Lesson Explainer: Motion of a Body on a Rough Plane | Nagwa Lesson Explainer: Motion of a Body on a Rough Plane | Nagwa

Lesson Explainer: Motion of a Body on a Rough Plane Mathematics • Third Year of Secondary School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

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, 𝑊, and the normal reaction force from the surface, 𝑅, as shown in the following figure.

The magnitude of 𝑊 is given by𝑊=𝑚𝑔, where 𝑚 is the mass of the body and 𝑔 is the acceleration due to gravity.

The force 𝑊 acts vertically downward. The force 𝑅 acts normally to the surface. For a body on a horizontal surface, 𝑅 acts vertically upward. For an unsuspended body in equilibrium on a horizontal surface, the magnitudes of 𝑊 and 𝑅 are equal.

A horizontal force 𝐻 can act on a body. If the magnitude of 𝐻 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, 𝐹, that acts in the opposite direction to 𝐻. When the magnitude of 𝐻 is greater than that of the limiting friction, however, the body will accelerate in the direction of 𝐻.

The magnitude of the frictional force, 𝐹friction, between a body moving along a surface and the surface that it moves along is given by 𝐹=𝜇𝑅,friction where 𝜇 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, 𝑅=𝑚𝑔, and so 𝐹=𝑚𝑔𝜇.friction

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 𝑔=9.8/ms.

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, 𝑅. 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 𝐹=𝑚𝑔155(45)=28(9.8)1552=𝑅.surfacesin

The magnitude of 𝑅 is therefore 𝑅=274.41552=164.8.N

Using the formula 𝐹=𝜇𝑅,friction we see that the frictional force has a magnitude of 𝐹=𝜇×164.8.frictionN

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, 𝐹=28×2.2=61.6.netN

Taking the direction of motion to be the positive direction, the net force on the body is given by 155(45)𝐹=𝐹.cosfrictionnet

Therefore,𝐹=155261.6=48.frictionN

Knowing the frictional force, we can find the coefficient of friction, 𝜇: 48=𝜇×164.8.

The value of 𝜇 can be determined by making it the subject of the equation: 𝜇=48164.8.

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, 𝜇k, and the coefficient of static friction, 𝜇s, are related by the following inequality:𝜇>𝜇.sk

The frictional force on a body on a rough surface is given by 𝐹=𝜇𝑅.friction

If we had 𝜇>𝜇,ks then for a horizontal force with a magnitude of 𝐹, applied to a body such that 𝐹 slightly exceeds 𝜇𝑅s and the body consequently starts to move, it would be the case that 𝐹𝜇𝑅<0,k 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 113 kg is placed on a rough horizontal plane. The coefficient of static friction between the body and the plane is 33 and the coefficient of kinetic friction is 37. A force is acting on the body where its line of action makes an angle of 60 to the horizontal. The force causes the body to be on the point of moving. If the magnitude of the force is increased from 𝐹 to 𝐹, the body would start moving and accelerate at 533 m/s2. Find 𝐹 and 𝐹. Take 𝑔=9.8/ms.

Answer

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

When the body is in equilibrium with the force 𝐹 acting on it, the magnitude of the frictional force on the body equals the horizontal component of 𝐹. We have, therefore, that𝐹=𝐹(60)=𝐹2.frictioncos

The body is on the point of moving, so it is also the case that 𝐹=𝜇𝑅=33𝑅,frictions where 𝑅 is the magnitude of the normal reaction force on the body.

An expression for 𝑅 can be found by rearranging the expression for the frictional force expressed in terms of 𝐹 as follows: 𝐹2=33𝑅𝑅=𝐹323.𝑅 does not equal the weight of the body in this case, as 𝐹 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 𝐹 upward. Equating the vertical downward forces with the vertical upward forces gives us 113×9.8=𝑅+𝐹60=𝑅+32𝐹.sin

Substituting the expression for 𝑅 in terms of 𝐹, we find that 107.83=323𝐹+32𝐹107.83=32𝐹+32𝐹107.83=𝐹3𝐹=107.8.N

When 𝐹 increases in magnitude to 𝐹, 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 107.83=𝑅+32𝐹.𝑅 can be expressed in terms of 𝐹 by rearranging as follows:𝑅=107.8332𝐹.

When 𝐹 increases in magnitude to 𝐹, the body accelerates horizontally at 533 m/s2.

Due to Newton’s second law of motion, 𝐹=𝑚𝑎; the magnitude of the horizontal (net) force on the body is now given by 113×533.N

Also, the frictional force is given by 37𝑅.

The magnitude of the net horizontal force equals the magnitude of the horizontal component of 𝐹 minus the magnitude of the frictional force on the body; hence,113×533=𝐹237𝑅.

This expression can be rearranged to determine the horizontal component of 𝐹, given by 𝐹2=55+37𝑅.

Substituting the value of 𝑅 obtained, we find that 𝐹2=55+37107.8×332𝐹𝐹2=55+17107.8×332𝐹𝐹2=55+17323.432𝐹𝐹2=55+323.47314𝐹𝐹2+314𝐹=55+323.47𝐹714+314=55+46.2=101.2𝐹=1410101.2=141.68.N

If the body is on an inclined plane, then the normal reaction force, 𝑅, 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 𝑅 equals the component of the weight perpendicular to the plane, 𝑚𝑔𝜃cos:𝑅=𝑚𝑔𝜃,cos where 𝜃 is the angle from the horizontal at which the surface slopes.

The resultant of 𝑅 and the weight of the body is shown in the following figure as the force 𝐹.

The direction of 𝐹 is downward along the surface. The following figure shows how the magnitude of 𝐹 can be determined.

We can see that 𝐹=𝑚𝑔𝜙=𝑚𝑔𝜃.cossin

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 𝑚 kg was placed on a plane inclined at 45 to the horizontal. A force of magnitude 3922 N was acting on the body along the line of greatest slope up the plane. As a result, the body accelerated uniformly at 𝑎 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 𝑎 m/s2. Given that the resistance of the plane to the body’s movement is 382 N in both cases, determine the values of 𝑚 and 𝑎, rounding the results to the nearest two decimal places. Take 𝑔=9.8/ms.

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 𝐹=3922382𝑚𝑔(45)𝐹=3922382𝑚𝑔2.upupsin

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 by𝐹=𝑚𝑔21962382.down

In both cases, we are told that the acceleration is the same, 𝑎. As 𝐹=𝑚𝑎, this means that the upward and downward forces have the same magnitude but act in opposite directions; therefore, we have 𝐹=𝐹,3922382𝑚𝑔2=𝑚𝑔21962382.updown

The following is added to both sides of the equation:1962+382+𝑚𝑔2, from which we obtain 5882=2𝑚𝑔25882=2𝑚𝑔588=𝑚𝑔𝑚=588𝑔=5889.8=60.kg

We have seen that 𝐹=3922382𝑚𝑔2.up

Substituting the value for 𝑚 obtained, we have that 𝐹=35425882=602.up

The value of 𝑎 is given by 𝑎=𝐹𝑚=60260=2,

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, 𝐹friction, between a body moving along a surface and the surface that it moves along is given by 𝐹=𝜇𝑅,friction where 𝜇 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, 𝜇k, and the coefficient of static friction, 𝜇s, are related by the following inequality: 𝜇>𝜇.sk
  • A body on an inclined plane has a net force on it due to its weight and the reaction force on it given by 𝐹=𝑚𝑔𝜃,sin where 𝑚 is the mass of the body, 𝑔 is the acceleration due to gravity, and 𝜃 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.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy