### Video Transcript

A car with a mass of 660 kilograms
drives at constant speed along a smooth circular track, as shown in the diagram. The car follows a path along the
center of the track, maintaining a constant distance π equals 22 meters to the
center of its circular path. The angle of the track above the
horizontal π equals 28 degrees. What is the magnitude of the force
that acts on the car along π? Answer to the nearest newton.

This question asks what the
magnitude of the force acting along the radius π of a circular path that a car
moves along is. This force on the car is due to the
normal reaction force on the car due to the contact between the car and an inclined
surface.

Letβs begin by recording the
information weβre given. The mass of the car, which weβll
call π, is 660 kilograms. The car travels on a circular track
with radius π equals 22 meters. And the track is angled at π
equals 28 degrees above the horizontal.

Clearing some space on screen,
letβs recall that when an object moves in a circle, it is acted on by a force toward
the center of that circle known as centripetal force. Represented by πΉ sub π,
centripetal force equals an objectβs mass π times its linear speed π£ squared
divided by the radius of the circular path the object follows π. In our example, the force on the
car towards the center of the track is centripetal. So weβll label it with a subscript
π.

Letβs consider viewing the car as
though it is moving directly toward us. Two forces act on the car: the
weight of the car downward and the normal reaction force that weβll call π
. The slope is smooth, so friction
forces up or down the slope of the track can be ignored. Notice that there is no force weβve
called πΉ sub π acting on the car. Nonetheless, there is a
center-seeking force on the car.

From our perspective, the center of
the track is somewhere off to the right. And a component of the normal
reaction force points this way. Since it always points toward the
trackβs center, it is a centripetal force. Centripetal forces are always
caused by real forces acting on an object. In this case, we see that force is
the normal reaction force. This is the physical reason for πΉ
sub π.

Rather than solving for πΉ sub π
using the equation above, weβre going to do it using our free body diagram. First, letβs express πΉ sub π in
terms of the normal reaction force π
. To do this, we can set up
coordinate axes like this so that πΉ sub π is entirely in the positive
π₯-direction. If we divide π
into its π₯- and
π¦-components, we see that a right triangle is formed by these vectors. This right triangle is actually
similar to the right triangle formed by the track, the ground, and the vertical line
connecting them. Therefore, because the track is
inclined at an angle π from the horizontal, the topmost angle in the triangle of
force vectors is also π.

We can say then that πΉ sub π is
equal to the force vector π
times the sin of π. We know π; itβs 28 degrees. But we donβt yet know π
. To solve for it, weβll compare the
normal reaction force with the weight force π. Weβll be helped in doing this by
resetting our coordinate axes so that now positive π₯ points down the track and
positive π¦ points in the direction of π
, perpendicularly away from the track.

If we separate π into its π₯- and
π¦-components using these new axes, we find that, first, we once again have a right
triangle similar to the one formed by the track, where here π is the angle between
π and π sub π¦. And second, the component π sub π¦
is equal in magnitude to the normal reaction force π
.

Using our right triangle of weight
force vectors, we see that π sub π¦ equals π times the cos of π. Combining these equations, we see
that π
equals π cos π. So we can substitute the right side
of this equation in for π
in our earlier expression. In general, the weight force π of
an object equals its mass π times the acceleration due to gravity π. We can make that substitution and
now have an expression for πΉ sub π in terms of known quantities. π is 660 kilograms, π is 9.8
meters per second squared, and π is 28 degrees.

Calculating this expression gives a
result, to the nearest newton, of 2681 newtons. This is the magnitude of the
centripetal force on the car, the force which acts in the direction of π
.