### Video Transcript

In this video, we will learn how to
solve problems on the motion of a system of two bodies suspended vertically by a
string passing over a smooth pulley. We will solve the problems using
Newton’s second law of motion, calculating resultant forces, and using the equations
of motion or SUVAT equations.

Newton’s second law states that the
net force on an object is equal to its mass multiplied by its acceleration. This is written as 𝐹 equals
𝑚𝑎. In this video, we will measure the
mass in kilograms or grams. The acceleration of the objects
will be measured in meters per square second or centimeters per square second. In order to convert between these
units, we recall that there are 1,000 grams in one kilogram. There are 100 centimeters in one
meter. Therefore, 100 centimeters per
square second is equal to one meter per square second. The forces that are acting on the
object will be measured in newtons or dynes. When the mass is in kilograms and
acceleration in meters per square second, the force will be in newtons. Likewise, multiplying a mass in
grams by an acceleration in centimeters per square second will give us a force in
dynes.

We will now consider how we can use
Newton’s second law to solve problems involving a system of bodies and a smooth
pulley. In the system modeled in the
diagram, we have a light inextensible string passing over a smooth pulley. At each end of the string is
attached an object of mass lowercase 𝑚 and capital 𝑀 kilograms, respectively. These will each have forces acting
in the downward direction equal to the mass multiplied by gravity. As the string has no mass and the
pulley has no friction, there will be equal tensions acting upward from the
masses. When the system is released, the
heavier body will accelerate downwards and the lighter body will accelerate
upwards. As the string is inextensible,
these accelerations will have the same magnitude.

We can now use Newton’s second
law. The sum of the net forces will be
equal to the mass multiplied by the acceleration. The left-hand body is moving
upwards. If we assume that this is the
positive direction for this body, the sum of the forces is equal to 𝑇 minus
𝑚𝑔. This is equal to the mass
multiplied by the acceleration. We get the equation 𝑇 minus 𝑚𝑔
is equal to 𝑚𝑎. We can now repeat this process for
the second body. This body is accelerating
downwards, which we will assume is the positive direction. The sum of the net forces is equal
to capital 𝑀𝑔 minus 𝑇. This is equal to capital 𝑀, the
mass of the body, multiplied by the acceleration 𝑎. We now have a pair of simultaneous
equations.

In the questions that follow, we
will be able to solve these to calculate any unknowns. In our first question, we need to
calculate the acceleration.

Two bodies of masses 12 kilograms
and 18 kilograms are attached to the ends of a light inextensible string which
passes over a smooth pulley. Determine the acceleration of the
system. Take 𝑔 equal to 9.8 meters per
square second.

We can begin by sketching the
system as shown. The two bodies have masses of 12
kilograms and 18 kilograms, respectively. This means that they will have a
downward force of 12𝑔 and 18𝑔, where gravity is 9.8 meters per square second. As we have a light string which
passes over a smooth pulley, the vertical tensions will be equal. Once the system is released, the
heavier body of mass 18 kilograms will accelerate downwards and the 12-kilogram body
will accelerate upwards. As the string is inextensible, the
magnitude of these accelerations will be equal.

We can now use Newton’s second law,
𝐹 equals 𝑚𝑎, to create two simultaneous equations. Body A is accelerating vertically
upwards. Therefore, the sum of its forces is
𝑇 minus 12𝑔. We take the positive direction to
be vertically upwards. This is equal to 12𝑎, the mass
multiplied by the acceleration. As body B is accelerating
downwards, we take this to be the positive direction. This means that the sum of the
forces is equal to 18𝑔 minus 𝑇. This is equal to 18𝑎.

We now have two simultaneous
equations, and we can eliminate 𝑇 in order to calculate the acceleration 𝑎 by
adding equation one and equation two. Negative 12𝑔 plus 18𝑔 is equal to
six 𝑔, and 12𝑎 plus 18𝑎 is 30𝑎. We can then divide both sides of
this equation by 30 so that 𝑎 is equal to six 𝑔 divided by 30. Taking 𝑔 equal to 9.8, we can type
this into the calculator, giving us an answer of 1.96. The acceleration of the system is
equal to 1.96 meters per square second. We could substitute this value back
into equation one or equation two to calculate the tension, but this is not required
in this question.

In our next question, we need to
calculate the mass of one of the bodies.

Two masses 𝑚 and 88 grams are
attached to the ends of a light string passing over a smooth pulley. Determine the value of 𝑚, given
that when the system was released, the other mass descended 11.76 meters in two
seconds. Take the acceleration due to
gravity 𝑔 equal to 9.8 meters per square second.

We begin by modeling the system as
shown. We are told that the masses are
equal to 𝑚 and 88 grams. We recall that there are 1,000
grams in one kilogram, and as the gravity is given in meters per square second, we
need the mass to be in kilograms. 88 grams is equal to 0.088
kilograms. This means that the downward forces
of the two bodies are 𝑚𝑔 and 0.088𝑔. We have a light string and a smooth
pulley, which means that the vertical tensions will be equal. The entire system will move with
the same acceleration, and we are told that the 88-gram body descends. This means that the body of mass 𝑚
will accelerate upwards.

We can now use Newton’s second law,
𝐹 equals 𝑚𝑎, to create equations for body A and body B. Body A is accelerating
vertically upwards. Therefore, the sum of its forces is
equal to 𝑇 minus 𝑚𝑔. This is equal to 𝑚𝑎. Body B is accelerating
downwards. Therefore, the sum of its forces is
0.088𝑔 minus 𝑇. We assume the downward direction is
positive. This is once again equal to the
mass multiplied by the acceleration. At this stage, we have three
unknowns: the tension, the mass of body A, and the acceleration of the system. We will need to use the information
from the question that tells us that the body of 88 grams descends 11.76 meters in
two seconds.

