating the Value of Friction in a Case of Limiting Equilibrium

The figure shows a body of mass 46 N resting on a rough inclined plane. Given that the body is on the point of sliding down the plane and that the coefficient of static friction is , find the magnitude of the limiting friction.

The body is on the point of sliding, so the frictional force on it is the limiting friction. The body is at rest, so the magnitude of frictional force on the body must equal the magnitude of the net force on the body parallel to the plane, . The weight of the body is equal to , and the reaction force has a magnitude of , given by

These two forces sum to give the magnitude of , as shown in the following figure:

Knowing that the force acting downward parallel to the plane is given by we have that

The frictional force must have a magnitude equal to for the body to be in equilibrium.

The answer can also be obtained by considering the limiting friction on the body, which has a magnitude given by

And is given by

Hence,

If an applied force acts on a body in equilibrium on a rough inclined plane, parallel to the plane, the magnitude of the friction must equal the sum of the net force on the body due to its weight and the normal reaction on it and the applied force.

The following figure shows the directions and magnitudes of the forces that act on a body in equilibrium on a rough inclined plane, parallel to the plane, when a force of magnitude is applied to the body parallel to the plane, either downward or upward along the plane.

We can see from the diagram that, for a fixed magnitude of , the frictional force on the body is greater if acts downward parallel to the plane. The maximum value possible for the greater frictional force that maintains the equilibrium of the body is the limiting friction.

We can consider the applied force required for equilibrium of the body when the force acts either upward or downward parallel to the plane if the frictional force on the body equals the limiting friction for either direction of the force . This is shown in the following figure.

We can see from the figure that when the frictional force on the body equals the limiting friction, the applied force on the body parallel to the plane must be greater if it acts upward parallel to the plane than if the force acts downward parallel to the plane.

Let us look at an example where an applied force acts parallel to the surface on a body in equilibrium on a rough inclined surface.

### Example 2: Calculating the Coefficient of Friction for a Body on an Inclined Plane

A body weighing 60 N rests on a rough plane inclined to the horizontal at an angle whose sine is . The body is pulled upward by a force of 63 N acting parallel to the line of the greatest slope. Given that the body is on the point of moving up the plane, find the coefficient of friction between the body and the plane.

The forces acting on the body parallel to the surface are shown in the following figure.

The force equal to is the resultant of the weight of the body and the normal reaction force on the body. For the body to be in equilibrium and on the point of moving, it must be the case that

We are told that hence, a section of the surface of length is the hypotenuse of a right triangle with legs of lengths and , as shown in the following figure.

The value of can be found using the Pythagorean theorem as follows:

From this, we can see that

The values of the sine and cosine of can be substituted into the equation to give us

If an applied force does not act parallel to the surface supporting a body, the effects of the applied force on the equilibrium of the body are more complex. Consider the forces acting on the body shown in the following figure.

With no applied force acting, the magnitude of the normal reaction force is given by

The applied force has a component that acts not only against the frictional force on the body, but against the weight of the body. The applied force reduces the force that the weight of the body exerts on the surface, which reduces the normal reaction force on the body.

The magnitude of the normal reaction force on the body is given by where is the magnitude of the applied force.

As the frictional force depends on the normal reaction force, the frictional force is also changed if the applied force is not parallel to the surface.

Let us look at an example where an applied force acts on a body in equilibrium on a rough inclined surface and the applied force does not act parallel to the surface.

### Example 3: Calculating the Tension in a String Required for a Body to be at the Point of Moving up an Inclined Plane

A body weighing 56 N rests on a rough plane inclined at an angle of to the horizontal. The coefficient of friction between the body and the plane is . The body is pulled upward by a string making an angle of to the line of greatest slope of the plane. Determine the minimum tension in the string required to cause the body to be on the point of moving up the plane.

It is helpful in answering this question to use a free-body diagram, as shown in the following figure.

The frictional force, , acts parallel to the plane and the reaction force, , acts normally to the plane. The tension, , acts at an angle of from the plane and the 56 N weight of the body acts at an angle of from the line normal to the plane. The components of parallel and perpendicular to the plane, and , are shown in the following figure.

The angle that makes with the plane is , so we see that

The component of perpendicular to the plane acts in the direction of and in the opposite direction to the component of the weight of the body acting perpendicularly to the plane. The net force on the body perpendicular to the plane is zero.

The frictional force, , has zero component perpendicular to the plane.

Taking the direction of as positive perpendicular to the plane, the component of the weight of the body perpendicular to the plane has a magnitude given by

The magnitude of the tension in the string perpendicular to the plane is given by

Equating the magnitude of the tension in the string perpendicular to the plane with the magnitudes of the components of the weight of and reaction on the body perpendicular to the plane, we obtain

From this, we can obtain an expression for the magnitude of :

The angle that makes with the plane is , so, taking the upward parallel to the plane as positive, we see that

The reaction force, , has zero component parallel to the plane.

The component of parallel to the plane acts in the opposite direction to and the component of the weight of the body parallel to the plane. The net force on the body parallel to the plane is zero.

The component of the weight of the body parallel to the plane has a magnitude given by

The angle is , so

The magnitude of depends on the magnitude of :

Equating the tension in the string parallel to the plane with the components of the weight of and reaction on the body parallel to the plane, we obtain

It has been shown that

Substituting this expression for into the equation for the forces acting parallel to the plane, we find that

This expression can be rearranged to make the subject as follows.

