### Video Transcript

In this video, we will learn how to
analyze weight, friction, and normal reaction forces contributing to the motion of
objects on sloped surfaces. Before we move on to sloped
surfaces, letβs first refresh our memory about using these forces on a horizontal
surface.

Letβs begin by drawing all the
forces acting on our block which is sitting on a horizontal surface. The weight of the block, which we
will represent as π in this video, originates from the center of mass. And the normal reaction force,
which we will call π
for this video, which originates from the surface at a
position thatβs directly underneath the center of mass. We should remember that the normal
reaction force will always make a 90-degree angle with the surface or be normal to
the plane.

In this situation where an object
is sitting on a horizontal surface, the weight of the object will be in the opposite
direction to the normal reaction force. In fact, in the absence of any
other vertical forces or vertical component of forces, the weight of the object will
be equal to the negative of the normal reaction force, meaning they will have the
same magnitudes but will be acting in opposite directions. If we were to alter the situation
by trying to pull on our block, we would have to include our force of friction,
assuming that the surface is rough. Our force of friction will pull in
the opposite direction to our pull force. In this case, our force of friction
will be to the right of our screen and will be represented by πΉ subscript f.

If our object remains at rest, the
negative π will equal πΉ f, which means that the pull force and the force of
friction will have the same magnitude but will be heading in opposite directions, as
can be seen in the diagram with the pull force to the left and the force of friction
to the right. If we draw in black dotted lines to
represent the horizontal and vertical axes, we would see that all four of the forces
align with the coordinate system. In reference to the plane which
happens to be along the horizontal axis, we can see that the pull force and the
force of friction are parallel to the plane. Being perpendicular to the plane
would be in the same direction as the vertical axis such as the normal reaction
force and the weight of our object.

Letβs apply these concepts to an
object that is on a sloped surface. Being a sloped surface means that
the plane has been rotated with reference to the coordinate system. In this case, it was rotated by an
angle π. If we assume that we put the box on
an inclined plane with a smooth surface, the weight of the object will still be
pulling straight down along the vertical axis, starting at the center of mass of the
object. And the normal reaction force is
perpendicular to the surface directly below the center of mass. This can be drawn in one of many
ways. In this video, we have chosen a
common convention to draw the normal reaction force through the center of mass. Another way is to start the force
on the plane where the force of gravity intersects the plane and draw the normal
reaction force perpendicularly.

Many times we use a simple point to
represent our object to avoid confusion when drawing a free-body diagram. Unlike on a horizontal surface, on
a sloped surface, we can see that our normal reaction force does not align with the
vertical axis. The force would have to be split
into its two components. We could call them π
subscript V
for the vertical component and π
subscript H for the horizontal component. For this example, the vertical
component would be pointing to the top of the screen, which is in the opposite
direction as the weight of the object. And the horizontal component would
be pointing to the left of the screen.

By keeping the coordinate system as
vertical and a horizontal, it makes it difficult to analyze the motion of the
object. Instead, we rotate our coordinate
system so that an axis is perpendicular to the plane and an axis is parallel to the
plane. Now the weight needs to be broken
down to its components so that it is aligned appropriately with the coordinate
system instead of their reaction normal force. The two components are the weight
perpendicular, which is the component of the weight thatβs perpendicular to the
plane and opposes the normal reaction force, and the weight parallel or the
component of the force that is parallel to the plane. We have used subscripts that have
the symbol for perpendicular and parallel in order to tell these two components
apart.

If our block has a mass π, letβs
determine what the perpendicular and parallel components of the weight would be in
terms of known constants and variables. The weight of the object and its
components make a right triangle, with the angle at the top here being π, the same
as the angle that our surface was sloped at. The parallel component of the
weight is the opposite side of the triangle to the angle π, and the weight is the
hypotenuse of the triangle. Remembering trigonometry functions
SOHCAHTOA, we can use SOH as we have opposite and hypotenuse. The sin of the angle π will equal
the opposite side of the triangle divided by the hypotenuse.

Plugging in our variables, we have
the sin of π will equal the parallel component of the weight divided by the
weight. To isolate the parallel component,
we multiply both sides by π. The weight of the object times the
sine of the angle is equal to the parallel component of the weight. We should remember that the weight
of an object can be found by multiplying the objectβs mass by the acceleration due
to gravity. So we could replace π in our
equation with ππ, leaving us with the parallel component of the weight being equal
to ππ sin π.

