Video Transcript
In this video, we’ll learn how to
solve problems involving moving a particle on a smooth, inclined plane. This will involve studying the
forces that act on such particles and using Newton’s second law of motion, as well
as the equations of motion, to describe this movement.
As we get started, say that we have
a body at rest on a flat, smooth surface. If we sketch in the forces acting
on this body, we know there’s a weight force equal to the object’s mass times the
acceleration due to gravity that acts straight downward. And there’s also an equal and
opposite reaction force. This has to do with the surface the
body rests on pushing back up on the body. Now Newton’s second law of motion
tells us that the net force acting on some object is equal to that object’s mass
times its acceleration. Because the weight force acting on
our body is equal and opposite to the reaction force, that means the net force on
the body is zero. And by Newton’s second law, we
therefore know that this object is not accelerating.
So this is what we find when our
body is at rest on a flat, smooth surface. But now let’s imagine that we
incline this smooth surface so that it forms an angle, we’ll call it 𝜃, with the
horizontal. Now if we draw the forces acting on
our body, once again, there’s the weight force acting straight down. But the reaction force no longer
points straight upward. In fact, this force points
perpendicularly to the surface the body rests on. Now, if this incline wasn’t smooth,
then there will be friction between the body and the incline. And we would include a third force,
a frictional force, acting up the slope. But since we are working with a
smooth slope, there is no such frictional force.
We see then that the only two
forces acting on our body are no longer equal and opposite. Therefore, the net force is not
zero. And so our body’s acceleration is
not zero, either. To look into what that acceleration
would be, say that we set up a coordinate frame, where our positive 𝑦-direction is
perpendicularly away from our surface, and the positive 𝑥-direction is down the
incline, if we now divide our two forces, the reaction and the weight forces, up
into their 𝑥- and 𝑦-components, we can see that the reaction force lies entirely
in the positive 𝑦-direction, while the weight force has a 𝑦- and an
𝑥-component.
To see that more clearly, let’s
look at this expanded view of the weight force acting on our body and its
components. This orange arrow here is the
𝑦-component of the weight force, while this one is the 𝑥-component. The key to solving for the
magnitude of these components is to recognize that this angle right here is
identical to the angle of inclination of our surface. And in fact, this will always be
the case, regardless of the value of 𝜃. If we then consider a general right
triangle, where another of the interior angles we call 𝜃, we can say that the sides
of this triangle are the hypotenuse, the side opposite 𝜃, and the side adjacent to
it. If we want to know the sin of the
angle 𝜃, that’s the ratio of the opposite side length to the hypotenuse.
On the other hand, the cos of 𝜃
equals the adjacent side length 𝑎 over the hypotenuse. We bring all this up because
solving for the 𝑦- and 𝑥-components of our weight force will involve using the cos
and sin of 𝜃. Note that comparing to the angle
𝜃, this 𝑦-component of our weight vector is effectively the adjacent side of our
right triangle. Looking at the equation involving
that side, we see that if we multiplied both sides by the hypotenuse ℎ, canceling
that factor on the right, then the adjacent side length equals the hypotenuse times
the cos of 𝜃. Since the hypotenuse of our actual
triangle is 𝑚 times 𝑔, we can write that the adjacent side of this right triangle
is 𝑚 times 𝑔 times the cos of 𝜃.
Likewise, the 𝑥-component of this
vector is the opposite side of the triangle to 𝜃. Working with that equation, if we
once again multiply both sides by the hypotenuse, canceling that factor on the
right, we see we can write the opposite side length as ℎ sin 𝜃. The 𝑥-component then of our weight
force is 𝑚 times 𝑔 times the sin of 𝜃. Knowing this, we can start to get a
bit of clarity about this acceleration that our body on the smooth incline
undergoes. If we look at the forces acting on
it in the 𝑦-dimension, we have the reaction force, which is positive, minus 𝑚
times 𝑔 times the cos of 𝜃. Note that we call the reaction
force positive and the weight force negative because of our choice of sign
convention.
