Lesson Explainer: Applications of Newton’s Second Law: Horizontal 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, where one is on a horizontal table.

Consider two bodies connected by a string that passes over a pulley that is attached to the edge of a horizontal surface, where one of the bodies is on the surface and the other is suspended freely from the string, as shown in the following figure.

If the surface is smooth and the mass of the string and the force required to produce rotation of the pulley are negligible, the acceleration of the bodies depends only on the weight of the suspended body, the tension in the string, and the masses of the bodies, as shown in the following figure.

The acceleration of a body can be determined from Newton’s second law of motion. Let us define this.

Definition: Newton’s Second Law of Motion

When a net force acts on a body, the body accelerates in the direction of the force. The magnitude of the acceleration depends on the magnitude of the force and on the mass of the body, according to the formula 𝐹=π‘šπ‘Ž, where π‘š is the mass of the body and π‘Ž is the acceleration of the body.

Assuming that the string is taut and cannot deform, the acceleration of the bodies must be equal. From the forces and masses shown in the figure, we can see that the acceleration of the body on the surface, π‘ŽοŠ§, is given by π‘Ž=π‘‡π‘š.

The acceleration of the suspended body, π‘ŽοŠ¨, is given by π‘Ž=π‘Šβˆ’π‘‡π‘š, where π‘Ž=π‘Ž=π‘ŽοŠ§οŠ¨ and π‘Š=π‘šπ‘”.

The tension in the string is, therefore, given by 𝑇=π‘šπ‘Ž=π‘š(π‘”βˆ’π‘Ž).

Let us look at an example of such an arrangement.

Example 1: Finding the Tension in the String Connecting a Mass on a Table to a Vertically Hanging Mass through a Pulley

Two bodies of masses 15 and 16.5 kilograms were attached to the opposite ends of a light inextensible string that passed over a smooth pulley fixed to the edge of a smooth horizontal table. The body of larger mass was placed on the smooth table while the smaller one was hanging vertically below the pulley. Determine the tension in the string, given that the acceleration due to gravity 𝑔=9.8/ms.

Answer

If the mass of the body on the surface is given by π‘šοŠ§ and the mass of the suspended body is given by π‘šοŠ¨, the formula 𝑇=π‘šπ‘Ž=π‘š(π‘”βˆ’π‘Ž) gives the value of 𝑇, as the bodies have equal accelerations.

The forces acting are shown in the following figure.

The tension in the string can be determined using either 𝑇=16.5π‘Ž or 𝑇=15(9.8βˆ’π‘Ž); but, in either case, it is first necessary to determine the value of π‘Ž. The value of a can be determined by equating the two equations above: 16.5π‘Ž=15(9.8βˆ’π‘Ž)16.5π‘Ž=147βˆ’15π‘Žπ‘Ž(16.5+15)=147π‘Ž=143/.ms

Substituting π‘Ž into 𝑇=16.5π‘Ž gives us 𝑇=16.5ο€Ό143=77.N

Let us now look at an example of determining the force acting on a pulley due to the tension in a string attached to two bodies.

Example 2: Finding the Force Exerted on a Pulley Connecting a Body on a Smooth Surface to a Vertically Hanging Body

A body is placed on a smooth horizontal table. It is connected, by a light inextensible string passing over a smooth pulley fixed at the edge of the table, to another body hanging freely vertically below the pulley. If the tension in the string was 1.04 N, find the force exerted on the pulley.

Answer

It is worth noting that neither the masses of either body nor the accelerations of the bodies are given. These values are not required to determine the force acting on the pulley. The forces acting on the pulley are shown in the following figure.

The only forces acting on the pulley are the tensions in the horizontal and vertical parts of the string. The weight of the suspended mass acts on the suspended mass only, not on the pulley. The lines of action of the tensions in the horizontal and vertical parts of the string act along these parts of the string, and the resultant force 𝐹R due to these tensions is shown in the following figure.

The point at which the forces act is actually a point on the pulley, as is shown in the figure above. The point at which the forces act does not, however, change the value of 𝐹R.

The value of 𝐹R depends on the values of 𝑇 and πœƒ. The value of 𝑇 is stated to be 1.04 N. The value of πœƒ is determined by the fact that the two tensions are of equal magnitude and are perpendicular, so πœƒ must be 45∘. The value of 𝐹R is, therefore, given by 𝐹=1.04((45)+(45))𝐹=2(1.04)ο€Ώβˆš22𝐹=2ο€Ό2625οˆο€Ώβˆš22𝐹=26√225.RRRRcossinN

The acceleration of a body connected to another body by a string that runs over a pulley can be calculated, and so the velocity and displacement of such a body can be determined. Let us look at an example where the velocity of such a body due to its acceleration is determined.

