### Video Transcript

Two bodies of masses π grams and
π 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 π equal to 9.8 meters per square second.

Letβs begin by modeling the
situation as shown. The two bodies have masses π grams
and π 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 ππ will
accelerate upwards, and the body of mass π plus 56π 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 π is
equal to π’π‘ plus a half ππ‘ squared. Substituting in our values gives us
64 is equal to zero multiplied by one plus a half multiplied by π multiplied by one
squared. The right-hand side simplifies to a
half π. We can then multiply both sides of
the equation by two, giving us π is equal to 128. The acceleration of the system is
equal to 128 centimeters per square second.

Our next step will be to use
Newtonβs second law, force equals mass multiplied by acceleration. As body A is accelerating upwards,
the sum of the forces is equal to π minus ππ. This is equal to ππ. We can substitute in our values for
π and π. Adding 980π to both sides of this
equation gives us π is equal to 1,108π. As body B is accelerating
downwards, we take this as the positive direction. This gives us the equation π plus
56π minus π is equal to π plus 56π. Substituting in our values for π
and π and replacing π with 1,108π gives us 980π plus 54,880 minus 1,108π is
equal to 128π plus 7,168. Simplifying by collecting like
terms gives us 256π is equal to 47,712.

We can then divide both sides by
256, giving us π is equal to 186.375 grams. We can now use this value to
calculate the tension π. 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 isnβt the final answer,
however, as we want the magnitude of the force exerted on the pulley. This is equal to two π 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.