Video Transcript
A body of mass 𝑚 grams rests on a
smooth plane inclined at an angle of 𝜃 to the horizontal. It is connected by a light string
passing over a smooth pulley fixed at the top of the plane to a mass of 40 grams
hanging freely vertically below the pulley. The system was released from rest,
and the body descended a distance of 686 centimeters down the plane in the first two
seconds of motion. Given that the force exerted on the
pulley was 54 root three gram-weight, determine the value of 𝑚. Take 𝑔 is equal to 9.8 meters per
second squared.
Firstly, let’s think carefully
about some of the terminology in this question. We’re told that the body has a mass
of 𝑚 g. Being a mass, the g stands for
grams in this case. At the end of the question, we’re
told to use a value of 9.8 meters per second squared for 𝑔. It’s important to recognize that
this is a different use of the letter 𝑔. In this case, it refers to the
acceleration due to gravity.
Next, we’re given that the plane is
smooth. And this tells us that there’s no
friction force resisting the movement of the body down the plane. The fact that we have a light
string means that we can assume it has a mass of zero, which will simplify our
calculations. The fact that the pulley is smooth
again means that we don’t need to account for any frictional force resisting the
movement of the string around the pulley.
Because the 40-gram body is said to
be hanging freely vertically below the pulley, we can assume that it will fall
straight downward or move straight upward unhindered by any obstructions when the
system is released. And when we talk about the force
exerted on the pulley, we’re talking about the resultant force of the two ends of
the string pulling down and to the left on it. We’re told that this force is equal
to the force of 54 root three grams acting under gravity. We’ll look at how to calculate the
resultant force later.
Now we’ve picked our way through
the language and the assumptions we can make. We can see that we’ve been asked to
find the value of 𝑚, the number of grams of mass that the body on the plane
has. The next thing we need to do is
draw a diagram to record all of the information we’ve been given to help us
visualize the problem. The plane is inclined at an angle
of 𝜃 to the horizontal. There’s a pulley fixed to the top
of the plane. And a light string hangs over the
pulley with a body at each end, one on the plane and one hanging directly beneath
the pulley. Each body has a weight force acting
directly downward due to gravity.
We can calculate the size of each
weight force by multiplying the mass by the acceleration due to gravity, Newton’s
second law. We will use standard SI units in
our calculations, so we will convert grams into kilograms. Looking at the freely hanging body
first, it has a mass of 40 grams, which is 0.04 kilograms. The acceleration due to gravity is
9.8 meters per second squared. The weight force then is equal to
0.04 times 9.8 newtons. That’s 0.392 newtons. The body on the plane has a mass of
𝑚 grams, which is 𝑚 over one thousand kilograms. So the weight force for this body
in newtons is 9.8𝑚 over a thousand.
The two bodies are pulling the
string in different directions, creating tension in the string. Let’s call the tension force 𝑇
newtons. Since the pulley is smooth and the
string can move freely over it, the tension either side of it will be equal to 𝑇
newtons. We can mark these forces on our
diagram. Now, there is also a normal
reaction between the inclined plane and the body resting on it. This acts perpendicularly to the
plane. Let’s call this force 𝑅 sub 𝐵,
with newtons as its unit.
Let’s take a closer look at the
normal reaction force of the body on the plane. As we said, it acts perpendicularly
to the plane. But we also know that the body
slides down the plane when released. This movement is perpendicular to
the line of the normal reaction. And the body doesn’t jump up off
the plane or sink down into the plane. This means that forces acting on
this body perpendicular to the plane are in equilibrium. The normal reaction force is
therefore equal to the component of the weight force acting in the opposite
direction.
We need to find an expression for
the component of the weight force acting directly opposite the normal reaction
force. We know that the weight force acts
directly downward. Therefore, we know that the angle
between the plane and the downward vertical is 90 minus 𝜃. And since we’re looking for the
component of the weight force acting perpendicularly downward from the plane, we can
see that this angle must be 𝜃. It’s 90 minus 90 minus 𝜃.
We know that the weight force is
9.8𝑚 over 1000 newtons. So the component of this acting to
balance the normal reaction against the plane is 9.8𝑚 over 1000 times cos 𝜃. And so the normal reaction force,
𝑅 sub 𝐵, is equal to 9.8𝑚 over 1000 cos 𝜃 newtons. Let’s add that to our diagram. From the point of view of the
pulley, the tension forces are pulling it downward and to the left. Let’s take a look at that more
closely.
Recall that the tension in the
string is constant, and we call it 𝑇 newtons. The string hangs over the
pulley. So there’s a tension force of 𝑇
newtons pulling it vertically downward and another force of 𝑇 newtons pulling it
down the line of the inclined plane. Because the tension forces are
equal in magnitude either side of the smooth pulley, the resultant force on the
pulley will act along the line that bisects the angle between the two ends of the
string. Let’s call each half of that angle
𝛼.
Now, the question told us that the
resultant force exerted on the pulley was equal to 54 root three gram-weight. Let’s convert that to newtons. A unit of gram-weight is the force
that would be exerted by one gram due to gravity. To convert to newtons, we’ll first
convert to kilogram-weight, as kilograms are the standard SI unit of mass. There are 1000 grams in a
kilogram. So 54 root three gram-weight is
equal to 54 root three over 1000 kilogram-weight. The force exerted by 54 root three
over 1000 kilograms due to gravity can be expressed as 54 root three over 1000 times
the acceleration due to gravity, which we are given is 9.8 meters per second
squared.
