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>> In this video,
we're going to take
a Closer Look at
advanced application of
Newton's 2nd law of motion.
Newton's 2nd law states that
the net force of an object
equals the object's mass
times its acceleration.
F sub net equals m times a.
Three unknown masses,
m1, m2 and m3,
hang together on thin wires.
If the wires have no mass
and the tension in the string
at each point is given as shown,
then what are the masses
of these objects?
From looking at this diagram,
we see that the tension force
in the top wire, T1,
supports all the masses.
T2 supports the masses
m2 and m3.
And T3 supports m3 only.
Let's focus on m3
and draw a free body diagram
of the forces
acting on this mass.
The forces acting on m3
are tension pulling up
and gravity pulling down.
Since m3 is at rest,
the net force acting on it
is zero.
We've made our
positive direction up,
which means that T3 minus m3
times g equals zero.
If we add m3 times g to
both sides of the equation
and then divide both sides
by g, we find that
m3 equals the tension
in the lowest wire, T3,
divided by the acceleration
due to gravity, g.
Plugging in 21 newtons for
T3 and 9.8 meters per second
squared for g,
we see that m3
equals 2.1 kilograms.
Now let's look at
a free body diagram on m2.
M2 has a tension force, T2,
pulling up,
a tension force, T3,
pulling down,
and the force of gravity,
m2 times g, also pulling down.
The mass is in equilibrium.
So the total force
of m2 is zero.
This means T2 minus T3,
minus m2 times g, equals zero.
If we add m2 times g
to both side of the equation
and then divide both sides
by g, we find that
m2 equals T2 minus T3,
all divided by g.
Plugging in 50 newtons for T2,
21 newtons for T3 and
9.8 meters per second squared
for g, we find that
m2 equals three kilograms.
Lastly, to solve for m1,
let's look at a free body
diagram of the forces
acting on this mass.
Just like with the other masses,
m1 is not moving.
So the net force
acting on it is zero.
That means T1 minus T2,
minus m1 times g, equals zero.
If we add m1 times g
to both sides of the equation
and then divide both sides
by g, we see that
m1 equals T1 minus T2,
all divided by the acceleration
due to gravity, g.
Plugging in 78 newtons for T1,
50 newtons for T2 and
9.8 meters per second squared
for g, we find that
m1 equals 2.9 kilograms.
Let's do another exercise
of Newton's 2nd law.
A rope is hung over
a frictionless tree branch.
One end is tied to
a 23 kilogram crate
sitting on the ground.
And an 18 kilogram monkey
climbs up the other side.
If the monkey climbs with
an acceleration of 1.7 meters
per second squared,
how fast does the crate
accelerate?
What is the highest acceleration
the monkey can have
without lifting the crate up?
Let's begin by drawing
a free body diagram of the
forces acting on the monkey.
There's the force of gravity
pulling down and
the tension of the rope
pulling the monkey up.
The tension force arrow is
longer because the problem
told us the monkey has
an upward acceleration
as it climbs the rope.
The net force on the monkey
equals its mass times
its acceleration.
The positive direction
we have chosen is upward.
The tension force, F sub T,
minus the weight of the monkey,
the mass of the monkey times g,
the acceleration due to gravity,
equals the monkey's mass
times its acceleration.
We know all the terms here
except the tension force,
F sub T.
We want to solve for that force
because it acts on the crate
lifting it up.
So if we add the mass
of the monkey times g
to both sides of the equation,
and then factor out
the monkey's mass
from the right side,
we see that the tension force
equals the mass of the monkey
times its acceleration,
plus g, the acceleration
due to gravity.
Plugging in 18 kilograms
for the monkey's mass,
1.7 meters per second squared
for its acceleration,
and 9.8 meters per second
squared for gravity's
acceleration, we find that
the tension in the rope
is 207 newtons,
or 210 newtons when using
significant figures.
Knowing that, now let's solve
for the acceleration
of the crate.
Draw a free body diagram
of the crate.
The same tension force that
pulls up on the monkey,
also pulls up on the crate.
And the crate also has
the force of gravity
acting downward on it.
Also, when the crate
is resting on the ground,
a normal force acts upward
on the crate.
It may turn out that
this normal force is zero.
That the tension force
is enough to counteract
the downward force of gravity
on the box,
and the ground isn't
pushing up on the box at all.
We can figure that out by
comparing the tension force
with the force of gravity.
The tension force
acting on the crate
is the same as the tension
force on the rope
as the monkey moves up it.
F sub T equals 210 newtons.
The weight force of the crate
equals its mass times the
acceleration due to gravity.
23 kilograms times 9.8 meters
per second squared.
That equals 225.4,
or with sig figs, 230 newtons.
Since the weight force is
greater than the tension force,
that means the crate continues
to rest on the ground.
So when the monkey's upward
acceleration is 1.7 meters
per second squared,
the crate remains at rest
on the ground.
Its acceleration is zero.
Now let's solve for
the highest acceleration
the monkey can have without
lifting up the crate.
Draw a free body diagram
on the crate.
The same tension force that
pulls up on the monkey
also pulls up on the crate.
