Lesson Explainer: Applications of Newton’s Second Law: Inclined Pulley Mathematics

In this explainer, we will learn how to solve problems on the motion of two bodies connected by a string passing over a smooth pulley with one of them on an inclined plane.

Consider two bodies connected by a light inextensible string, where the body of mass π‘šοŠ§ is supported by a smooth horizontal surface and the body of mass π‘šοŠ¨ is suspended from the string. The string runs over a smooth pulley that requires negligible force to turn. The following figure shows the forces acting on the bodies, where force 𝑇 is the tension in the string and 𝑅 is the normal reaction force.

The acceleration of each body is determined by its mass and the net force acting on it, according to Newton’s second law of motion. The tension in the string is constant.

Each body has the same acceleration, given by π‘Ž=π‘‡π‘š=(π‘šπ‘”)βˆ’π‘‡π‘š, where 𝑔 is the acceleration due to gravity.

If the surface that supports the body of mass π‘šοŠ§ is an inclined plane, then the forces acting on the bodies are those shown in the following figure.

The resultant of the weight of the body of mass π‘šοŠ§, π‘šπ‘”οŠ§, and the normal reaction force on it, π‘šπ‘”πœƒοŠ§cos, is π‘šπ‘”πœƒοŠ§sin, where πœƒ is the angle above the horizontal of the incline of the plane. The net force on the body parallel to the plane is given by 𝐹=π‘‡βˆ’(π‘šπ‘”πœƒ).sin

Let us look at an example in which the acceleration of such a system is determined.

Example 1: Finding the Acceleration of a System Involving a Smooth Inclined Plane and a Pulley

A body of mass 5 kg rests on a smooth plane inclined at an angle of 35∘ to the horizontal. It is connected by a light inextensible string passing over a smooth pulley, fixed at the top of the plane, to another body of mass 19 kg hanging freely vertically below the pulley. Given that the acceleration due to gravity 𝑔=9.8/ms, determine the acceleration of the system.

Answer

The following figure shows the forces acting on the bodies and what the forces on the 5 kg mass body due to its weight and the normal reaction on it sum to.

The accelerations of both bodies are equal. The acceleration of the 19 kg mass body is given by π‘Ž=19(9.8)βˆ’π‘‡19=186.2βˆ’π‘‡19.

Multiplying the expression by 19, we obtain

19π‘Ž=186.2βˆ’π‘‡.(1)

The net force on the supported body can be expressed as 𝐹=π‘‡βˆ’(5(9.8)(35)).sin

Force 𝐹 can also be expressed as 𝐹=5π‘Ž.

Hence, we have

5π‘Ž=π‘‡βˆ’(49(35)).sin(2)

We now have two equations, (1) and (2), that can be added to give 19π‘Ž+5π‘Ž=186.2βˆ’π‘‡+π‘‡βˆ’(49(35)).sin

This simplifies to 24π‘Ž=186.2βˆ’(49(35))π‘Ž=86.2βˆ’(49(35))24.sinsin

To two decimal places, this is 6.59 m/s2.

Let us now look at an example in which the tension in the string is determined.

Example 2: Finding the Force Acting on the Pulley in a System with an Inclined Plane

Two bodies of equal masses of 7.4 kg are connected by a light inelastic string. One of the bodies rests on a smooth plane inclined at 60∘ to the horizontal. The string passes over a smooth pulley fixed at the top of the plane, and the other body is left to hang freely vertically below the pulley. Find the force acting on the pulley when the system is released from rest. Take the acceleration due to gravity to be 𝑔=9.8/ms.

Answer

The force acting on the pulley is the resultant of the tension forces in the strings. The net force on a pulley due to two equal vertical tension forces, βƒ‘π‘‡οŠ§ and βƒ‘π‘‡οŠ¨, is shown in the following figure.

In this case, however, one tension force acts vertically downward and the other acts parallel to the inclined plane. The net force due to these two tension forces is, therefore, equivalent to force ⃑𝐹 shown in the following figure.

The angle of inclination of the plane is 60∘, hence, the forces on the pulley due to the tensions in the string act as shown in the following figure.

The tension is constant throughout the string, so βƒ‘π‘‡οŠ§ and βƒ‘π‘‡οŠ¨ have the same magnitude. For the purposes of determining βƒ‘π‘‡οŠ§ and βƒ‘π‘‡οŠ¨, this system of forces is equivalent to the system of forces shown in the following figure.

We can define a tension of magnitude 𝑇, where 𝑇=𝑇=𝑇.