Now let us look at an example where an applied force acts on a body in equilibrium on a rough horizontal surface and the applied force does not act horizontally.

### Example 4: Calculating an Unknown Force and the Normal Reaction Force for a Body on an Inclined Plane

When a body of weight 262.5 N was resting on a rough plane inclined to the horizontal at an angle whose tangent is , it was on the point of moving. The same body was later placed on a horizontal surface of the same roughness. A force acted on the body pulling upward at an angle of to the horizontal where . Given that under these conditions the body was on the point of moving, determine the magnitude of and the normal reaction .

There are two distinct sections involved in determining and . The coefficient of static friction of the surface must be determined for the body on the inclined surface, and when this is determined, the forces acting on the body when it is on the horizontal surface can be determined.

If we let the angle of the incline of the surface be , we are told that

This is equal to the coefficient of static friction between the body and the surface, as the object is on the point of moving; therefore,

When the body is in equilibrium on the horizontal surface, the forces acting on the body are shown in the following figure, where is the limiting friction.

If we let the angle made by with the surface be , we are told that hence,

The magnitudes of the vertical and the horizontal components of , and , are given by and

Equating vertical forces on the body gives us

From this, we can see that

Equating horizontal forces on the body gives us

This equation can be rearranged to make the subject as follows:

As we know that we can see that

The magnitude of the resultant of the weight of a body on a rough inclined plane and the normal reaction on the body can be greater than the limiting friction between the body and the plane. If no external force acts on the body, the body cannot be in equilibrium. An external force acting upward parallel to the plane can maintain the equilibrium of the body. There are two possible values for the magnitude of this force.

Let us define a force that is equal to the resultant of the weight, , of a body on a rough inclined plane at an angle from the horizontal and the normal reaction on the body. The magnitude of is given by

The limiting friction between the body and the plane has a magnitude .

The maximum force parallel to the plane, , that can act on a body in equilibrium on a rough inclined plane is a force that satisfies the condition and has a magnitude given by

The minimum force parallel to the plane, , that can act on a body in equilibrium on a rough inclined plane is a force that satisfies the condition and has a magnitude given by

Both and act upward parallel to the plane. The direction of the frictional force on the body is opposite when and act, as shown in the following figure.

Let us now look at an example where an applied force acts on a body in equilibrium on a rough inclined surface, the applied force does not act parallel to the surface, and the limiting friction acts in alternate directions.

### Example 5: Finding the Maximum Force Required to Keep a Body in a State of Equilibrium on a Rough Inclined Plane

A body weighing 75 N rests on a rough plane inclined at an angle of to the horizontal under the action of a horizontal force. The minimum horizontal force required to maintain the body in a state of equilibrium is 45 N. Find the maximum horizontal force that will also maintain the equilibrium.

It is helpful in answering this question to use a free-body diagram, as shown in the following figure showing the forces acting on the body when the horizontal force is minimum.

The 45 N and 75 N forces can be resolved into their components parallel and perpendicular to the plane. As each component of each force has a magnitude equal to the magnitude of the force divided by . The parallel- and perpendicular-to-the-plane components of the forces acting on the body are shown in the following figure.

Equating forces perpendicular to the plane, we see that the magnitude of the normal reaction force is given by

Equating forces parallel to the plane, we see that

Substituting the value of obtained, we find that

Having determined , the magnitude of the maximum horizontal force that maintains equilibrium can be determined. The following figure shows the forces acting on the body when the horizontal force is maximum. The components of the horizontal force parallel and perpendicular to the plane are shown as dashed red arrows.

The perpendicular- and parallel-to-the-plane components of the forces acting on the body when the maximum horizontal force acts are shown in the following figure.

Equating forces perpendicular to the plane, we see that

Equating forces parallel to the plane, we see that

As we have determined that is , we have that

Substituting the expression for obtained, we find that

All the terms in the expression can be multiplied by the square root of 2 to give us

Expanding the bracketed term, we have

Let us now summarize what has been learned in these examples.

### Key Points

• For a body in equilibrium on a rough inclined surface that is acted on only by the weight of and the normal reaction force on the body, it is the case that where is the angle the surface makes with the horizontal, is the coefficient of static friction between the body and the surface, is the mass of the body, and is acceleration due to gravity.
• When an applied force acts on a body in equilibrium on a rough inclined plane, parallel to the plane, the magnitude of the friction must equal the sum of the net force on the body due to its weight and the normal reaction on it and the applied force.
• A force that acts on a body in equilibrium on a rough inclined surface that does not act parallel to the surface changes the normal reaction force and frictional force on the body. Changing the normal reaction force necessitates that the perpendicular components of the applied force be equated with perpendicular components of the weight of the body, the normal reaction force on the body, and the frictional force on the body to determine any of these forces other than the weight of the body.
• The magnitude of the resultant of the weight of a body on a rough inclined plane and the normal reaction on the body can be greater than the limiting friction between the body and the plane. If no external force acts on the body, the body cannot be in equilibrium.
• An external force acting upward parallel to the plane can maintain the equilibrium of the body. There are two possible values for the magnitude of this force. These are given by and where is the magnitude of the resultant of the weight of the body and the normal reaction force on it and is the magnitude of the limiting friction.