Lesson Explainer: Motion of a Body on a Smooth Inclined Plane Mathematics

In this explainer, we will learn how to solve problems involving moving a particle on a smooth 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, π‘Š, is given by π‘Š=π‘šπ‘”, where π‘š is the mass of the body and 𝑔 is the 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=π‘šπ‘”πœƒ,cos where πœƒ is the angle from the horizontal at which the surface slopes.

Let us look at an example in which the reaction force on a body on an inclined surface is determined.

Example 1: Finding the Reaction Force of a Body Moving on an Inclined Plane

A body of mass 0.7 kg was placed on a smooth plane inclined at 66∘ to the horizontal and it was left to move freely under the effect of gravity, where the acceleration due to gravity is 9.8 m/s2. Find, to the nearest two decimal places, the magnitude of the reaction of the plane to the body.

Answer

The reaction force on the body can be found using the formula 𝑅=π‘šπ‘”πœƒ,cos which gives 𝑅=0.7(9.8)(66).cos

To two decimal places, the reaction force is 2.79 N.

The resultant force on a body on an inclined plane, 𝐹, is the sum of 𝑅 and π‘Š. We can express this as 𝐹=𝑅+π‘šπ‘”.

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

where |𝑅|=π‘šπ‘”πœƒ.cos

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

We see that 𝐹 and βˆ’π‘… are perpendicular components of π‘šπ‘”, and hence |𝐹|=π‘šπ‘”πœƒ.sin

Let us look at an example of a body on an inclined surface in which the acceleration of the body parallel to the surface is determined.

Example 2: Finding the Acceleration of a Body on an Inclined Plane

A body of mass 1.4 kg was placed on a smooth plane inclined 45∘ to the horizontal. If a force of 59 N is acting on the body upward along the line of greatest slope of the plane, determine the acceleration of the body rounded to the nearest two decimal places. Take 𝑔=9.8/ms.

Answer

If no external force acted on the body, the weight of and reaction on the body would result in a net force on the body downward parallel to the plane given by 𝐹=π‘šπ‘”πœƒ=1.4(9.8)(45)=13.72√2.sinsin

In addition to this force, a force of 59 newtons acts upward parallel to the plane, as shown in the following figure.

The resultant force on the body upward parallel to the plane is given by 𝐹=59βˆ’13.72√2.

The acceleration of the body upward parallel to the plane is given by Newton’s second law of motion, rearranged to make a the subject: π‘Ž=πΉπ‘š=59βˆ’1.4.οŠ§οŠ©οŽ–οŠ­οŠ¨βˆšοŠ¨

To two decimal places, the acceleration is 35.21 m/s2.

Now let us look at an example of a body on an inclined surface in which an applied force acts on the body in a direction not parallel to the surface.

Example 3: Finding a Missing Force Acting on a Body Moving on an Inclined Plane

A body of mass 205 kg was left to slide down a plane inclined at 45∘ to the horizontal. A force started acting on the body causing the magnitude of its acceleration to halve. Given that the line of action of the force made an angle of 45∘ with the line of greatest slope of the plane and that they both lie in the same vertical plane, find the magnitude of this force. Consider the acceleration due to gravity to be 9.8 m/s2.

Answer

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

The following figure shows the components of these forces acting parallel and perpendicularly to the plane.

The force at an angle of 45∘ to the plane has a vertical line of action. When this force does not act, the magnitude of the force on the body parallel to the slope is given by π‘šπ‘”(45)=π‘šπ‘”ο€Ώβˆš22.sin

When the vertical force acts on the body, the acceleration of the body is halved in magnitude. The acceleration of the body is halved due to the action of the vertical force. The net force, 𝐹, on a body of mass π‘š can be determined by Newton’s second law: 𝐹=π‘šπ‘Ž.

The magnitude of the net force on the body is, therefore, half that of the net force on the body before the vertical force acted and so is given by π‘šπ‘”ο€½ο‰2=π‘šπ‘”ο€Ώβˆš24.√

The magnitude of the resultant force acting perpendicularly to the plane when the vertical force is included, taking upward parallel to the plane as positive, can be expressed as βˆ’π‘šπ‘”ο€Ώβˆš24=𝐹(45)βˆ’π‘šπ‘”(45),cossin which gives βˆ’π‘šπ‘”ο€Ώβˆš24=πΉο€Ώβˆš22ο‹βˆ’π‘šπ‘”ο€Ώβˆš22.

This expression can be rearranged to make 𝐹 the subject: πΉο€Ώβˆš22=π‘šπ‘”ο€Ώβˆš22ο‹βˆ’π‘šπ‘”ο€Ώβˆš24.

The expression can be divided by √2: 𝐹2=π‘šπ‘”2βˆ’π‘šπ‘”4=π‘šπ‘”2.

The expression can be multiplied by 2: 𝐹=π‘šπ‘”2=205(9.8)2=1004.5.N

If 𝐹 is allowed to increase without limit, the body would eventually no longer slide down the plane as there would be a nonzero net force perpendicular to the plane; hence, the body would be lifted off the plane.