The magnitude of the resultant of βƒ‘π‘‡οŠ§ and βƒ‘π‘‡οŠ¨ is, therefore, given by 𝐹=2𝑇(15).cos

The tension in the string can be determined by equating the accelerations of the bodies and, hence, the magnitudes of the forces producing these accelerations, as the masses of the bodies are equal. The forces acting on the bodies are shown in the following figure.

We see from the figure that π‘‡βˆ’(60)π‘šπ‘”=π‘šπ‘”βˆ’π‘‡.sin

This can be rearranged to give 2𝑇=π‘šπ‘”+(60)π‘šπ‘”.sin

It has been shown that the force on the pulley is given by 𝐹=2𝑇(15),cos which can be expressed as 𝐹=(π‘šπ‘”(1+(60)))(15)=ο€Ώο€Ώ72.52+√32(15).sincoscos

To two decimal places, this is 130.71 N.

Let us now look at an example in which kinematic equations are used.

Example 3: Solving an Inclined Plane Pulley System Using Newton’s Second Law and the Equations of Motion

A body of mass 2.4 kg rests on a smooth plane inclined at an angle of 30∘ to the horizontal. It is connected by a light inextensible string passing over a smooth pulley, fixed at the top of the plane, to another body of mass 1.6 kg hanging freely vertically below the pulley. When the system was released from rest, the two bodies were on the same horizontal level. Then, 10 seconds later, the string broke. Determine the time taken for the first body to start moving in the opposite direction after the string broke. Take 𝑔=9.8/ms.

Answer

The initial state of the system is shown in the following figure.

The figure shows that the resultant of the weight of the body of mass π‘šοŠ§ and the normal reaction force on the body acts downward parallel to the plane. The force parallel to the plane is the sum of the weight of the body and the normal reaction on the body.

The net force on the supported body is given by 𝐹=π‘‡βˆ’(π‘šπ‘”πœƒ).sin

The net force on the suspended body is given by 𝐹=(π‘šπ‘”)βˆ’π‘‡.

The acceleration of each body equals the net force on the body divided by its mass. The accelerations of the bodies are equal; hence, the acceleration of the system is given by π‘Ž=π‘‡βˆ’(π‘šπ‘”πœƒ)π‘š=(π‘šπ‘”)βˆ’π‘‡π‘š.sin

We see that π‘šπ‘Ž=(π‘šπ‘”)βˆ’π‘‡οŠ¨οŠ¨ and π‘šπ‘Ž=π‘‡βˆ’(π‘šπ‘”πœƒ).sin

Adding these expressions, we obtain π‘šπ‘Ž+π‘šπ‘Ž=(π‘šπ‘”)βˆ’π‘‡+π‘‡βˆ’(π‘šπ‘”πœƒ)π‘Ž=(π‘šπ‘”)βˆ’(π‘šπ‘”πœƒ)π‘š+π‘šπ‘Ž=𝑔(π‘šβˆ’π‘šπœƒ)π‘š+π‘š.sinsinsin

Substituting known values, we have π‘Ž=9.8(1.6βˆ’1.2)4=0.98/.ms

The magnitude of the tension in the string is given by 𝑇=π‘š(π‘Ž+π‘”πœƒ)=2.4(0.98+4.9)=14.112.sinN

The magnitude of the sum of the weight of the body and the normal reaction force on it is given by π‘šπ‘”πœƒ=2.4(9.8)2=11.76<𝑇.sinN

So, we see that the supported body accelerates upward parallel to the surface and the suspended body descends vertically. We should expect this as the question states that when the tension ceases to act on the body on the surface, it eventually reverses direction. If the body was initially moving downward parallel to the surface, removing the tension force acting on it would not make the body start to move upward parallel to the surface.

In a time of 10 seconds, the velocity of the body is upward along the slope with a magnitude given by 𝑣=𝑒+π‘Žπ‘‘=0+0.98(10)=9.8/.ms

When the tension in the string ceases to act on the body, the body accelerates downward parallel to the surface. With upward parallel to the surface taken as positive, the acceleration is given by π‘Ž=βˆ’π‘”πœƒ=βˆ’9.82=βˆ’4.9/.sinms

The body starts to move downward parallel to the surface at the instant that it has an instantaneous velocity of zero upward parallel to the surface. The time taken for this velocity change can be determined using the formula 𝑣=𝑒+π‘Žπ‘‘.

Rearranging to make 𝑑 the subject, we have 𝑑=π‘£βˆ’π‘’π‘Ž.

Substituting known values, we obtain 𝑑=0βˆ’9.8βˆ’4.9=2.s

Let us now look at an example in which the friction coefficient of a rough inclined surface needs to be found.

