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

Video Transcript

In this video, we will learn how to solve problems on the equilibrium of a body on a rough inclined plane. We will begin by recalling some key definitions.

In many problems in mathematics, it is assumed that the surface is smooth. This means there will be no frictional force, whereas when dealing with a rough surface, there will be some frictional resistance. When the frictional force is at its maximum, friction is said to be limiting. If the body at this point is still stationary, it is said to be in limiting equilibrium. At this point, the sum of the forces parallel to the plane and those perpendicular to the plane are equal to zero.

The coefficient of friction, denoted by the Greek letter 𝜇, can be calculated by dividing the frictional force by the normal reaction force. This means that the frictional force 𝐹 r is equal to 𝜇 multiplied by the normal reaction force 𝑅. The normal reaction force is sometimes written as 𝑁 or 𝐹 sub n. The value of the coefficient of friction 𝜇 lies between zero and one inclusive. We will now model what this will look like in a free body diagram and resolve parallel and perpendicular to the plane.

Let’s consider a body resting on an inclined plane as shown in the diagram. The angle of inclination is 𝛼. If the body is of mass 𝑚 kilograms, there will be a vertical downward force equal to its weight. This will be the mass multiplied by gravity. In this video, we will assume that 𝑔 is equal to 9.8 meters per second squared. The normal reaction force 𝑅 will be perpendicular to the plane as shown. If the body is in limiting equilibrium and is on the point of sliding down the plane, there will be a frictional force 𝐹 r parallel to the plane acting in an upward direction. When a body is in limiting equilibrium, the sum of the net forces is equal to zero. We can therefore resolve parallel and perpendicular to the plane in the knowledge that the sum of the forces will equal zero. We may sometimes see the shorthand notation for parallel and perpendicular.

As the weight force is acting vertically downwards, we firstly need to find the components of this parallel and perpendicular to the plane. We do this using our knowledge of right angle trigonometry. We know that the sin of angle 𝜃 is equal to the opposite over the hypotenuse and the cos of angle 𝜃 is equal to the adjacent over the hypotenuse. We have labeled the side opposite angle 𝛼 𝑥, and the side adjacent is 𝑦. Therefore, sin 𝛼 is equal to 𝑥 over 𝑚𝑔 and cos of 𝛼 is equal to 𝑦 over 𝑚𝑔. We can then multiply both sides of both of these equations by 𝑚𝑔. This means that the component of the weight force parallel to the plane is equal to 𝑚𝑔 sin 𝛼 and that perpendicular to the plane is equal to 𝑚𝑔 cos 𝛼.

Resolving parallel to the plane with downwards being the positive direction gives us 𝑚𝑔 sin 𝛼 minus 𝐹 r is equal to zero. Resolving perpendicular to the plane, the sum of the net forces is equal to 𝑅 minus 𝑚𝑔 cos 𝛼. This is also equal to zero. We also know that in limiting equilibrium, the frictional force 𝐹 r is equal to 𝜇 multiplied by the normal reaction force. We will now look at some questions where we need to resolve parallel and perpendicular to the plane as well as using the equation 𝐹 r is equal to 𝜇𝑅 to calculate unknowns.

A body weighing 60 newtons rests on a rough plane inclined to the horizontal at an angle whose sin is three-fifths. The body is pulled upward by a force of 63 newtons acting parallel to the line of 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.

We will begin by sketching a diagram to model the problem. We are told that the body weighs 60 newtons. Therefore, there will be a downward vertical force equal to 60 newtons. The plane is inclined to the horizontal at an angle 𝛼, where sin of 𝛼 is equal to three-fifths. We are told that the body is pulled upwards with a force of 63 newtons acting parallel to the line of greatest slope. There will be a normal reaction force 𝑅 acting perpendicular to the plane. And as the body is on the point of moving up the plane, there will be a frictional force parallel to the plane acting downwards.

