### Video Transcript

A point mass ๐ is on a slope, as
shown in the diagram. The weight of the mass, ๐, and the
normal reaction force, ๐
, act on the point mass. What is the relationship between ๐
and ๐ if the angle of the slope above the horizontal is zero? What is the magnitude of ๐
if the
angle of the slope above the horizontal is 90 degrees?

In the first question, it asked us
to find the relationship between ๐
and ๐, where ๐
is the normal reaction force
and ๐ is the weight of the object when the angle of the slope above the horizontal
is zero. Letโs 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 Newtonโs second law to the
situation using only the vertical forces as those are the only forces acting on our
point, we get that ๐
, the normal reaction force, minus ๐, 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, thatโs why that ๐
has a positive value, and that down is negative,
which is why the weight ๐ 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
๐
and ๐ is that they are equal. For the second question, we are
asked to find the magnitude of ๐
if the angle of the slope above the horizontal is
90 degrees. Letโs once again draw our diagram
with an angle of 90 degrees this time.

To solve for this question, letโs
find a relationship for ๐
in terms of ๐. 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 ๐ and the weight ๐ is the
hypotenuse of the triangle. So we can use the equation the cos
of the angle ๐ is equal to the adjacent side of a triangle divided by the
hypotenuse of the triangle. Plugging in our values, we have cos
๐ 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 ๐, canceling out the ๐ on the
right side of the equation, leaving us with the weight of the object times the cos
of the angle ๐ is equal to the perpendicular component of the weight. We said earlier that the magnitude
of ๐
will be equal to the magnitude of the perpendicular component of the weight,
so we can replace the perpendicular component with ๐
. We should also recall that weight
is ๐๐, so we could replace the weight ๐ with ๐๐. So the expression for the normal
reaction force is ๐๐ cos ๐. The cos of 90 degrees is zero,
which means that the normal reaction force of point mass ๐ when the angle of the
slope is 90 degrees is zero. The magnitude of ๐
as the angle of
the slope above the horizontal becomes 90 degrees is zero.