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, 𝑅, 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, 𝑊, is given by 𝑊=𝑚𝑔, where 𝑚 is the mass of the body and 𝑔 is acceleration due to gravity.

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

The resultant force on a body on an inclined plane, 𝐹, is the sum of 𝑅 and 𝑊. We can express this as 𝐹=𝑅+𝑚𝑔.

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

where 𝑅=𝑚𝑔𝜃.cos

The direction of 𝐹 is downward, parallel to the plane.

We see that 𝐹 and 𝑅 are perpendicular components of 𝑚𝑔; hence, 𝐹=𝑚𝑔𝜃.sin

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 𝐹. 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, 𝐿, on a body at rest on a surface is given by 𝐿=𝜇𝑅, where 𝜇 is the coefficient of static friction between the body and the surface and 𝑅 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 𝑚𝑔𝜃=𝑚𝑔𝜇𝜃sincos and hence that sincostan𝜃𝜃=𝑚𝑔𝜇𝑚𝑔=𝜇=𝜃.

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

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 3, find the magnitude of the limiting friction.

Answer

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, 𝐹. The weight of the body is equal to 𝑚𝑔, and the reaction force has a magnitude of 𝑅, given by 𝑅=𝑚𝑔60.cos

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

𝐹=𝑚𝑔60.sin

Knowing that the force acting downward parallel to the plane is given by 𝐹=𝑚𝑔60,sin we have that 𝐹=46(60)=4632=233.sinN

The frictional force must have a magnitude equal to 𝐹 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 𝐿=𝜇𝑅=3𝑅.

And 𝑅 is given by 𝑅=46(60)=23.cosN

Hence, 𝐿=𝜇𝑅=233.N

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 𝐹 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 𝐹, the frictional force on the body is greater if 𝐹 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 𝐹. 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 35. 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.

Answer

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

The force equal to 𝑚𝑔𝜃sin 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 60𝜃+60𝜇𝜃=63.sincos

We are told that sin𝜃=35; hence, a section of the surface of length 5𝑙 is the hypotenuse of a right triangle with legs of lengths 3𝑙 and 𝑥𝑙, as shown in the following figure.

The value of 𝑥 can be found using the Pythagorean theorem as follows: 𝑥=53=16𝑥=16=4.

From this, we can see that cos𝜃=45.

The values of the sine and cosine of 𝜃 can be substituted into the equation 60𝜃+60𝜇𝜃=63sincos to give us 6035+60𝜇45=6336+48𝜇=6348𝜇=27𝜇=2748=916.

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 𝑅=𝑚𝑔𝜃.cos

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 𝑅=𝑚𝑔𝜃𝐹𝜙,cossin where 𝐹 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 30 to the horizontal. The coefficient of friction between the body and the plane is 36. The body is pulled upward by a string making an angle of 30 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.

Answer

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

The frictional force, 𝐹, acts parallel to the plane and the reaction force, 𝑅, acts normally to the plane. The tension, 𝑇, acts at an angle of 30 from the plane and the 56 N weight of the body acts at an angle of 30 from the line normal to the plane. The components of 𝑇 parallel and perpendicular to the plane, 𝑇 and 𝑇, are shown in the following figure.

The angle that 𝑇 makes with the plane is 30, so we see that 𝑇=𝑇(30)=𝑇2.sin

The component of 𝑇 perpendicular to the plane acts in the direction of 𝑅 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, 𝐹, has zero component perpendicular to the plane.

Taking the direction of 𝑅 as positive perpendicular to the plane, the component of the weight of the body perpendicular to the plane has a magnitude given by 𝑊=56(30)=283.cosN

The magnitude of the tension in the string perpendicular to the plane is given by 𝑇=𝑇(30)=𝑇2.sin

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 𝑇2+𝑅=283.

From this, we can obtain an expression for the magnitude of 𝑅: 𝑅=283𝑇2.

The angle that 𝑇 makes with the plane is 30, so, taking the upward parallel to the plane as positive, we see that 𝑇=𝑇(30)=𝑇23.cos

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

The component of 𝑇 parallel to the plane acts in the opposite direction to 𝐹 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 𝑊=56(𝜃).sin

The angle 𝜃 is 30, so 𝑊=56(30)=28.sinN

The magnitude of 𝐹 depends on the magnitude of 𝑅: 𝐹=𝜇𝑅=36𝑅.

