Hello, everyone.
I'm Jesse Mason and in this episode of Teach Me we'll tackle our first pulley problem.
A pulley, by the way, is a wheel on an axle that's used to facilitate and direct the movement
of a rope or cable.
In this problem, a 100 kilogram crate is pulled along a horizontal surface with constant velocity
by a rope of negligible mass which is parallel to the surface.
"Negligible mass" means we can ignore the rope's inertial contribution to the system's
motion, thereby simplifying the problem.
Next, a massless, frictionless pulley is mounted to the end of the surface.
Again, this "massless, frictionless" pulley doesn't exist in the real world but in Physicsland...
situations can be idealized so that we're able to focus our attention on the underlying
fundamentals.
Okay, the rope in our problem runs over the pulley and then supports a hanging body (so
that the rope is vertical).
Our task is to determine the mass of the hanging body... given the coefficient of kinetic friction
between the surface and crate, of course.
The first step, as always, is to draw a picture.
So here's our surface.
And our crate, which we'll label m-sub-c.
Attached to the crate is our special rope, which runs over our special pulley and down
to a hanging body; we'll label that m-sub-b.
Now we identify our knowns and our unknowns.
We're given the mass of the crate, 100 kilograms, and we're looking for the mass of the hanging
body so we'll note that with a question mark.
We know how sticky the interface is between surface and crate, that is to say, we know
its coefficient of friction: 0.5.
Next, we'll indicate the directions of the velocity of the crate and the velocity of
the hanging body.
Since these bodies are connected via a rope, we know that the magnitude of their velocities
are the same.
And we're told in the problem that the crate's velocity is constant so we finish by writing
equals constant.
Of course, no picture or diagram is complete without an assigned coordinate system, so
we'll indicate our choice here: positive x-axis, to the right and positive y-axis, straight
up.
Okay, we've got forces at play here, which means it's time for -- you guessed it -- a
free body diagram.
Since our two bodies are connected in some fashion, we'll need an FBD for each of them.
First the crate: we've got the weight of the crate, pulling it downward.
Opposite the weight is the normal force, due to the surface along which our crate is moving.
The rope is pulling our crate to the right, so we'll indicate this tension force with
a T. Kinetic friction is present and that always opposes motion so we'll draw its force
vector opposite the velocity of the crate, to the left.
Oops, forgot the k.
Let's also not forget to draw our coordinate system.
Next we'll draw a free body diagram for the hanging body.
Here we have the weight of the body, directed downward and the tension of the rope pulling
upward.
Notice that we're using the same T for both crate and hanging body.
Since the tension of our rope is uniform -- this is always the case for massless ropes - we
know that the magnitude of the tension applied to the crate and to hanging body are equal.
So we can use the same notation for both tensions.
Now, normally we indicate somewhere on our free body diagrams the direction of the acceleration
of the body or bodies in question.
But since the velocities of both crate and hanging body are constant, their accelerations
are zero.
This implication is very important to us.
So a-sub-c equals zero, and a-sub-b equals zero.
Good!
Now we're ready to use equations to analyze the motions in our problem.
We'll start by using the FBD of the hanging body.
Applying Newton's Second Law of Motion to the body in the y-direction yields: the sum
of the forces in the y-direction equals the mass of the body times the acceleration of
the body in the y-direction.
As mentioned before, the body's acceleration is zero.
Now we sum the forces along the y-direction that are acting on the hanging body.
In the positive y-direction we have the tension force and in the negative y-direction we have
the body's weight.
Set that equal to zero.
Adding the weight to both sides and substituting our definition for weight, we get: T equals
the product of the m-sub-b and g.
Solving for m-sub-b, the unknown quantity for which we're looking, we get m-sub-b equals
T over g.
Now, tension is not a known quantity so to get the mass of the body, we need a value
or expression for T. Since we've learned all we can from the hanging body we now turn our
attention to the crate.
Using the FBD of the crate and applying N2 in the x-direction yields: the sum of the
forces in the x-direction equals the mass of the crate times the acceleration of the
crate in the x-direction.
Again, we know the acceleration is zero so we write that here and proceed to sum the
forces in the x-direction.
In the positive x-direction we have the tension force.
And in the negative x-direction we have the friction force.
Equals zero.
Adding the friction force to both sides and substituting our definition for kinetic friction
we get: T equals the product of the coefficient of kinetic friction and the normal force.
We've yet to determine a value or expression for the normal force.
So to get T, we first need N. So let's try N2 in the y-direction for the crate.
The sum of the forces along the y-direction equal the mass of the crate times the acceleration
of the crate in the y-direction.
You know what we're going to do next: set the acceleration to zero and then sum the
forces in the y-direction.
Okay, here we have the normal force in the positive y-direction and the weight of the
crate in the negative y-direction.
Set that equal to zero.
Next we move the weight to the right and substitute mg for the weight.
And now we have an expression for the normal force comprised of known values: m-sub-c and
g.
Time to hopscotch our way back to m-sub-b.
Substituting our expression for N into our expression for T, we get: T equals mu-sub-k
times m-sub-c times g.
And substituting this expression into our expression for m-sub-b, we get: m-sub-b equals
mu-sub-k times m-sub-c times g divided by g.
The g's divide out and we're left with an elegantly simple solution.
Lastly, we insert our values for the coefficient of kinetic friction and the mass of the crate
and we find that the mass of the hanging body is 50 kilograms.
And we're done with our first pulley problem.
I'm Jesse Mason, I hope you found this video helpful.
If you have any suggestions for future Teach Me videos, or just wanna say hello from your
part of the world, please do so in the comments below.
And until next time, happy learning!