### 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.