### 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 π 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 0.203π. This is its mass multiplied by
gravity. As π 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, Newtonβs 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 ππ
, where π is the coefficient of friction, in this
case, 0.2. Letβs 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. π
is equal to 1.9894. Resolving horizontally, we have π
minus πΉ 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 π.

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 π. This is equal to 0.493π. Using the equation friction is
equal to ππ
, 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 π and acceleration
π. In this question, we are only
interested in the acceleration, so it makes sense to eliminate π from these
equations. We can do this by adding equation
one and equation two as the πβs would cancel. The left-hand side becomes 4.8314
plus negative 0.39788, and the right-hand side is equal to 0.696π.

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