### 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, with one of them on an inclined plane. We will study problems where the
plane is both smooth and rough. To begin this video, we will recall
Newton’s second law of motion.

Newton’s second law states that the
net force on an object is equal to its mass multiplied by its acceleration. In this video, we will be dealing
with scalar quantities and use the equation 𝐹 equals 𝑚𝑎, where 𝐹 is the sum of
the net forces, 𝑚 is the mass, and 𝑎 the acceleration. We will measure acceleration in
meters per square second. The mass will be measured in
kilograms. Multiplying these gives us kilogram
meters per square second. One kilogram meter per square
second is equal to one newton. And this is the unit we’ll use to
measure our forces.

We will now consider what the
problem will look like when the plane is smooth. We will have two bodies 𝐴 and 𝐵
connected by a light inextensible string passing over a smooth pulley. This means that the tension in the
string will be equal throughout. Both bodies will have a force going
vertically downwards equal to its weight. This will be equal to the mass of
the body multiplied by gravity. There will be a reaction force
perpendicular to the plane. However, as the plane is smooth,
there will be no frictional force on it.

As the string is inextensible, when
the system is released, it will accelerate uniformly. We will then use Newton’s second
law to resolve for body 𝐴 and body 𝐵. In the case of body 𝐵, we will
resolve vertically. For body 𝐴, we will resolve
parallel to the plane. In order to do this, we will need
to know the angle of inclination, in this case 𝛼. Using our knowledge of right angle
trigonometry, we can identify the component of the weight of body 𝐴 parallel and
perpendicular to the plane. The force acting parallel to the
plane is equal to 𝑀𝑔 multiplied by sin 𝛼, and the force perpendicular to the
plane is equal to 𝑀𝑔 multiplied by cos 𝛼.

If body 𝐵 is accelerating
downwards and we consider this to be the positive direction, the sum of the forces
are equal to 𝑀𝑔 minus 𝑇. This is equal to the mass of body
𝐵 multiplied by the acceleration. Body 𝐴 is moving up the plane. If this is the positive direction,
the sum of the forces acting parallel to the plane are equal to 𝑇 minus 𝑚𝑔 sin
𝛼. This is equal to the mass of body
𝐴 multiplied by the acceleration.

This will give us two simultaneous
equations that we can use to calculate any unknowns. We will now look at a couple of
examples.

A body of mass five kilograms rests
on a smooth plane inclined at an angle of 35 degrees to the horizontal. It is connected by a light
inextensible string passing over a smooth pulley fixed at the top of the plane to
another body of mass 19 kilograms hanging freely vertically below the pulley. Given that the acceleration due to
gravity 𝑔 is equal to 9.8 meters per square second, determine the acceleration of
the system.

We will begin by sketching the
system. We are told that the smooth plane
is inclined at an angle of 35 degrees. The masses of the two bodies are
five kilograms and 19 kilograms. This means that they will have a
vertical downward force equal to five 𝑔 and 19𝑔, respectively, where the gravity
𝑔 is equal to 9.8 meters per square second.

The pulley is smooth, and the
string is light and inextensible. This means that the tension in the
string will be equal throughout. When the system is released, the
magnitude of acceleration will also be constant. As the plane is smooth, there will
be no frictional force. In order to solve this problem, we
will use Newton’s second law, which states that the sum of the net forces is equal
to the mass multiplied by the acceleration. For the body on the plane, we will
resolve parallel to the plane. And for the body hanging freely, we
will resolve vertically.

The five 𝑔 force is neither
parallel nor perpendicular to the plane. Therefore, we need to calculate
these two components. Using our knowledge of right angle
trigonometry, we see that the perpendicular component is equal to five 𝑔 multiplied
by cos of 35 degrees and the component parallel to the plane is equal to five 𝑔
multiplied by sin of 35 degrees.

Body 𝐴 is moving up the plane. If we assume this is the positive
direction, the sum of its forces is equal to 𝑇 minus five 𝑔 multiplied by sin
35. This is equal to five 𝑎 as the
mass of the object is five kilograms. As object 𝐵 is accelerating
downwards, the sum of its forces is equal to 19𝑔 minus 𝑇. This is equal to 19𝑎. We now have two simultaneous
equations that we can solve to determine the acceleration of the system. By adding equation one and equation
two, we eliminate the tension 𝑇.

The left-hand side becomes 19𝑔
minus five 𝑔 multiplied by sin of 35 degrees. And the right-hand side becomes
24𝑎. We can then divide both sides of
this equation by 24. Typing this into the calculator
gives us a value of 𝑎 equal to 6.5872 and so on. Rounding this to two decimal places
gives us an answer of 6.59 meters per square second. We could substitute this value back
into equation one or equation two to calculate the tension 𝑇. However, this is not required in
this question.

We will now consider what happens
when the surface is rough. Recall in the diagram we saw
earlier for a smooth plane we used Newton’s second law to create two equations. We resolved vertically for the
freely hanging object and resolved parallel to the plane for the object on the
inclined plane. Let’s now consider what happens
when the plane is rough.

When body 𝐴 is accelerating up the
plane, there will be a frictional force acting downwards parallel to the plane. This means that we have a tension
force acting in the positive direction and two forces acting in the negative
direction. The sum of the net forces parallel
to the plane is equal to 𝑇 minus 𝑚𝑔 multiplied by sin 𝛼 minus the frictional
force 𝐹 𝑟. This is equal to the mass
multiplied by the acceleration.

We know that when dealing with a
frictional force, it is equal to 𝜇, the coefficient of friction, multiplied by the
normal reaction force 𝑅, when 𝜇 is a constant that lies between zero and one
inclusive. If we resolve perpendicular to the
plane, we see that the sum of the net forces is equal to 𝑅 minus 𝑚𝑔 multiplied by
cos 𝛼. However, the body is not moving in
this direction. Therefore, the acceleration is
equal to zero. 𝑅 minus 𝑚𝑔 multiplied by cos 𝛼
is equal to zero.