Now, since these forces are the
only forces acting on our body in the 𝑦-dimension, Newton’s second law tells us
that their sum equals our body’s mass times its acceleration in this dimension. But then 𝑎 sub 𝑦 is zero. Our body isn’t accelerating either
into or away from the plane surface. So that means the whole right-hand
side of this expression is zero, implying that these forces are equal in
magnitude. So the acceleration of our body is
not in the 𝑦-direction. It must therefore be in the
𝑥-dimension. If we look at the forces along this
axis, we find that there’s just one. It’s the 𝑥-component of the weight
force. And since this force points in what
we’ve called the positive 𝑥-direction, we report it as a positive value.
Again, using Newton’s second law,
we can say that the net force on our object in the 𝑥-direction, 𝑚 times 𝑔 times
the sin of 𝜃, equals our object’s mass times its acceleration in the
𝑥-direction. Notice that the mass of our body 𝑚
appears on both sides of this equation. This means we can divide each side
by 𝑚 and cancel it out. We now know how our body
accelerates; it’s purely in the 𝑥-direction. And the acceleration has a
magnitude of 𝑔 times the sin of 𝜃. And notice that the acceleration
due to gravity and 𝜃 are both constant values. Therefore, the body’s acceleration
𝑎 stays constant in time. This is important because it means
a set of equations, called the equations of motion, apply for describing the motion
of this body.
The equations of motion are often
written this way. And every time we see an 𝑎
representing acceleration, we know that that value must be constant. That’s because these equations only
apply under that condition. Since we’ve established that our
body moving on a smooth incline is accelerating at a constant rate, we know its
motion can be described using these relationships. Knowing all this about the motion
of a body on a smooth inclined plane, let’s get a bit of practice now through an
example.
A body of mass 0.7 kilograms was
placed on a smooth plane inclined at 66 degrees to the horizontal and it was left to
move freely under the effect of gravity, where the acceleration due to gravity is
9.8 meters per second squared. Find, to the nearest two decimal
places, the magnitude of the reaction of the plane to the body.
Alright, let’s say that this is our
plane, and this is our body that starts out at rest on the plane. We know, though, that because of
the weight force acting on the body equal to its mass times the acceleration due to
gravity and the fact that we’re working with a smooth plane, meaning that there’s no
friction between the plane and our body, the body will start to move. When it does, it does not follow
the direction of the weight force but rather moves down the incline like this. The reason the body doesn’t move
straight down is because of a force called the reaction force. This acts on the body
perpendicularly from the incline, and it’s this force magnitude that we want to
solve for.
To start doing this, let’s note
that the mass of our body, we’ll call it 𝑚, is 0.7 kilograms, and that it’s moving
on a plane inclined to the horizontal at an angle of 66 degrees. Now, if we set up a coordinate
frame so that 𝑥 points down the incline and 𝑦 points perpendicularly away from it,
we can see, first of all, that our body’s acceleration is entirely in the
𝑥-direction. And this means that if we divide
the weight force acting on our object into its 𝑥- and 𝑦-components, then it must
be the case that the 𝑦-component of that force represented by this orange vector is
equal and opposite to the reaction force. This is because our body
experiences no acceleration in the 𝑦-dimension.
So if we can solve for the
𝑦-component of our weight force, then we’ll know the reaction force magnitude
𝑅. If we sketch this triangle at a
little bit bigger scale, we see we have a right triangle, the length of whose
hypotenuse is 𝑚 times 𝑔. To solve for this side length, the
𝑦-component of the weight force, we’ll need to recognize that this angle here of
our triangle is identical to the angle of inclination of our plane, what we’ve
called 𝜃. Knowing that, we can recognize that
this 𝑦-component then equals 𝑚 times 𝑔 times the cos of this angle. And since we’re given 𝑚 and 𝜃 and
we know 𝑔, the acceleration due to gravity, we can substitute those values into
this equation, leaving out units for now. And to two decimal places, we find
that this equals 2.79.