Example 3: Finding the Velocity at Which a Mass Collides with the Pulley

A body 𝐴 of mass 180 g is resting on a smooth horizontal table. It is connected by a light inelastic string that passes over a smooth pulley, fixed to the edge of the table, to another body 𝐡 of mass 120 g hanging freely vertically below the pulley. When body 𝐴 is 90 cm away from the pulley, the system is released from rest. Determine the speed at which body 𝐴 collides with the pulley. Take 𝑔=9.8/ms.

Answer

The velocity of body 𝐴 when it has a displacement of 90 cm from its initial position depends on the acceleration of body 𝐴 over that displacement. The acceleration of body 𝐴 can be determined using the formula π‘šπ‘Ž=π‘š(π‘”βˆ’π‘Ž).

The displacement is converted from 90 cm to 0.9 m, and the masses of the bodies in grams are converted to the SI base unit kilograms.

Substituting the values of 𝑔 and the masses of the bodies, we obtain 0.18π‘Ž=0.12(9.8βˆ’π‘Ž)0.18π‘Ž=βˆ’0.12π‘Ž+1.1760.3π‘Ž=1.176π‘Ž=117.60.3=3.92/.ms

The velocity of a body that accelerates over a displacement 𝑠 can be found using the formula 𝑣=𝑒+2π‘Žπ‘ , where in this case 𝑒 is zero as body 𝐴 is initially at rest. Substituting the values of π‘Ž and 𝑠 gives us 𝑣=2(3.92)(0.9)=7.056.

This allows 𝑣 to be determined exactly, as follows: 𝑣=7.056=0.84(10)𝑣=√7.056=0.84√10/.ms

In units of centimetres per second, this is given by 𝑣=√0.7056=84√10/.cms

Sometimes the horizontal surface in our system is rough instead of smooth. This creates a frictional force that resists the motion of the body on the surface.

The magnitude of the frictional force 𝐹f between a moving body and a surface is given by 𝐹=πœ‡π‘…,f where 𝑅 is the normal reaction force of the surface on the body, which for a horizontal surface has a magnitude equal to the weight of the body, π‘šπ‘”.

The constant πœ‡ is the coefficient of friction between the surface and the body. The frictional force acts in the opposite direction to the direction of the motion of the body. If a body is accelerated across a rough horizontal surface by a force ⃑𝐹, the magnitude of the net force on the body along the line of ⃑𝐹 is given by 𝐹=πΉβˆ’π‘šπ‘”πœ‡.net

Let us consider a system of connected masses and a pulley, as shown in the following figure.

The value of 𝐹 is given by 𝐹=π‘šπ‘”πœ‡, where πœ‡ is the coefficient of friction.

The value of π‘Š is given by π‘Š=π‘šπ‘”.

The relation of the acceleration of the horizontally moving body to the forces acting on it is given by π‘‡βˆ’πΉ=π‘šπ‘Žπ‘‡βˆ’π‘šπ‘”πœ‡=π‘šπ‘Ž.

The relation of the acceleration of the vertically moving body to the forces acting on it is given by π‘Šβˆ’π‘‡=π‘šπ‘Žπ‘šπ‘”βˆ’π‘‡=π‘šπ‘Ž.

These terms can be combined as follows: π‘Ž(π‘š+π‘š)=π‘‡βˆ’(π‘šπ‘”πœ‡)+π‘šπ‘”βˆ’π‘‡π‘Ž(π‘š+π‘š)=π‘šπ‘”βˆ’π‘šπ‘”πœ‡π‘Ž=π‘šπ‘”βˆ’π‘šπ‘”πœ‡(π‘š+π‘š)π‘Ž=𝑔(π‘šβˆ’π‘šπœ‡)(π‘š+π‘š).

Let us look at an example where friction between a body and a surface affects the motion of the body.

Example 4: Finding the Acceleration of a System Involving a Rough Horizontal Table and a Pulley

A body of mass 203 g rests on a rough horizontal table. It is connected by a light inextensible string, passing over a smooth pulley fixed to the edge of the table, to a body of mass 493 g hanging freely vertically below the pulley. Given that the coefficient of friction between the first body and the table is 0.2, find the acceleration of the system. Take 𝑔=9.8/ms.