The body is in limiting equilibrium, so the sum of the net forces will equal zero. There is no acceleration. We will need to resolve parallel and perpendicular to the plane. Before doing this, we will need to find the component of the weight force acting parallel and perpendicular to the plane. We will do this using our knowledge of right angle trigonometry. We know that the sine of an angle is equal to the opposite over the hypotenuse and the cosine is equal to the adjacent over the hypotenuse. This means that the force acting parallel to the plane is equal to 60 sin 𝛼, and the force acting perpendicular to the plane is equal to 60 cos 𝛼.

By considering a three, four, five Pythagorean triple, we know that if sin 𝛼 is equal to three-fifths, then cos 𝛼 is equal to four-fifths. Resolving parallel to the plane, we have 63 minus 60 sin 𝛼 minus the frictional force is equal to zero. Resolving perpendicular to the plane, we have 𝑅 minus 60 cos 𝛼 equals zero. We can now solve these two equations to calculate the frictional force and the normal reaction force. Adding 𝐹 r to both sides of equation one and replacing sin 𝛼 with three-fifths, we get 63 minus 60 multiplied by three-fifths is equal to 𝐹 r. The left-hand side simplifies to 63 minus 12 multiplied by three, which is equal to 27. The frictional force is equal to 27 newtons. Substituting in our value for cos 𝛼, equation two becomes 𝑅 is equal to 60 multiplied by four-fifths. This gives us 𝑅 is equal to 48. The normal reaction force is 48 newtons.

We are asked to calculate the coefficient of friction. And we know that the frictional force is equal to 𝜇, the coefficient of friction, multiplied by the normal reaction force. Substituting in our values gives us 27 is equal to 𝜇 multiplied by 48. Dividing both sides of this equation by 48 gives us 𝜇 is equal to 27 over 48. As both the numerator and denominator are divisible by three, this simplifies to nine over 16 or nine sixteenths. This is the coefficient of friction between the body and the plane.

In our next question, we will need to calculate the tension in a string that is attached to a body on an inclined plane.

A body weighing 56 newtons rests on a rough plane inclined at an angle of 30 degrees to the horizontal. The coefficient of friction between the body and the plane is root three over six. The body is pulled upward by a string making an angle of 30 degrees to the line of greatest slope of the plane. Determine the maximum tension in the string required to cause the body to be on the point of moving up the plane.

We will begin by drawing a free body diagram. The angle of the plane is 30 degrees, and the body has weight 56 newtons. Therefore, this force will act vertically downwards. There will be a normal reaction force 𝑅 acting perpendicular to the plane and a frictional force 𝐹 r acting parallel to the plane in a downward direction, as the body is on the point of moving up the plane. We are also told that the body is pulled upward by a string acting at an angle of 30 degrees to the line of greatest slope. As the body is on the point of moving, we know that the sum of the net forces is equal to zero. There is no acceleration, and we will need to resolve parallel and perpendicular to the plane.

Both the tension and weight force will have components in these directions. We will need to use our knowledge of right angle trigonometry to calculate these. As the sine of an angle is equal to the opposite over the hypotenuse and the cosine of an angle is equal to the adjacent over the hypotenuse, the four forces will be as shown. We have three forces acting parallel to the plane. If we consider up the plane to be the positive direction, we have 𝑇 cos 30 minus 56 sin 30 minus 𝐹 r is equal to zero. There are also three forces acting perpendicular to the plane, giving us the equation 𝑅 plus 𝑇 sin 30 minus 56 cos 30 is equal to zero.

We can now simplify these two equations using the fact that sin of 30 degrees is one-half and cos of 30 degrees is root three over two. Equation one becomes root three over two 𝑇 minus 28 minus 𝐹 r is equal to zero. And equation two becomes 𝑅 plus a half 𝑇 minus 28 root three is equal to zero. We also know that the frictional force 𝐹 r is equal to 𝜇, the coefficient of friction, multiplied by the normal reaction force 𝑅. The coefficient of friction is root three over six, so 𝐹 r is equal to root three over six multiplied by 𝑅. We will now clear some space so we can solve the equations to calculate the maximum tension 𝑇.

