 # Determining the Force Normal on a Toy Car moving up a Curved Hill

and Billy, please translate. ♫ Flipping Physics ♫ – A 0.453-kilogram toy car
moving at 1.15 meters per second is going up a semicircular
hill with a radius of 0.89 meters. – Mass equals 0.453 kilograms. Tangential velocity equals
1.15 meters per second, and radius equals 0.89 meters. – When the hill makes
an angle of 32 degrees with the horizontal, what is the magnitude of the force normal on the car? – Theta equals 32 degrees and force normal equals question mark. – Mr. P, could you please
clarify that angle? I’m not really sure what it
means it says, “when the hill makes an angle of 32
degrees with the horizontal.” – Sure, Bobby. Let me clarify that angle. As the car goes up the
hill, the angle of the hill where the car is located increases. If we draw a line which
intersects the center of the two wheels, you can see how the
angle of the hill increases. – [Bobby] Sure. – [Mr. P] And if we pause
the car at the point we are talking about, you can see
how the hill makes an angle of 32 degrees with the horizontal. – [Billy] You know, that looks
just like the diagram we used when broke the force of
gravity into components on an incline and determined the equations for the force of gravity parallel and perpendicular to the incline. – [Mr. P] Yes, Billy, it does,
and we are going to use that fact in a few moments, thank you. – [Billy] You are welcome!
– [Bo] Nice. – Okay, Bobby, please start the problem. – Yeah, um. Let’s draw a free-body diagram. The force of gravity is straight
down and the force normal is up and perpendicular
to the hill. That’s it. So let’s break forces into components. The car is moving in a circle,
so we need to break forces into components in the in
direction and a direction which is perpendicular to the in
direction, which is, in this case, tangent to the circle or in
the tangential direction. The force normal is in the in
direction, so we don’t need to break that into components. That means we need to break
the force of gravity into components in the in direction
and the tangential direction. However, because theta is
just like an incline angle, we have our equations for
the force of gravity parallel and perpendicular, which use
the angle of the incline. We don’t have a numbers
dependency, so let’s not plug in numbers just yet, let’s just
redraw our free-body diagram. The force normal is still
there, but now we have the force of gravity perpendicular,
which is in the out direction, and the force of gravity
parallel, which is in the tangential direction and down. – Billy, please finish the problem. – Now we sum the forces
in the in direction. The net force in the in
direction equals force normal minus force of gravity
perpendicular, and the net force in the in direction equals
mass times centripetal acceleration, add force
of gravity perpendicular to both sides to solve for force normal, and now we can substitute in equations. Force of gravity perpendicular
equals mass times acceleration due to
gravity times cosine theta, and centripetal acceleration
equals tangential velocity squared, divided by radius. Next, let’s plug in numbers. So that’s 0.453 times 9.81
times the cosine of 32 degrees, plus 0.453 times 1.15 squared,
divided by 0.89, which equals 4.4418, or 4.4 newtons with
two significant digits. – Very nice. A couple of things to point out. First, the tangential velocity of the car, 1.15 meters per second, is
an instantaneous velocity at a specific point in time,
because the car slows down as it goes up the hill. Second, the magnitude of
the force normal decreases as the car goes up the hill. Both because the instantaneous
tangential velocity of the car is decreasing,
but also because the force of gravity perpendicular
decreases as theta, the angle of the hill, increases. Thank you very much for
learning with me today. I enjoyed learning with you.