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 𝑇23=28+36𝑅.

It has been shown that 𝑅=283𝑇2.

Substituting this expression for 𝑅 into the equation for the forces acting parallel to the plane, we find that 𝑇23=28+36283𝑇2.

This expression can be rearranged to make 𝑇 the subject as follows. 𝑇23=28+2836312𝑇𝑇23=42𝑇1236𝑇123=42𝑇123𝑇7123=42𝑇3=72𝑇=243.N

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 34, it was on the point of moving. The same body was later placed on a horizontal surface of the same roughness. A force 𝐹 acted on the body pulling upward at an angle of 𝜃 to the horizontal where sin𝜃=35. Given that under these conditions the body was on the point of moving, determine the magnitude of 𝐹 and the normal reaction 𝑅.

Answer

There are two distinct sections involved in determining 𝐹 and 𝑅. 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 𝜃, we are told that tan𝜃=34.

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, 𝜇=34.

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

If we let the angle made by 𝐹 with the surface be 𝜙, we are told that sin𝜙=35; hence, cos𝜙=45.

The magnitudes of the vertical and the horizontal components of 𝐹, 𝐹V and 𝐹H, are given by 𝐹=35𝐹V and 𝐹=45𝐹.H

Equating vertical forces on the body gives us 35𝐹+𝑅=262.5.

From this, we can see that 𝑅=262.535𝐹.

Equating horizontal forces on the body gives us 45𝐹=34𝑅=34262.535𝐹.

This equation can be rearranged to make 𝐹 the subject as follows: 45𝐹=196.875920𝐹1620𝐹=196.875920𝐹2520𝐹=54𝐹=196.875𝐹=196.87545=157.5.N

As we know that 𝑅=262.535𝐹, we can see that 𝑅=262.535157.5=168.N

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 𝐹 that is equal to the resultant of the weight, 𝑊, of a body on a rough inclined plane at an angle 𝜃 from the horizontal and the normal reaction on the body. The magnitude of 𝐹 is given by 𝐹=𝑊𝜃.sin

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

The maximum force parallel to the plane, 𝐹max, that can act on a body in equilibrium on a rough inclined plane is a force that satisfies the condition 𝐹>𝐹max and has a magnitude given by 𝐹=𝐹+𝐿.max

The minimum force parallel to the plane, 𝐹min, that can act on a body in equilibrium on a rough inclined plane is a force that satisfies the condition 𝐹<𝐹min and has a magnitude given by 𝐹=𝐹𝐿.min

Both 𝐹max and 𝐹min act upward parallel to the plane. The direction of the frictional force on the body is opposite when 𝐹max and 𝐹min 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 45 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.

Answer

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 sincos45=45=12, each component of each force has a magnitude equal to the magnitude of the force divided by 2. 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 𝑅=452+752=1202.

Equating forces parallel to the plane, we see that 𝑅𝜇+452=752.

Substituting the value of 𝑅 obtained, we find that 1202𝜇+452=7521202𝜇=302𝜇=30120=14.

Having determined 𝜇, 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 𝑅=(𝐹+75)2.max

Equating forces parallel to the plane, we see that 𝐹2=𝑅𝜇+752.max

As we have determined that 𝜇 is 4, we have that 𝐹2=𝑅4+752.max

Substituting the expression for 𝑅 obtained, we find that 𝐹2=𝐹+7542+752.maxmax

All the terms in the expression can be multiplied by the square root of 2 to give us 𝐹=𝐹+754+75.maxmax

Expanding the bracketed term, we have 𝐹=𝐹4+754+7534𝐹=5475𝐹=544375=5375=125.maxmaxmaxmaxN

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 𝑚𝑔𝜃=𝑚𝑔𝜇𝜃,sincos where 𝜃 is the angle the surface makes with the horizontal, 𝜇 is the coefficient of static friction between the body and the surface, 𝑚 is the mass of the body, and 𝑔 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 𝐹=𝐹+𝐿max and 𝐹=𝐹𝐿,min where 𝐹 is the magnitude of the resultant of the weight of the body and the normal reaction force on it and 𝐿 is the magnitude of the limiting friction.

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