We can rearrange equation two so that 𝑅 is equal to 28 root three minus a half 𝑇. The frictional force in equation one is equal to root three over six multiplied by this. This gives us root three over two 𝑇 minus 28 minus root three over six multiplied by 28 root three minus a half 𝑇 is equal to zero. Distributing our parentheses gives us negative 14 plus root three over 12 𝑇. We can then collect like terms. This gives us seven root three over 12 𝑇 minus 42 is equal to zero. We can then add 42 to both sides of this equation. Finally, we can divide both sides by seven root three over 12 so that 𝑇 is equal to 24 root three. The maximum tension in the string required to cause the body to be on the point of moving up the plane is 24 root three newtons.

In our last question, we will calculate the maximum force required to maintain equilibrium.

A body weighing 75 newtons rests on a rough plane inclined at an angle of 45 degrees 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 newtons. Find the maximum horizontal force that will also maintain the equilibrium.

In this question, we are looking at two different scenarios: when the force is at its minimum and maximum. In both of these situations, the body is in equilibrium. Therefore, the sum of the net forces must equal zero. We have a vertical downward force of 75 newtons equal to the weight. The plane is inclined at an angle of 45 degrees. When we have the minimum horizontal force equal to 45 newtons, the body will be on the point of sliding down the plane. This means that the frictional force will act up the plane.

We will resolve parallel and perpendicular to the plane. Before doing this, we will need to work out the components of the 75- and 45-newton forces in these directions. We do this using right angle trigonometry, giving us the four forces shown. Parallel to the plane, we have three forces. As the body is on the point of sliding down the plane, we will use this as the positive direction. This gives us the equation 75 sin 45 minus 45 cos 45 minus 𝐹 r is equal to zero. The sin of 45 degrees and cos of 45 degrees are both equal to root two over two. Adding the frictional force to both sides of this equation gives us 𝐹 r is equal to 15 root two.

There are also three forces acting perpendicular to the plane. This gives us the equation 𝑅 minus 75 cos 45 minus 45 sin 45 is equal to zero. This simplifies to 𝑅 minus 60 root two equal zero, which means that 𝑅 is equal to 60 root two. We now have the frictional force and normal reaction force in newtons. And we know that 𝐹 r equals 𝜇, the coefficient of friction, multiplied by 𝑅. 𝜇 is equal to 15 root two divided by 60 root two. This simplifies to one-quarter.

We now need to consider the case where the horizontal force is at its maximum. In this situation, most of our forces remain the same. However, the body will be on the point of moving up the plane, so the frictional force acts downwards. We need to calculate the unknown force 𝐹. Resolving parallel to the plane, we have 𝐹 cos 45 minus 75 sin 45 minus 𝐹 r is equal to zero. Simplifying this and using the fact that 𝜇 equals one-quarter gives us root two over two 𝐹 minus 75 root two over two minus one-quarter 𝑅 is equal to zero. In the perpendicular direction, we have 𝑅 minus 75 cos 45 minus 𝐹 sin 45 is equal to zero. Simplifying this, we get 𝑅 is equal to 75 root two over two plus root two over two 𝐹.

We now have two simultaneous equations that we can solve to calculate 𝐹. We will replace the value of 𝑅 in equation two into equation one. We can then divide through this whole equation by root two over two. Distributing our parentheses gives us 𝐹 minus 75 minus 75 over four minus a quarter 𝐹 is equal to zero. This in turn simplifies to three-quarters 𝐹 is equal to 375 over four. Dividing both sides of the equation by three-quarters gives us 𝐹 is equal to 125. The maximum force is equal to 125 newtons. This means that if the force is between 45 and 125 newtons inclusive, the body will remain in equilibrium.

We will now summarize the key points from this video. In all three questions, we saw that when a body is in limiting equilibrium, the sum of the forces parallel and perpendicular to the plane are equal to zero. We can resolve parallel and perpendicular to the plane as well as using the equation 𝐹 r is equal to 𝜇𝑅 to calculate any unknowns. 𝜇 is the coefficient of friction and lies between zero and one inclusive. We can also use right angle trigonometry to calculate the relevant components.