In this explainer, we will learn how to study the motion of a particle on horizontal and inclined rough planes against friction force.
A body that is in equilibrium on a horizontal surface has zero resultant force acting on it. Two forces act on the body: its weight, , and the normal reaction force from the surface, , as shown in the following figure.
The magnitude of is given by where is the mass of the body and is the acceleration due to gravity.
The force acts vertically downward. The force acts normally to the surface. For a body on a horizontal surface, acts vertically upward. For an unsuspended body in equilibrium on a horizontal surface, the magnitudes of and are equal.
A horizontal force can act on a body. If the magnitude of is below or equal to a certain value, known as the limiting friction, the body will remain at rest. This is due to a static frictional force, , that acts in the opposite direction to . When the magnitude of is greater than that of the limiting friction, however, the body will accelerate in the direction of .
The magnitude of the frictional force, , between a body moving along a surface and the surface that it moves along is given by where is called the coefficient of kinetic friction between the body and the surface.
This is very similar to the coefficient of static friction but applies to moving bodies.
For a horizontal surface, and so
Let us look at an example where the coefficient of kinetic friction between a body and a horizontal surface is determined.
Example 1: Calculating the Coefficient of Friction for a Body Moving on a Rough Horizontal Plane
A body of mass 28 kg moves on a horizontal plane with an acceleration of magnitude 2.2 m/s2. It is subjected to an applied force of 155 N whose direction is given in the figure. Calculate the coefficient of kinetic friction rounding the result to the nearest two decimal places. Take the acceleration due to gravity .
Answer
The following figure shows the forces acting on the body.
The frictional force on the body depends on the normal reaction force on the body, . The normal reaction force on the body depends on the force that the body exerts on the surface. The force that the body exerts on the surface is due to its weight. However, some component of the 155 N applied force acting on the body acts in the opposite direction to the weight of the body, vertically upward.
Taking the force vertically downward as positive, the sum of the weight of the body and the vertically upward component of the applied force has a magnitude given by
The magnitude of is therefore
Using the formula we see that the frictional force has a magnitude of
The magnitude of the frictional force can be determined from those of the horizontal component of the applied force on the body and the net force on the body.
The magnitude of the net force on the body is the product of the mass of the body and the magnitude of acceleration of the body; hence,
Taking the direction of motion to be the positive direction, the net force on the body is given by
Therefore,
Knowing the frictional force, we can find the coefficient of friction, :
The value of can be determined by making it the subject of the equation:
To two decimal places, this is 0.29.
A frictional force is also exerted on a body moving across a rough surface. A frictional force is exerted on a body at rest on a rough surface. The coefficient of kinetic friction, , and the coefficient of static friction, , are related by the following inequality:
The frictional force on a body on a rough surface is given by
If we had then for a horizontal force with a magnitude of , applied to a body such that slightly exceeds and the body consequently starts to move, it would be the case that meaning that the body would accelerate from rest in the opposite direction to the direction of the applied force. This is never observed to occur. The coefficient of kinetic friction is always less than the coefficient of static friction.
The relationship between coefficients of static and kinetic friction can be intuitively appreciated from the perceived difficulty of pushing an object that is at rest to the point at which it begins to move compared to the perceived difficulty of pushing the object to maintain its motion.
Let us now look at an example involving both static and kinetic coefficients of friction.
Example 2: Calculations Involving Static and Kinetic Friction on a Rough Horizontal Plane
A body of mass kg is placed on a rough horizontal plane. The coefficient of static friction between the body and the plane is and the coefficient of kinetic friction is . A force is acting on the body where its line of action makes an angle of to the horizontal. The force causes the body to be on the point of moving. If the magnitude of the force is increased from to , the body would start moving and accelerate at m/s2. Find and . Take .
Answer
The forces acting on the body are shown in the following figure.
When the body is in equilibrium with the force acting on it, the magnitude of the frictional force on the body equals the horizontal component of . We have, therefore, that
The body is on the point of moving, so it is also the case that where is the magnitude of the normal reaction force on the body.