Example 4: Finding the Coefficient of Friction in a System with a Pulley and a Rough Inclined Plane

A body 𝐴 of mass 240 g rests on a rough plane inclined to the horizontal at an angle whose sine is 35. It is connected, by a light inextensible string passing over a smooth pulley fixed to the top of the plane, to another body 𝐡 of mass 300 g. If the system was released from rest and body 𝐡 descended 196 cm in 3 seconds, find the coefficient of friction between the body and the plane. Take 𝑔=9.8/ms.

Answer

The resultant of the weight of body 𝐴 and the normal reaction on it acts downward parallel to the plane. This resultant force is given by 𝐹=π‘šπ‘”πœƒ.sin

The weight of the body depends on the mass of the body. To make the units of mass consistent with the base SI units used for distance and time, the masses of the bodies in grams are converted to masses in kilograms. Then, we have 𝐹=0.24(9.8)ο€Ό35=1.4112.N

The system is shown in the following figure, where 𝐹 is the unknown frictional force on body 𝐴 and 𝑇 is the unknown tension in the string.

The plane is inclined at an angle whose sine is 35, so we see that the distances vertical and parallel to the plane are in the ratio 3∢5.

The acceleration of the system can be determined from the motion of body 𝐡. Body 𝐡 accelerates from rest and, in a time of 3 seconds, has a displacement of 196 cm. As the value of 𝑔 in the question is given as 9.8 m/s2, distance in centimetres is converted to distance in metres. So, the displacement is 1.96 m. The formula 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ can be rearranged to make π‘Ž the subject, taking 𝑒 as zero: π‘Ž=2𝑠𝑑=3.929/.ms

The net force on body 𝐴 is the sum of 𝐹, 𝐹, and 𝑇. Body 𝐴 moves upward parallel to the plane, so the frictional force on it acts downward parallel to the plane. The magnitude of the frictional force, 𝐹, is given by 𝐹=πœ‡π‘…=πœ‡π‘šπ‘”πœƒ.cos

We are told that sinπœƒ=35.

Hence, when body 𝐡 moves along a section of the plane that has a length of 5 metres, it ascends 3 metres vertically. From the Pythagorean theorem, 5=3+𝑑𝑑=√25βˆ’9=4, where 𝑑 is the horizontal distance moved by body 𝐡.

Hence, cosπœƒ=45 and 𝐹=πœ‡(0.24)(9.8)ο€Ό45=1.8816πœ‡.

The net force on body 𝐡 is given by

𝐹=π‘šπ‘ŽπΉ=π‘‡βˆ’πΉβˆ’πΉ=0.24ο€Ό3.929.(3)

The net force on body 𝐴 is given by

𝐹=π‘šπ‘”βˆ’π‘‡=π‘šπ‘ŽπΉ=0.3(9.8)βˆ’π‘‡=0.3ο€Ό3.929.(4)

Adding equations (3) and (4) to remove 𝑇, we obtain 0.3(9.8)βˆ’π‘‡+π‘‡βˆ’πΉβˆ’πΉ=0.3ο€Ό3.929+0.24ο€Ό3.9290.3(9.8)βˆ’πΉβˆ’πΉ=0.54ο€Ό3.929οˆβˆ’πΉβˆ’πΉ=0.54ο€Ό3.929οˆβˆ’0.3(9.8)βˆ’πΉβˆ’πΉ=0.2352βˆ’2.94=βˆ’2.7048𝐹+𝐹=2.7048.

Substituting the values of the forces, we get 1.4112+1.8816πœ‡=2.70481.8816πœ‡=1.2936πœ‡=1.29361.8816=1116.

Now, let us look at another example involving a rough inclined plane.

Example 5: Solving a Rough Inclined Plane Pulley System Using Newton’s Second Law and the Equations of Motion

A body of mass 162 g rests on a rough plane inclined to the horizontal at an angle whose tangent is 43. It is connected, by a light inextensible string passing over a smooth pulley fixed to the top of the plane, to another body of mass 181 g hanging freely vertically below the pulley. The coefficient of friction between the first body and the plane is 12. Determine the distance covered by the system in the first 7 seconds of its movement, given that the bodies were released from rest. Take 𝑔=9.8/ms.

Answer

We are told that tanπœƒ=43.

Hence, when the body supported by the plane moves along the plane, it ascends or descends 4 metres vertically for each 3 metres it travels horizontally. From the Pythagorean theorem, β„Ž=4+3β„Ž=√25=5, where β„Ž is the distance traveled along the surface by the body.

Hence, sinπœƒ=45 and cosπœƒ=35.