Letβs use the same approach to
determine the perpendicular component of the weight. The perpendicular component of the
weight is the adjacent side of the triangle to π, and once again the weight is the
hypotenuse. From our trigonometry functions, we
can use the cosine. cos of π is equal to the adjacent
side of the triangle divided by the hypotenuse. Plugging in our variables, we have
the cos of π is equal to the perpendicular component of the weight divided by the
weight. To isolate the perpendicular
component, we multiply both sides by π. The weight of the object times the
cos of the angle π is equal to the perpendicular component of the weight. Once again, since weight is equal
to ππ, we can replace π with ππ and get ππ cos π is equal to the
perpendicular component of the weight.

Letβs analyze the perpendicular
forces and parallel forces to determine the motion of our block. To do this, weβll apply Newtonβs
second law πΉ net equals ππ to the perpendicular axis and to the parallel axis
separately. The two forces that we add together
along the perpendicular axis to get the net force are the normal reaction force and
the perpendicular component of the weight. We have chosen the common
convention as into the plane of being negative, which is why π
is positive and ππ
cos π is negative. Our block is not accelerating into
or away from the surface of the plane, and therefore ππ will be zero. So the magnitude of the normal
reaction force is equal to ππ cos π.

Letβs look at the extreme cases of
a sloped surface, a horizontal surface and a vertical surface. We did the horizontal surface
earlier, and we saw that π
is equal to π, assuming there are no other vertical
forces. Looking back at our formula, that
should make sense as the cos of zero degrees is one. When our plane makes an angle π of
zero degrees to the horizontal, then the reaction normal force is equal to the
weight or ππ. When our block is sliding along the
vertical surface, the angle π that the plane makes with the horizontal would be 90
degrees, and the cos of 90 degrees is zero. So the normal reaction force in
this case is equal to zero. Assuming a smooth surface, the only
force acting on the object would be the weight, so the object would essentially be
in free fall.

Next, weβll look at the forces that
are parallel to the plane. With the object being on a smooth
surface and no external forces, the only force thatβs acting parallel to the plane
is the parallel component of the weight. So ππ sin π is the net force
parallel to the plane, and that is equal to ππ. Since thereβs an π on both sides
of our equation, they can cancel each other out, leaving us with an equation for the
acceleration of our box of π sin π. As our slope goes to 90 degrees,
sin π becomes closer and closer to one. And the acceleration of the object
approaches free fall or π.

But what if the box is on a rough
surface instead of a smooth surface? The force of friction will be
parallel to the plane and, in this case, will be going up the incline as the box
will be trying to slide down the incline. Because the force of friction is
parallel to the plane, we must include it in the net force for the forces that are
parallel. Our Newtonβs second law equation
becomes the weight parallel to the plane minus the force of friction is equal to
ππ. We have chosen to make down the
incline a positive number as that is the motion of the box, thereby making up the
incline β in this case, the force of friction β to be negative.

To isolate the friction force, we
must add the force of friction to both sides of the equation, thereby canceling out
the force of friction on the left side. Then we must subtract ππ from
both sides of the equation, thereby canceling out ππ on the right side, leaving us
with an expression for the force of friction of ππ sin π minus ππ. Now that we have learned how to
analyze weight, friction, and normal reaction forces for objects on sloped surfaces,
letβs do a few example problems.

A point mass π is on a slope, as
shown in the diagram. The weight of the mass, π, and the
normal reaction force, π
, act on the point mass. What is the relationship between π
and π if the angle of the slope above the horizontal is zero? What is the magnitude of π
if the
angle of the slope above the horizontal is 90 degrees?

In the first question, it asked us
to find the relationship between π
and π, where π
is the normal reaction force
and π is the weight of the object when the angle of the slope above the horizontal
is zero. Letβs redraw our diagram where we
have an angle of zero. If the angle is zero degrees, that
implies that we have a horizontal surface. Applying Newtonβs second law to the
situation using only the vertical forces as those are the only forces acting on our
point, we get that π
, the normal reaction force, minus π, the weight, is equal to
zero, as our point mass is not accelerating up and down.

