Lesson Video: Applications of Newton’s Second Law: Horizontal Pulley Mathematics

In this video, 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.

17:59

Video Transcript

In this video, 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. This will involve the table being both smooth and rough. We will therefore solve problems with and without friction. We will begin by recalling Newton’s second law and seeing how we can use it in this type of problem.

Newton’s second law of motion states that the acceleration of an object is dependent on two variables: firstly the net force acting upon the object and secondly the mass of the object. This leads us to the equation force is equal to mass multiplied by acceleration. For all the questions in this video, we will be dealing with a smooth pulley. This means there will be no friction. We will also be dealing with a light inextensible string. This means that the string has no mass and is a fixed length. As the string is inextensible, the acceleration of the connected objects will be the same. Finally, when dealing with problems involving a rough table, we must consider the frictional force. This frictional force is equal to 𝜇 multiplied by the reaction force 𝑅, where 𝜇 is the coefficient of friction which has a value that lies between zero and one inclusive. In practical terms, this is how rough the table is. The greater the value of 𝜇, the greater the frictional force.

In our first question, we need to calculate the tension in the string.

Two bodies of masses 15 and 16.5 kilograms were attached to the opposite ends of a light inextensible string, which 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 is equal to 9.8 meters per square second.

In any question of this type, we begin by modeling the situation with a diagram. We are told that the masses of the two bodies are 15 and 16.5 kilograms. Their corresponding downward forces will therefore be equal to 15𝑔 and 16.5𝑔, respectively, where 𝑔 is gravity equal to 9.8 meters per square second. Both the pulley and table are smooth, so we do not need to consider friction in this question. The tension 𝑇 in the string will be constant throughout. As the string is inextensible, when the system is released, both bodies will move with the same acceleration. On release, we assume that the free hanging body accelerates downwards.

We can now use Newton’s second law, force equals mass times acceleration, to set up some equations. If we consider the body on the table, this will move in a horizontal direction. The only force acting in this direction is the tension 𝑇. Therefore, 𝑇 is equal to 16.5 multiplied by 𝑎. The free hanging body B will accelerate in a downwards vertical direction. The force 15𝑔 is acting in this direction, whereas the tension is acting in the opposite direction. This means that the sum of the forces is equal to 15𝑔 minus 𝑇. This will be equal to 15𝑎. As we are taking 𝑔 equal to 9.8 and 15 multiplied by 9.8 is 147, this equation simplifies to 147 minus 𝑇 is equal to 15𝑎.

We now have two simultaneous equations that we can solve to calculate the acceleration 𝑎 and the tension in the string 𝑇. Whilst we’re only interested in calculating the tension in this question, it is easier to work out the value of 𝑎 first. This is because we already have 𝑇 as the subject of equation one. Substituting this into equation two gives us 147 minus 16.5𝑎 is equal to 15𝑎. We can add 16.5𝑎 to both sides so that 147 is equal to 31.5𝑎. Finally, dividing by 31.5 gives us 𝑎 is equal to fourteen-thirds or 14 over three. We can now substitute this value back in to equation one. We need to multiply fourteen-thirds by 16.5. This gives us an answer of 77. We can therefore conclude that the tension in the string is 77 newtons.

In our next question, we will deal with a problem on a rough horizontal table.

A body of mass 203 grams 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 grams 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 equal to 9.8 meters per square second.

We begin by sketching the scenario. As there are 1000 grams in a kilogram, the body resting on the table has a downward force of 0.203𝑔. This is its mass multiplied by gravity. As 𝑔 is equal to 9.8 meters per square second, the downward force is equal to 1.9894 newtons. The freely hanging body has a mass of 493 grams, which is equal to 0.493 kilograms. Multiplying this by 9.8 gives us a downward force of 4.8314 newtons. We know that the tension in the string is constant as there is no friction in the pulley. As the horizontal table is rough, there will be a frictional force. We also need to consider the reaction force going vertically upwards from the body on the table. When the system is released, the freely hanging body will accelerate downwards and the body on the table will accelerate towards the pulley. As the string is inextensible, these bodies will have the same acceleration.

We will be able to solve this problem using two equations, firstly, Newton’s second law. This states that the sum of the forces is equal to the mass multiplied by the acceleration. We will also use the fact that the frictional force is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction, in this case, 0.2. Let’s begin by considering the body on the table A. There is no acceleration in the vertical direction. Therefore, the forces up will be equal to the forces down. 𝑅 is equal to 1.9894. Resolving horizontally, we have 𝑇 minus 𝐹 r. This is because the tension is acting in the positive direction and the frictional force in the negative direction. This is equal to the mass of the body 0.203 multiplied by the acceleration 𝑎.