Let us now look at an example in which the acceleration of a body is deduced from its motion along a plane of unknown inclination.

Example 4: Using the Equations of Motion to Solve a Problem Involving a Body Moving on an Inclined Plane

A body of mass 9 kg is released from rest on a smooth inclined plane. It moves a distance of 25.2 m in the first 4 seconds of its motion. If the body is projected upward along a line of greatest slope on the same plane, with an initial velocity of 12.6 m/s, how far does it travel before coming to instantaneous rest? Take 𝑔=9.8/ms.

Answer

When the body is released from rest, it accelerates downward parallel to the plane. With this acceleration, the body is displaced by 25.2 metres in 4 seconds. Taking displacement upward parallel to the plane as positive, the acceleration required to produce a displacement of 25.2 metres in a time of 4 seconds can be determined using the formula 𝑠=𝑒𝑑+12π‘Žπ‘‘, where 𝑒=0/ms, 𝑑=4s, and 𝑠=βˆ’25.5m.

Rearranging to make a the subject, we obtain π‘Ž=2𝑠𝑑=βˆ’50.416=βˆ’3.15/.ms

When the body is projected upward parallel to the plane, it has an initial velocity in the opposite direction to its acceleration; hence, if the direction in which the body is projected is taken as positive, the displacement of the body is related to its acceleration and its final velocity by the equation 𝑣=𝑒+2π‘Žπ‘ , where 𝑣 is zero at the point where the body is instantaneously at rest. Rearranging to make 𝑠 the subject, we obtain 𝑠=βˆ’π‘’2π‘Ž.

Substituting known values, we find that 𝑠=βˆ’12.62(βˆ’3.15)=25.2.m

It is worth noting that the displacement was determined without knowing the angle of inclination of the plane or taking into account any of the forces acting on the body, as the body accelerates uniformly and so its motion can be modeled only using kinematic equations for constant acceleration.

Let us now look at an example involving modeling the motion of a body on an inclined plane that is acted on by a force that does not act parallel to the plane.

Example 5: Using the Equations of Motion to Solve a Problem Involving a Body Moving on an Inclined Plane

A body of mass 10√3 kg was placed on a smooth plane inclined at 30∘ to the horizontal. A horizontal force of 126 N directed toward the plane was acting on the body such that the line of action of the force and the line of greatest slope of the plane both lie in the same vertical plane. After moving for 7 seconds, the body reached a speed 𝑣, at which point the force stopped acting, and the body continued moving until it momentarily came to rest 𝑑 seconds after the force stopped acting. Find 𝑣 and 𝑑. Take 𝑔=9.8/ms.

Answer

The following figure shows the forces acting on the body.

Nonzero components of the weight of the body and of the 126-newton force exist parallel to the plane, whereas the component of the reaction force perpendicular to the plane is zero.

The following figure shows the components of these forces acting parallel and perpendicularly to the plane.

The magnitude of the net force upward parallel to the plane is given by the following: 𝐹=126(30)βˆ’10√3(9.8)(30)𝐹=126√32βˆ’10√3ο€Ό9.82𝐹=63√3βˆ’49√3=14√3.netnetnetcossin

The magnitude of the acceleration of the body due to 𝐹net is given by π‘Ž=14√310√3=1.4/.ms

The magnitude of the velocity of the body after 7 seconds of this acceleration is given by 𝑣=𝑒+π‘Žπ‘‘=0+1.4(7)=9.8/.ms

When the 136-newton force stops acting on the body, the acceleration of the body downward parallel to the plane is given by π‘Ž=π‘”πœƒ=9.82=4.9/.sinms

The initial velocity of the body is 9.8 m/s upward parallel to the plane, so the time taken for the body to reach an instantaneous velocity of zero is given by 𝑑=π‘£βˆ’π‘’π‘Ž=0βˆ’9.8βˆ’4.9=2.seconds

Let us summarize what we have learned in these examples.

Key Points

  • For a body on which the only forces acting are the weight of the body and the normal reaction on the body, the normal reaction force, 𝑅, on the body when on an inclined surface is given by 𝑅=π‘šπ‘”πœƒ,cos where πœƒ is the angle from the horizontal at which the surface slopes, π‘š is the mass of the body, and 𝑔 is the acceleration due to gravity.
  • The net force on a body on a smooth inclined surface due to the weight of and the reaction force on the body is given by 𝐹=π‘šπ‘”πœƒ,sin where πœƒ is the angle from the horizontal at which the surface slopes, π‘š is the mass of the body, and 𝑔 is the acceleration due to gravity.
  • If a force acts on a body on a smooth inclined surface in a direction that is not parallel to the surface, the force changes the reaction force on the body. This necessitates that the perpendicular components of the applied force be equated with the perpendicular components of the weight of the body and the normal reaction force on the body to determine any of these forces other than the weight of the body.

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