Since this is a force, the units
involved are newtons. Now technically, since the positive
𝑦-direction points perpendicularly away from our plane, the 𝑦-component of the
weight force will be a negative value. However, in this question, we want
to solve for the magnitude of this component because that is equal to the magnitude
of the reaction of the plane to the body. And that’s just what we’ve
calculated here. So we can say that the magnitude of
the reaction of the plane to the body is 2.79 newtons.
Let’s look now at a second
example.
A body of mass 1.4 kilograms was
placed on a smooth plane inclined at 45 degrees to the horizontal. If a force of 59 newtons is acting
on the body upward along the line of greatest slope of the plane, determine the
acceleration of the body rounded to the nearest two decimal places. Take 𝑔 to equal 9.8 meters per
second squared.
Okay, so if this is our plane
inclined at 45 degrees to the horizontal, we’re told there’s a body on this
plane. And it’s subjected to a force of 59
newtons acting up the incline. Knowing this, as well as the mass
of our body that we’ll call 𝑚, the acceleration due to gravity, and the fact that
we’re working with a smooth plane, which means there’s no friction involved, we want
to determine the acceleration of our body. Newton’s second law tells us that
the acceleration a body experiences multiplied by its mass is equal to the net force
that acts on it. To understand acceleration then,
let’s look at the forces acting on our body.
In addition to the 59-newton force,
there’s also a weight force acting straight downward equal to the body’s mass times
𝑔. And then there’s a reaction or
normal force 𝑅. This acts at 90 degrees to the
incline. To analyze the effect of these
forces, let’s set up a coordinate frame, where we say that the positive 𝑥-direction
is up the incline and positive 𝑦 is perpendicularly away from it. Dividing up the weight force into
its 𝑦- and 𝑥-components, we can see that our body won’t experience any
acceleration in what we’ve called the 𝑦-dimension. All of its acceleration will be
either up or down the incline in the 𝑥-direction.
Let’s consider then the
𝑥-direction forces acting on our body. There are two of these. One is the 59-newton force acting
in the positive 𝑥-direction, and the other is this component here of our weight
force, which acts in the negative 𝑥. To solve for this component, we’ll
want to recognize that this interior angle of the right triangle we’ve made is
exactly equal to the angle of inclination of our slope. Therefore, this 𝑥-component has a
magnitude of 𝑚 times 𝑔 times the sin of 45 degrees. When we add this to our 59-newton
force, we use a minus sign because the 𝑥-component of the weight force is in what
we’ve called the negative 𝑥-direction. By Newton’s second law, this sum is
equal to our body’s mass times its acceleration.
Dividing both sides by 𝑚,
canceling that factor out on the right, we have that 𝑎 is equal to 59 newtons minus
𝑚𝑔 sin 45 all divided by the mass of our body 𝑚. If we substitute in for 𝑚 and 𝑔
and leave out units at this step, when we calculate this fraction to the nearest two
decimal places, we get a result of 35.21. Remembering that this is an
acceleration, we know it will have units of meters per second squared. And so our final answer is that the
acceleration of our body to two decimal places is 35.21 meters per second
squared.
Let’s now look at one last
example.
A body of mass nine kilograms is
released from rest on a smooth inclined plane. It moves a distance of 25.2 meters
in the first four seconds of its motion. If the body is projected upward
along a line of greatest slope on the same plane, with an initial velocity of 12.6
meters per second, how far does it travel before coming to instantaneous rest? Take 𝑔 to equal 9.8 meters per
second squared.
