Question Video: Finding the Acceleration of a System of a Mass on a Rough Horizontal Table Connected to a Vertically Hanging Body by a String through a Pulley Mathematics

Video Transcript

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.