An expression for can be found by rearranging the expression for the frictional force expressed in terms of as follows: does not equal the weight of the body in this case, as has an upward vertical component that decreases the force that acts on the surface that supports the body, decreasing the normal reaction force on the body.
The vertical forces that act on the body are the weight of the body downward, the reaction on the body upward, and the vertical component of upward. Equating the vertical downward forces with the vertical upward forces gives us
Substituting the expression for in terms of , we find that
When increases in magnitude to , the normal reaction force changes magnitude as the upward force on the body increases, as shown in the following figure.
Equating the vertical downward forces with the vertical upward forces on the body, we find that can be expressed in terms of by rearranging as follows:
When increases in magnitude to , the body accelerates horizontally at m/s2.
Due to Newtonβs second law of motion, ; the magnitude of the horizontal (net) force on the body is now given by
Also, the frictional force is given by
The magnitude of the net horizontal force equals the magnitude of the horizontal component of minus the magnitude of the frictional force on the body; hence,
This expression can be rearranged to determine the horizontal component of , given by
Substituting the value of obtained, we find that
If the body is on an inclined plane, then the normal reaction force, , on the body is shown in the following figure.
If there is no motion in the direction perpendicular to the plane, then the magnitude of the reaction force equals the component of the weight perpendicular to the plane, : where is the angle from the horizontal at which the surface slopes.
The resultant of and the weight of the body is shown in the following figure as the force .
The direction of is downward along the surface. The following figure shows how the magnitude of can be determined.
We can see that
If the inclined surface is rough, then a frictional force acts on the body parallel to the surface, in the opposite direction to the net force on the body.
Let us now look at an example where a body moves along a rough inclined surface.
Example 3: Calculations Involving a Body on a Rough Inclined Plane
A body of mass kg was placed on a plane inclined at to the horizontal. A force of magnitude N was acting on the body along the line of greatest slope up the plane. As a result, the body accelerated uniformly at m/s2 up the plane. If the magnitude of the force acting on the body is halved while maintaining its original direction, the body will move down the plane at m/s2. Given that the resistance of the plane to the bodyβs movement is N in both cases, determine the values of and , rounding the results to the nearest two decimal places. Take .
Answer
The following figure shows the forces acting on the body before the change in the magnitude of the applied force on the body. The acceleration of the body is shown in red.
When the body is moving upward parallel to the plane, the magnitude of the net force, which is parallel to the plane, is given by
The following figure shows the forces acting on the body after the change in the magnitude of the applied force on the body. The acceleration of the body is shown in red.
When the body is moving downward parallel to the plane, the frictional force acts upward parallel to the plane as the frictional force acts in the opposite direction to the motion of the object. The magnitude of the net force acting downward parallel to the plane is given by
In both cases, we are told that the acceleration is the same, . As , this means that the upward and downward forces have the same magnitude but act in opposite directions; therefore, we have
The following is added to both sides of the equation: from which we obtain
We have seen that
Substituting the value for obtained, we have that
The value of is given by
To two decimal places, this 1.41 m/s2.
Let us summarize what we have learned in these examples.
Key Points
- The magnitude of the frictional force, , between a body moving along a surface and the surface that it moves along is given by where is the coefficient of kinetic friction between the body and the surface.
- The force due to kinetic friction on a moving body acts in the opposite direction to the motion of the body.
- The force due to static friction on a body at rest acts in the opposite direction to the net force on the body.
- The coefficient of kinetic friction, , and the coefficient of static friction, , are related by the following inequality:
- A body on an inclined plane has a net force on it due to its weight and the reaction force on it given by where is the mass of the body, is the acceleration due to gravity, and is the angle that the plane makes with the horizontal.
- When a body moving along an inclined plane is subjected to an applied force that is not parallel to the plane, the sum of the component of the applied force perpendicular to the plane, the weight of the body, and the normal reaction force is zero. The reaction force, therefore, has a different magnitude than it would have if the applied force acted parallel to the plane.