Alright, so here we have a smooth
inclined plane. And we’re told that a body with a
mass we’ll call 𝑚 of nine kilograms is released from rest on the plane. Since there’s no frictional force
opposing its motion, the body starts to slide downward. And we’re told that after four
seconds, it’s moved a distance on the plane of 25.2 meters. This information is given to us so
we can solve for the body’s acceleration when it’s on the plane. We’ll need to know that to answer
this question of how far the body moves up the plane projected at an initial
velocity before coming to rest.
To solve then for this body’s
acceleration as it slides down the plane, it’s important to realize that this
acceleration is constant. If we call the angle of inclination
of our smooth plane 𝜃, then 𝑎, the acceleration, is equal to 𝑔 times the sin of
𝜃. The point here is that acceleration
is constant because 𝑔 and 𝜃 are constant. Therefore, we can calculate the
acceleration of our body using this given information by referring to the equations
of motion. We won’t list all four of those
equations. Here, we’ll just write down
one. This equation of motion says that
the displacement of a body is equal to its initial velocity times the time elapsed
plus one-half its acceleration times that elapsed time squared.
Now, in our case, the distance
traveled is 𝑑, and the change in time is Δ𝑡. And because our body is released
from rest, that means that 𝑣 sub 𝑖 in this equation, the body’s initial velocity,
is zero. Therefore, 𝑑 equals one-half 𝑎
times Δ𝑡 squared. If we multiply both sides of this
equation by two and divide both sides by Δ𝑡 squared, we get an equation where 𝑎 is
the subject. We won’t yet solve for 𝑎, but we
will come back to this equation in a moment.
For now, let’s consider this second
stage of motion for our body. We’re to imagine it’s projected up
the plane at 12.6 meters per second. We know that thanks to the effects
of gravity, the body will slow down over time and eventually it will come to a stop
just instantaneously.
If we call the distance the body
travels as it slows down from 12.6 to zero meters per second capital 𝐷, then we can
see that it’s this distance we want to solve for to answer our question. To do this, we’ll once again refer
to the equations of motion. To see what that equation is, let’s
clear away our problem statement, and we’ll now refer to this equation of
motion. It says that the final velocity of
an object squared equals its initial velocity squared plus two times its
acceleration times its displacement. Written in terms of our variables,
we could say that 𝑣𝑓 squared equals 𝑣𝑖 squared plus two times 𝑎 times capital
𝐷.
And notice this: because our object
ends up at rest, that means that 𝑣 sub 𝑓 is equal to zero. So we can write that zero equals 𝑣
sub 𝑖 squared plus two 𝑎𝐷. Since it’s 𝐷 we want to solve for,
let’s rearrange this expression. We get that 𝐷 equals negative 𝑣
sub 𝑖 squared over two 𝑎. And we can now substitute in the
expression we have for the acceleration 𝑎. Now, regarding this minus sign, if
we consider 𝑣 sub 𝑖, the initial velocity of our body, to be positive, then that
means we’re saying that motion up the incline is in the positive direction. Therefore, motion the opposite way
is in the negative direction. And that’s the way our acceleration
𝑎 will point.
By this sign convention, 𝑎 is a
negative quantity, and therefore the negative sign in numerator and denominator will
cancel out. We’re finally ready to plug in for
𝑣 sub 𝑖, 𝐷, and Δ𝑡. Leaving out units, we use a value
of 12.6 for 𝑣 sub 𝑖, 25.2 for 𝐷, and four for Δ𝑡. When we compute this fraction, we
actually get a result of exactly 25.2. This distance is in meters. So we can say that if our body is
projected up the incline at 12.6 meters per second, the distance that will move
before it comes to rest is 25.2 meters.
Let’s finish up this lesson now by
summarizing a few key points. In this lesson, we’ve seen that a
body on a smooth, inclined plane experiences no friction. Moreover, the only forces present
are the weight force and the reaction force. We also saw that Newton’s second
law of motion that the net force acting on a body equals that body’s mass times its
acceleration can be used to solve for a body’s acceleration. And lastly, we saw that when a
body’s acceleration is constant, the equations of motion apply.