We use the common convention that
up is positive, thatβs why that π
has a positive value, and that down is negative,
which is why the weight π has a negative value. This means the magnitude of the
normal reaction force is equal to the magnitude of the weight. Therefore, the relationship between
π
and π is that they are equal. For the second question, we are
asked to find the magnitude of π
if the angle of the slope above the horizontal is
90 degrees. Letβs once again draw our diagram
with an angle of 90 degrees this time.

To solve for this question, letβs
find a relationship for π
in terms of π. If we align our coordinate system
to be parallel and perpendicular to the plane, then we must break down our weight
vector into parallel and perpendicular components. We can see the normal reaction
force will have the same magnitude as the perpendicular component of the weight. To find the perpendicular component
of the weight, we can use trigonometry as the perpendicular component is the
adjacent side of a right triangle to the angle π and the weight π is the
hypotenuse of the triangle. So we can use the equation the cos
of the angle π is equal to the adjacent side of a triangle divided by the
hypotenuse of the triangle. Plugging in our values, we have cos
π is equal to the perpendicular component of the weight divided by the weight.

To isolate the perpendicular
component of the weight, we multiply both sides by π, canceling out the π on the
right side of the equation, leaving us with the weight of the object times the cos
of the angle π is equal to the perpendicular component of the weight. We said earlier that the magnitude
of π
will be equal to the magnitude of the perpendicular component of the weight,
so we can replace the perpendicular component with π
. We should also recall that weight
is ππ, so we could replace the weight π with ππ. So the expression for the normal
reaction force is ππ cos π. The cos of 90 degrees is zero,
which means that the normal reaction force of point mass π when the angle of the
slope is 90 degrees is zero. The magnitude of π
as the angle of
the slope above the horizontal becomes 90 degrees is zero.

An object has a mass π is equal to
1.3 kilograms. The object is on a smooth slope
with an angle π equals 33 degrees, as shown in the diagram. What is the acceleration π of the
object parallel to the surface of the slope?

Letβs begin to solve the problem by
drawing in the forces on our object. The weight of the object pulls
straight down, and the normal reaction force acts perpendicularly to the
surface. Since the slope is smooth, we do
not need to worry about friction. The question asks us for the
acceleration of the object parallel to the surface of the slope. Therefore, we need to look at any
forces that also happen to be parallel. Only the weight can have a
component that is parallel as a normal reaction force is perpendicular to the
surface. Letβs break down the weight into
its components. The only component that we care
about is the parallel component of the weight.

Looking at the right triangle that
we drew between the weight and its components, we can see that the parallel
component of the weight is the opposite side of the triangle to the angle π and the
weight π is the hypotenuse. We can use trigonometry to
determine the parallel component of the weight using sin of π is equal to the
opposite side of the triangle divided by the hypotenuse. Plugging in our variables, weβd
have sin π is equal to the parallel component divided by the weight. To isolate the parallel component,
we multiply both sides of the equation by π. This cancels out the π on the
right-hand side. Our expression for the parallel
component of the weight becomes π sin π.

We should remember that the weight
of an object is equal to the mass of the object times the acceleration due to
gravity, so we can replace the weight with ππ. Now we can plug in the values from
our problem. We replace mass with 1.3 kilograms,
acceleration due to gravity with 9.8 meters per second squared, and the sin of the
angle π with the sin of 33 degrees, which gives us a value for the parallel
component of our weight of 6.94 newtons. How does that help us find the
acceleration in the parallel direction?

We should recall that Newtonβs
second law is a relationship between the net force and the acceleration of an
object, πΉ net equals ππ. The net force parallel to the
surface of the slope, we just found to be 6.94 newtons. And the mass of the object was
given to us as 1.3 kilograms. To isolate the acceleration, we
divide both sides of the equation by 1.3 kilograms, thereby canceling out 1.3
kilograms on the right-hand side. When we divide our two numbers, we
get 5.3 meters per second squared. The acceleration of the object
parallel to the surface of the slope is 5.3 meters per second squared.

Key Points

Use Newtonβs second law of motion,
πΉ net equals ππ, to find the normal reaction force on an object on a sloped
surface. Break down the weight of an object
into parallel and perpendicular components for an object on a sloped surface. When a smooth sloped surface is at
90 degrees above the horizontal, the normal reaction force is zero.