And the crate also has
the force of gravity
acting downward on it.
Also, when the crate is
resting on the ground,
a normal force acts upward
on the crate.
When the tension force
on the crate equals
but does not exceed
the weight force,
then the normal force
on the crate will be zero,
and the crate will be
as near to being in motion
as it can be
without actually moving.
That's a condition we'll
assume that we can solve for
the monkey's top acceleration
without moving the crate.
By looking at this
free body diagram,
we see that the net force
acting on the crate is zero.
F sub T minus the mass
of the crate, times g,
equals zero.
Rearranging our equation,
we see that F sub T equals
the mass of the crate times g.
Plugging in our values
for the crate's mass
of 23 kilograms,
and the acceleration
due to gravity, g,
which is 9.8 meters
per second squared,
we see that the tension force
is 225.4 newtons,
or 230 newtons
when taking into account
significant figures.
Now let's switch back to
the monkey's free body diagram
with our new tension force
being applied.
By Newton's 2nd law,
with the positive direction
upward, the tension force
minus the monkey's weight
equals the monkey's mass
times its acceleration.
We want to know
the monkey's acceleration,
so we divide both sides
of the equation
by the monkey's mass.
The monkey's acceleration
equals the tension force
minus the monkey's mass,
times g, all divided by
the mass of the monkey.
Plugging in 230 newtons
for the tension force,
18 kilograms for
the monkey's mass,
and 9.8 meters per second
squared for g,
we find that the monkey's
maximum acceleration
without lifting up the crate
is 3.0 meters
per second squared.
Now let's solve another
2nd law problem involving
inclined planes.
Two blocks,
where m1 equals 13 kilograms
and m2 equals 27 kilograms,
sit on two frictionless planes.
The first elevated
at 47 degrees,
and the second at 28 degrees.
The masses are connected
by a rope strung over
a frictionless pulley.
What is the acceleration
of the masses?
Assume that motion to
the right is positive.
Take a minute, press pause,
and try this problem
out for yourself.
Did you find that
the system of blocks
accelerates to the right at
0.8 meters per second squared?
If so, great work.
If not, let's go through
the exercise together.
The key to this problem
is understanding that
the two blocks are a system.
If m1 moves, m2 moves
just as much and vice versa.
We'll start by drawing
a free body diagram
of each block separately.
Each one experiences
a gravitational force,
a normal force,
and the tension force
due to the rope.
We want to find the component
of the weight force
acting down each incline,
so we can compare them
and see which way
the system moves
or if it stays still.
Let's look more closely at
block's one free body diagram.
The normal force acts on
the box perpendicular
to the surface.
Tension acts parallel to,
and of the incline,
and gravity acts at an angle
to the incline perpendicular
to the surface of the earth.
Here, the components of
the force due to gravity
are drawn in as dotted lines.
One component of the force
due to gravity
acts normal to the plane.
And one component
is parallel to it.
That's the component
that acts on the block
down the incline
in the negative X direction.
The angle between the force
due to gravity
and the normal component of
that force is theta one.
From our experience
with vector analysis,
we know that the component
of the weight force
pushing block one
down the incline in
the negative X direction,
which we'll call F sub 1Gx,
is the block's mass
times the acceleration
due to gravity,
times the sine of 1 theta.
Now let's take a closer look
at block's two
free body diagram.
Just like with block one,
the normal force acts on
the box perpendicular
to the surface.
Tension acts parallel
and of the incline,
and gravity acts at an angle
to the incline perpendicular
to the surface of the earth.
There's a component of gravity
that acts down the incline
in the positive X direction.
In this case, the component
which we'll call F sub 2Gx,
equals m2 times the acceleration
due to gravity,
times the sine of theta 2.
When we consider the system of
blocks one and two together,
these forces along the plane
are the only ones that cause
the system to accelerate
in the X direction.
Since all other forces
acting on them are acting
in the vertical direction
and don't affect acceleration
in the X direction,
the acceleration
we are solving for.
applying Newton's 2nd law
to the system, we write that
the net force equals
the combined mass, m1, plus m2,
times the acceleration
of the system.
So, F sub 2Gx
minus F sub 1Gx
equals the combined mass and
accelerations of m1 and m2.
We want to solve for
the system's acceleration,
so we divide both sides
of the equation
by the combined mass.
Then replace both forces
with the values we arrived at
earlier in terms of mass,
the acceleration due to gravity
and the angle of the incline.
F sub 2Gx with
m2 g sine theta 2.
And F sub 1Gx with
m1 g sine theta 1.
We can now plug in the numbers
given for these variables.
The acceleration of the system
equals 27 kilograms
times 9.8 meters
per second squared,
times the sine of 28 degrees,
minus 13 kilograms,
times 9.8 meters
per second squared,
times the sine of 47 degrees,
all divided by 40 kilograms.
The net acceleration
of the system of blocks is
0.8 meters per second squared
to the right.
That's a closer look at
advanced application of
Newton's 2nd law of motion.
For more "Physics in Motion",
go to our homepage
where the entire series
is available to you.
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