We can use the equations of uniform
acceleration to calculate the acceleration. These are often referred to as the
SUVAT equations, where 𝑠 is the displacement — in this case equal to 11.76
meters. 𝑢 is the initial velocity, 𝑣 is
the final velocity, 𝑎 is the acceleration, and 𝑡 is the time, in this question,
two seconds. We know that 𝑠 is equal to 𝑢𝑡
plus a half 𝑎𝑡 squared. Substituting in our values, we have
11.76 is equal to zero multiplied by two plus a half multiplied by 𝑎 multiplied by
two squared. The right-hand side simplifies to
two 𝑎, so this is equal to 11.76. Dividing both sides of this
equation by two gives us 𝑎 is equal to 5.88. The acceleration of the system,
once it is released, is 5.88 meters per square second.

Substituting this as well as 𝑔
equals 9.8 into our equation for body A gives us 𝑡 minus 9.8𝑚 is equal to
5.88𝑚. Adding 9.8𝑚 to both sides of this
equation gives us 𝑡 is equal to 15.68𝑚. We will call this equation one. Substituting the value for the
acceleration into the equation for body B gives us 0.8624 minus 𝑇 is equal to
0.51744. Rearranging this equation gives us
a value of 𝑇 equal to 0.34496. We can now substitute the tension,
which is equal to 0.34496 newtons, into equation one. This tension force is equal to
15.68𝑚. Dividing both sides by 15.68 gives
us 𝑚 is equal to 0.022. This is the mass of the body in
kilograms. By multiplying this answer by
1,000, we can convert to grams, which is the same units as the other mass. The mass 𝑚 is equal to 22
grams.

In our final question, we will
calculate the force exerted on a pulley.

Two bodies of masses 𝑚 grams and
𝑚 plus 56 grams are connected to each other by a light string which passes over a
smooth fixed pulley. The system was released from rest
when the two bodies were at the same horizontal level. One second later, the vertical
distance between them was 128 centimeters. Find the magnitude of the force
exerted on the pulley while the bodies were in motion. Take the acceleration due to
gravity 𝑔 equal to 9.8 meters per square second.

Let’s begin by modeling the
situation as shown. The two bodies have masses 𝑚 grams
and 𝑚 plus 56 grams. This means that they will each
exert a downward force of mass multiplied by gravity. We notice that the gravity is given
in meters per square second. As the masses are in grams and the
distance in centimeters, we need to convert this to centimeters per square
second. As there are 100 centimeters in one
meter, the acceleration due to gravity is 980 centimeters per square second. We have a light string passing over
a smooth pulley. Therefore, the vertical tensions
will be equal. When the system is released, the
magnitude of the acceleration of both bodies will be equal. The body of mass 𝑚𝑔 will
accelerate upwards, and the body of mass 𝑚 plus 56𝑔 will accelerate downwards.

We are told that the bodies start
at the same horizontal level, and one second later, the distance between them is 128
centimeters. This means that each body has moved
a distance of 64 centimeters, one in the downward direction and one in the upward
direction. We can use our SUVAT equations or
equations of motion to calculate the acceleration. We know that the initial velocity
was zero meters per second as the body started at rest, and we are dealing with a
time of one second. We will use the equation 𝑠 is
equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values gives us
64 is equal to zero multiplied by one plus a half multiplied by 𝑎 multiplied by one
squared. The right-hand side simplifies to a
half 𝑎. We can then multiply both sides of
the equation by two, giving us 𝑎 is equal to 128. The acceleration of the system is
equal to 128 centimeters per square second.

Our next step will be to use
Newton’s second law, force equals mass multiplied by acceleration. As body A is accelerating upwards,
the sum of the forces is equal to 𝑇 minus 𝑚𝑔. This is equal to 𝑚𝑎. We can substitute in our values for
𝑔 and 𝑎. Adding 980𝑚 to both sides of this
equation gives us 𝑇 is equal to 1,108𝑚. As body B is accelerating
downwards, we take this as the positive direction. This gives us the equation 𝑚 plus
56𝑔 minus 𝑇 is equal to 𝑚 plus 56𝑎. Substituting in our values for 𝑔
and 𝑎 and replacing 𝑇 with 1,108𝑚 gives us 980𝑚 plus 54,880 minus 1,108𝑚 is
equal to 128𝑚 plus 7,168. Simplifying by collecting like
terms gives us 256𝑚 is equal to 47,712.

We can then divide both sides by
256, giving us 𝑚 is equal to 186.375 grams. We can now use this value to
calculate the tension 𝑇. 1,108 multiplied by 186.375 is
equal to 206,503.5. As our masses were in grams and the
acceleration in centimeters per square second, the tension will be measured in
dynes. This isn’t the final answer,
however, as we want the magnitude of the force exerted on the pulley. This is equal to two 𝑇 as there
are two tension forces acting on the pulley. The magnitude of the force exerted
on the pulley is therefore equal to 413,007 dynes.

We will now briefly summarize the
key points from this video. When solving problems on the motion
of a system of two bodies suspended vertically by a string, we use Newton’s second
law, 𝐹 equals 𝑚𝑎, where 𝐹 is the net force, 𝑚 is the mass, and 𝑎 is the
acceleration. We use simultaneous equations to
calculate any unknowns. We also saw it was important to use
the correct units, either kilograms, meters per square second, and newtons or grams,
centimeters per square second, and dynes. We also used the equations of
motion to help us solve more complicated problems. These are sometimes known as the
SUVAT equations.