When dealing with the freely hanging body, we see it is accelerating in a downwards direction. This means that the sum of the forces is equal to 4.8314 minus 𝑇. This is equal to 0.493𝑎. Using the equation friction is equal to 𝜇𝑅, we have a frictional force of 0.2 multiplied by 1.9894. This is equal to 0.39788. We now have two simultaneous equations that we can use to calculate the value of the tension 𝑇 and acceleration 𝑎. In this question, we are only interested in the acceleration, so it makes sense to eliminate 𝑇 from these equations. We can do this by adding equation one and equation two as the 𝑇’s would cancel. The left-hand side becomes 4.8314 plus negative 0.39788, and the right-hand side is equal to 0.696𝑎.

Simplifying the left-hand side gives us 4.43352. Dividing both sides of this equation by 0.696 gives us 𝑎 is equal to 6.37. The acceleration of the system is 6.37 meters per square second.

We could substitute this back in to equation one or two to calculate the tension. However, this is not required in this question.

In our final question, we will also need to consider the equations of uniform acceleration.

A body of mass 200 grams rests on a rough horizontal table. It is connected by a light inextensible string passing over a smooth pulley fixed to the edge of a table to another body of the same mass hanging freely below the pulley two centimeters above the ground. The coefficient of friction between the table and the body resting on it is one-third. 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 acceleration due to gravity 𝑔 to be equal to 9.8 meters per square second.

There is a lot of information in this question, so we will begin by drawing a diagram. We have two bodies of the same mass of 200 grams. The weight of these bodies is equal to the mass multiplied by gravity. As there are 1000 grams in one kilogram, these both have a downward force of 0.2𝑔. The tension throughout the string will be equal. And when the system is released, the freely hanging body will accelerate downwards and the body on the table will accelerate to the right. As the horizontal table is rough, there will be a frictional force acting against the body. There is also a reaction force going vertically upwards.

We know that the frictional force is equal to 𝜇 multiplied by the reaction force when 𝜇 is the coefficient of friction, in this case, one-third. Newton’s second law states that force is equal to mass multiplied by acceleration. We can use this to resolve vertically and horizontally. The body on the table is not moving vertically. Therefore, the forces up must equal the forces down. We have 𝑅 is equal to 0.2𝑔. The friction force must be equal to a third of this. This gives us a friction force of 49 over 75 newtons. Adding in the fact that the freely hanging body is two centimeters from the ground, we will now clear some space for the remainder of our calculations.

Resolving horizontally for the body on the table, we see that the sum of the forces is equal to 𝑇 minus 𝐹 r. This is equal to 0.2𝑎. We can substitute in the value of friction here. The freely hanging particle is accelerating downwards. This gives us the equation 0.2𝑔 minus 𝑇 is equal to 0.2𝑎. We have a pair of simultaneous equations that we can solve to calculate the acceleration 𝑎. Adding equation one and equation two gives us 98 over 75 is equal to 0.4𝑎. Dividing both sides by 0.4 gives us an acceleration of 49 over 15. Once released, the system moves with an acceleration of 49 over 15 meters per second squared.

After the freely hanging body hits the ground, the string becomes slack and its tension is equal to zero. We can then calculate the acceleration of the body on the table after this point by solving negative 49 over 75 is equal to 0.2𝑎. Dividing both sides of this equation by 0.2 gives us 𝑎 is equal to negative 49 over 15. We notice that the acceleration after the string becomes slack is the negative acceleration prior to it.

Let’s now consider what is happening to the body on the table throughout its motion. We know that it starts from rest and accelerates uniformly at 49 over 15 meters per square second. During this time, it travels a distance of two centimeters as it must travel the same distance as body B. After the freely hanging body hits the ground, the body on the table decelerates. Its acceleration is equal to negative 49 over 15 meters per square second. We need to calculate how far it travels before it comes to rest. The final speed of the first part of the journey is equal to the initial speed of the second part.

We can now use our equations of uniform acceleration or SUVAT equations to calculate 𝑠. We know the following values for the two parts of the journey. We can now substitute these values into the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. In the first part, as 𝑢 is equal to zero, we have 𝑣 squared is equal to two multiplied by 49 over 15 multiplied by two. In the second part, as 𝑣 is equal to zero, we have zero is equal to 𝑣 squared plus two multiplied by negative 49 over 15 multiplied by 𝑠.

We can solve these simultaneously to eliminate 𝑣 squared. We can then divide our new equation by two and 49 over 15. Two minus 𝑠 is equal to zero, which means that 𝑠 is equal to two. The body travels another two centimeters until it comes to rest. This is the same distance as in the first part of the journey as the acceleration and then deceleration are equal.

We will now summarize the key points from this video. In this video, we have used Newton’s second law of motion, 𝐹 equals 𝑚𝑎, to solve problems involving one body on a horizontal plane connected to another body suspended below a smooth pulley. When the horizontal surface is rough, we can calculate the frictional force by multiplying 𝜇 by the reaction force when 𝜇 is the coefficient of friction and takes a value between zero and one. For more complex problems, we saw that we could use the equations of motion or SUVAT equations. These involve constant acceleration.

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