Answer

The forces acting on the bodies are shown in the following diagram, where the friction force is represented by 𝐹.

The acceleration of the system can be determined using the formula π‘Ž=𝑔(π‘šβˆ’π‘šπœ‡)(π‘š+π‘š).

To obtain a value of π‘Ž in metres per second squared, the values of the masses are converted to the SI base unit of mass, the kilogram. The value of π‘šοŠ§ is 0.203 kg and the value of π‘šοŠ¨ is 0.493 kg. The acceleration due to gravity, 𝑔, is given as 9.8 m/s2. Substituting the known values gives us π‘Ž=9.8(0.493βˆ’0.203(0.2))(0.203+0.493)=6.37/.ms

Let us now look at an example where friction between a body and a surface affects the motion of the body and the displacement of the body is determined.

Example 5: Finding the Distance Traveled by a Mass on a Rough Plane Connected to a Vertically Hanging Mass

A body of mass 200 g rests on a rough horizontal table. It is connected by a light inextensible string passing over a smooth pulley, fixed to the edge of the table, to another body of the same mass hanging freely below the pulley 2 cm above the ground. The coefficient of friction between the table and the body resting on it is 13. Given that the system was released from rest, and the hanging body descended until it hit the ground, how much further did the body on the table travel until it came to rest? Take the acceleration due to gravity 𝑔=9.8/ms.

Answer

The forces acting on the bodies are shown in the following diagram, where the friction force is represented by 𝐹.

The acceleration of the system can be determined using the formula π‘Ž=𝑔(π‘šβˆ’π‘šπœ‡)(π‘š+π‘š).

For this system, π‘š=π‘š, and so the acceleration of the system is given by π‘Ž=π‘”π‘š(1βˆ’πœ‡)2π‘š.

Substituting the value of πœ‡ given in the question, we have π‘Ž=𝑔(1βˆ’πœ‡)2=𝑔1βˆ’ο‡2π‘Ž=𝑔3.

Both bodies are accelerated for the time that it takes the vertically suspended body to descend a distance of 2 centimetres, which is 0.02 metres. The velocity of the horizontally moving body at that time can be found using the formula 𝑣=𝑒+2π‘Žπ‘ , where 𝑒 is 0 m/s; hence, 𝑣=2𝑔30.02=0.04𝑔3𝑣=ο„ž0.04𝑔3/.ms

The displacement of the horizontally moving body when it comes to rest can be found using the formula 𝑣=𝑒+2π‘Žπ‘ , where 𝑣 is 0 m/s and 𝑒=ο„ž0.04𝑔3/.ms

Since the suspended body has reached the floor and stopped moving, the tension in the string will become zero. Therefore, the only force acting on the body sliding on the rough surface is the frictional force. We can use this to calculate the acceleration of the body as it slows down.

The magnitude of the frictional force is given by 𝐹=π‘šπ‘”πœ‡,N which produces an acceleration in the opposite direction to the direction of motion of πΉπ‘š=π‘šπ‘”πœ‡π‘š=π‘”πœ‡=𝑔3/.ms

By substitution into 𝑣=𝑒+2π‘Žπ‘ , we obtain 0=ο€Ώο„ž0.04𝑔3ο‡ο‹βˆ’2𝑔3𝑠0=0.04𝑔3ο‡βˆ’2𝑔3𝑠0.04𝑔3=2𝑔3𝑠0.04=2𝑠𝑠=0.02.m

The displacement is 2 cm.

Key Points

  • For two bodies connected by a light, inextensible string that runs over a pulley that can be turned by a negligible force, where body 𝐴 is on a horizontal surface and body 𝐡 is freely suspended from the string, the accelerations of the bodies are of equal magnitude and the forces on each body provided by the tension in the string are of equal magnitude.
  • For two bodies connected by a light, inextensible string that runs over a pulley that can be turned by a negligible force, where body 𝐴 is on a horizontal surface and body 𝐡 is freely suspended from the string, the resultant forces on the bodies are as follows: 𝐹=π‘šπ‘Ž=𝑇, and 𝐹=π‘šπ‘Ž=π‘šπ‘”βˆ’π‘‡, where π‘Ž is the acceleration of either body and 𝑇 is the tension in the string.
  • If the horizontal surface is rough, the net force on the horizontally moving body is given by 𝐹=π‘šπ‘Ž=π‘‡βˆ’π‘šπ‘”πœ‡, where πœ‡ is the coefficient of friction of body 𝐴 with the surface.

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