The following figure shows all but one of the forces acting on the supported body, where the weight of the body and the normal reaction force on it are shown by their resultant, which acts downward parallel to the plane. 𝑇 is the tension in the string.

The force not shown is the frictional force, 𝐹, on the supported body. The force is not shown as a frictional force acts in the opposite direction to the net force on a body, and the direction of the net force on the supported body has not been established. Let us now determine it.

The vertically downward force on the body suspended by the string is given by 𝐹=π‘”π‘š.suspended

The force acting parallel to the plane on the body supported by the plane is given by 𝐹=π‘”πœƒπ‘š=ο€Ό35οˆπ‘”π‘š.supportedsin

Dividing 𝐹suspended by 𝐹supported, we obtain 𝐹𝐹=π‘”π‘šο€»ο‡π‘”π‘š=5π‘š3π‘š.suspendedsupported

As π‘š>π‘š, we see that 𝐹>𝐹.suspendedsupported

Hence, the suspended mass descends and the supported mass moves upward parallel to the plane.

The frictional force on the supported body, 𝐹 is given by the product of the normal reaction force on the body and the coefficient of friction between the body and the surface that supports it. All the forces acting on the supported body act parallel to the plane; hence, we can express the frictional force on the body, 𝐹, as 𝐹=π‘šπ‘”πœ‡πœƒ.cos

The net force on the supported body is, therefore, given by

𝐹=π‘‡βˆ’π‘šπ‘”πœƒβˆ’π‘šπ‘”πœ‡πœƒπΉ=π‘‡βˆ’162(9.8)ο€Ό45οˆβˆ’162(9.8)ο€Ό12οˆο€Ό35𝐹=π‘‡βˆ’162(9.8)ο€Ό810οˆβˆ’162(9.8)ο€Ό310𝐹=π‘‡βˆ’162(9.8)ο€Ό1110=162π‘Ž,supportedsupportedsupportedsupportedsincos(5)

where π‘Ž is the acceleration of the system.

The net force on the suspended body is given by

𝐹=π‘šπ‘”βˆ’π‘‡πΉ=181(9.8)βˆ’π‘‡=181π‘Ž.suspendedsuspended(6)

Adding the net forces on the bodies given by equations (5) and (6), we obtain 162π‘Ž+181π‘Ž=π‘‡βˆ’162(9.8)ο€Ό1110+181(9.8)βˆ’π‘‡343π‘Ž=βˆ’162(9.8)ο€Ό1110+181(9.8)π‘Ž=27.44343=0.08/.ms

The displacement of the supported body can be determined using the formula 𝑠=𝑒𝑑+12π‘Žπ‘‘.

Substituting known values, we find that after 7 seconds of acceleration, 𝑠=0(7)+12(0.08)ο€Ή7=1.96.m

This can be expressed as an integer value by converting it to a value in centimetres, which gives 196 cm.

Let us summarize what we have learned in these examples.

Key Points

The following points apply to a system of two bodies connected by a light inextensible string, where the body of mass π‘šοŠ§ is supported by a smooth plane inclined at an angle πœƒ above the horizontal and the body of mass π‘šοŠ¨ is suspended from the string. The string runs over a smooth pulley that requires negligible force to turn, as shown in the following figure.

  • The net force on the suspended body equals the sum of the tension in the string and the weight of the body. It is given by 𝐹=(π‘šπ‘”)βˆ’π‘‡, where 𝑇 is the tension in the string and 𝑔 is the acceleration due to gravity.
  • The net force on the supported body equals the sum of the tension in the string and the component of the weight of the body acting parallel to the plane. This is given by 𝐹=π‘‡βˆ’(π‘šπ‘”πœƒ),sin where 𝑇 is the tension in the string and 𝑔 is the acceleration due to gravity.
  • The accelerations of the suspended and the supported bodies are both equal to the net forces on them divided by their masses, and both accelerations are equal. Then, we have π‘Ž=(π‘šπ‘”)βˆ’π‘‡π‘š=π‘‡βˆ’(π‘šπ‘”πœƒ)π‘š,sin where 𝑇 is the tension in the string and 𝑔 is the acceleration due to gravity.
  • If the inclined plane that supports the body of mass π‘šοŠ§ is rough, the acceleration of the system is given by π‘Ž=(π‘šπ‘”)βˆ’π‘‡π‘š=π‘‡βˆ’(π‘šπ‘”πœƒΒ±πœ‡π‘šπ‘”πœƒ)π‘š,sincos where πœ‡ is the coefficient of friction of the body with the plane. The frictional force on the body of mass π‘šοŠ§ may be either in the same direction as or in the opposite direction to the tension in the string, depending on whether the body moves upward or downward parallel to the plane.

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