Now, recall that this force is the
resultant of the tension forces from each end of the string. And since these each act at an
angle of 𝛼 from the line of resultant action, we can express their sum as 𝑇 cos 𝛼
plus 𝑇 cos 𝛼 newtons, which can be simplified to two 𝑇 cos 𝛼 newtons. This means that 54 root three over
1000 times 9.8 is equal to two 𝑇 cos 𝛼.
If we look back at our main
diagram, we can see that the plane was inclined at an angle of 𝜃 to the
horizontal. So the angle it makes with the
vertical is 90 minus 𝜃. This is the same angle as two
𝛼. We can write that two 𝛼 is equal
to 90 minus 𝜃. And hence 𝛼 is equal to 45 minus
𝜃 over two. And substituting this value for 𝛼,
we have that 54 root three over 1000 times 9.8 is equal to two 𝑇 cos 45 minus 𝜃
over two, which slightly simplifies to 1323 root three over 5000 is equal to 𝑇 cos
45 minus 𝜃 over two.
Now, that isn’t immediately helpful
because we don’t know the value of either 𝑇 or 𝜃. So let’s make a note of it and look
to see what else we can deduce from the question or diagram. Well, we know that the body on the
plane descends 686 centimeters in the first two seconds when the system is
released. We also know that in this
direction, a component of its weight is pulling it down the plane, resisted by the
tension force in the string. This means that there’s a constant
force acting on the body. And therefore it will travel down
the plane with a constant acceleration, which we will call 𝑎 meters per second
squared.
Let us convert the distance into
standard SI units of meters. 686 centimeters is 6.86 meters. So the distance traveled, which we
will call 𝑠, is 6.86 meters. And the constant rate of
acceleration is 𝑎 meters per second squared. The time interval, let’s call that
𝑡, is two seconds. And the system is released from
rest. So the initial velocity of the body
— let’s call that 𝑢 — is zero meters per second.
Now we can use one of Newton’s
equations of motion under constant acceleration to calculate the acceleration of the
body. This will be helpful in determining
the magnitude of the tension force in the string later. We’ll use distance 𝑠 equals the
initial velocity 𝑢 times the time 𝑡 plus a half the acceleration 𝑎 times the
square of the time 𝑡. Since all of these variables are
now in standard SI units, we can simply substitute them into the equation. And with a bit of simplifying, we
find that the acceleration of the body down the plane is 3.43 meters per second
squared. We can add this to our diagram.
Now let’s resolve the forces down
the plane from the point of view of the body on the plane. Using Newton’s second law, the
resultant force acting on this body is equal to the component of its weight acting
down the plane minus the tension force 𝑇. Since the angle between the weight
force and the plane is 90 minus 𝜃 degrees, the resultant force is 9.8 over 1000
times 𝑚 times cos of 90 minus 𝜃 minus 𝑇. The mass is 𝑚 over 1000 kilograms,
and the acceleration is 3.43 meters per second squared. We know that cos of 90 minus 𝜃 is
equal to sin of 𝜃, so let’s use that instead.
Next, we can resolve the forces
vertically upward from the point of view of the freely hanging mass. Since the string is not stretching,
we know that the freely hanging mass will accelerate upward at the same rate the
other body accelerates down the plane. That’s 3.43 meters per second
squared. So we can also add that information
to our diagram. The resultant force acting upward
is the tension force 𝑇 newtons minus the downward weight force, which we calculated
earlier was 0.392 newtons. The mass of this body is 40 grams,
which is 0.04 kilograms, and its acceleration is 3.43 meters per second squared.
We can add 0.392 to both sides and
simplify to find that 𝑇 is equal to 0.5292, which means that the tension force is
0.5292 newtons. We can replace 𝑇 on our diagram by
this number. But we can also replace it in our
other unsolved equations. Now this equation has only one
unknown in it; we can solve it to find the value of 𝜃. We will be able to divide both
sides by 0.5292 to isolate the cosine term involving 𝜃 on the right. Luckily, the left-hand side
simplifies to root three over two. And we know that cos of 30 degrees
is equal to root three over two. So we can deduce that 45 minus 𝜃
over two is equal to 30.
Now, because of the physical
constraints of the problem, we’re only looking for the value of 𝜃 between zero and
90. With some rearranging, we find that
𝜃 is equal to 30 degrees. Let’s make a note of that on our
diagram. Now, we can replace 𝜃 with 30
degrees in our remaining equation and solve it for the thing that we are ultimately
looking for, the value of 𝑚. And we know that sin of 30 degrees
is equal to a half. We can gather the terms involving
𝑚 on the left and add 0.5292 to each side. Then, we can factor 𝑚 on the left
and simplify the factored terms. Finally, we can divide both sides
by 0.0049 minus 0.00343 to find the value of 𝑚, which is 360.
So the mass of the object on the
plane is 360 grams, and the value of 𝑚 is 360.
It’s worth quickly mentioning that
we didn’t end up using the normal reaction force of the body on the plane in the
end. We resolved perpendicularly to this
down the plane, so we didn’t need it. You can save a bit of time if you
spot these things. But in a complicated question like
this, you can often end up calculating a few things you don’t need along the
way.