This can be rewritten as 𝑅 is
equal to 𝑚𝑔 multiplied by cos 𝛼. When solving any problem of this
type on a rough plane, we can use our three equations to calculate any unknowns,
together with the formula that links the frictional force and the normal reaction
force. We will now look at an example of
this type.

A body 𝐴 of mass 240 grams rests
on a rough plane inclined to the horizontal at an angle whose sine is
three-fifths. It is connected by a light
inextensible string passing over a smooth pulley fixed to the top of the plane to
another body 𝐵 of mass 300 grams. If the system was released from
rest and body 𝐵 descended 196 centimeters in three seconds, find the coefficient of
friction between the body and the plane. Take 𝑔 equal to 9.8 meters per
square second.

We begin by sketching the
system. We are told that the sin of angle
𝛼 is equal to three-fifths. Using our knowledge of right angle
trigonometry and the Pythagorean triple three, four, five, we know that cosine of
𝛼, or cos of 𝛼, is equal to four-fifths. The mass of the two bodies is given
in grams. We know that there are 1000 grams
in one kilogram. This means that 240 grams is equal
to 0.24 kilograms. We divide our value in grams by
1000.

Body 𝐴 will, therefore, have a
force acting vertically downwards equal to 0.24𝑔, where 𝑔 is equal to 9.8 meters
per square second. Body 𝐵 has a mass of 300 grams,
and this is equal to 0.3 kilograms. Therefore, this body has a downward
force of 0.3 multiplied by 𝑔.

We have a light inextensible string
passing over a smooth pulley. This means that the tension
throughout the string will be equal. It also means that when released,
the system will travel with uniform acceleration. Body 𝐴 has a normal reaction force
perpendicular to the plane. As the plane itself is rough, there
will be a frictional force acting down the plane.

We will now use Newton’s second
law, which states that the sum of the net forces is equal to the mass multiplied by
the acceleration, to resolve parallel and perpendicular to the plane for body 𝐴 and
vertically for body 𝐵. The weight of body 𝐴 is acting
vertically downwards. Therefore, we need to find the
components of this that are parallel and perpendicular to the plane. Once again, using our knowledge of
right angle trigonometry gives us a force of 0.24𝑔 multiplied by cos 𝛼
perpendicular to the plane and 0.24𝑔 multiplied by sin 𝛼 parallel to the
plane.

There are three forces acting on 𝐴
parallel to the plane: the tension force, the frictional force, and this weight
component. This gives us the equation 𝑇 minus
0.24𝑔 multiplied by sin 𝛼 minus the frictional force 𝐹 𝑟 is equal to 0.24𝑎. Perpendicular to the plane, the sum
of the net forces is equal to 𝑅 minus 0.24𝑔 multiplied by cos 𝛼. The body is not moving in this
direction. Therefore, this is equal to
zero.

Rearranging the equation, we see
that the normal reaction force 𝑅 is equal to 0.24𝑔 multiplied by cos 𝛼. Finally, resolving vertically for
body 𝐵, where downwards is the positive direction, gives us 0.3𝑔 minus 𝑇 is equal
to 0.3𝑎. We can substitute in our values for
sin 𝛼 and cos 𝛼. This means that the normal reaction
force is equal to 1176 over 625. 0.24𝑔 multiplied by sin 𝛼 is
equal to 882 over 625.

Our next step is to calculate the
acceleration of the system, given the fact that body 𝐵 descended 196 centimeters in
three seconds. In order to do this, we will use
the equations of motion, also known as the SUVAT equations. The displacement of the body was
196 centimeters. This is equal to 1.96 meters, as
there are 100 centimeters in a meter. The body was released from rest, so
the initial velocity is zero meters per second. We are trying to calculate the
acceleration, and we are told the time is three seconds.

This means that we can use the
equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values gives us
1.96 is equal to zero multiplied by three plus a half multiplied by 𝑎 multiplied by
three squared. The right-hand side simplifies to
4.5𝑎. Dividing both sides by 4.5 gives us
𝑎 is equal to 98 over 225. The acceleration of the system is
98 over 225 meters per square second. We can now substitute this value
into our equations.

We now have two unknowns left, the
tension 𝑇 and the frictional force 𝐹 𝑟. We know that the normal reaction
force is 1176 over 625 newtons. By adding 𝑇 and subtracting 49
over 375 to both sides of the bottom equation, we can calculate the tension force
𝑇. This gives us a value of 𝑇 equal
to 2107 over 750 newtons. We can substitute this into the top
equation. Rearranging this equation gives us
a frictional force 𝐹 𝑟 equal to 1617 over 1250 newtons.

We now need to calculate the
coefficient of friction. And we know that the frictional
force is equal to this coefficient of friction multiplied by the normal reaction
force. This means that the coefficient of
friction 𝜇 is equal to 𝐹 𝑟 divided by 𝑅. Typing this into the calculator
gives us a coefficient of friction 𝜇 equal to eleven sixteenths.

We will now summarize the key
points from this video. To solve problems involving pulleys
on an inclined plane, we used Newton’s second law, 𝐹 equals 𝑚𝑎. We resolve forces vertically as
well as parallel and perpendicular to the plane. If the plane is rough, we have a
frictional force 𝐹 𝑟 acting against the motion, where this frictional force is
equal to the coefficient of friction 𝜇 multiplied by the normal reaction force
𝑅. We can also use the equations of
motion, or SUVAT equations, to calculate unknowns and help us solve problems of this
type.