So previously we looked at how can we solve kinetics problems using Newton's second law, which states
that the force is equal to the mass, multiplied by the acceleration in the direction of the force.
That is why we've broken the force up into its components.
And when we've used rectangular components in two dimensions, that would be an X component and a Y
component or an eye and a J component.
And we could calculate the force in each one of those components.
If we have the mass of the particle that accelerates and we have the acceleration of the particle using
it in second law.
So we looked at an example like that.
Now we can also apply Newton's second law within in T coordinates that'll say normal and tangential
coordinates.
We looked at rectangular coordinates, but we can also do it in any coordinates and that becomes handy
when we are faced with problems of circular motion later vehicle going around a curved road, something
like a roller coaster going around a hoop.
Those kind of problems are easier to solve with empty coordinates.
So.
What I've drawn here is a curved path and there is a particle with a certain mass moving on the spot.
We could say it's a vehicle.
And as it moves the direction of the tangential component of the velocity changes and the normal component
points into the radius of curvature.
So if you look at this point, that tangential direction is tangent to the curvature and the normal
direction points to the center of curvature.
Now, what we want to do is we want to apply Newton's second law within this coordinate system.
So what we do is we say that there's a certain force in the tangential direction and a certain force
in the normal direction, and we apply Newton's second law to that so we can say the sum of the forces
is equal to the mass, multiplied by the acceleration with the forces, the acceleration, those are
vectors.
And we can break it up in the tangential and the normal direction.
So the force in a tangential direction is equal to mass, multiplied by the acceleration in the tangential
direction, or we can say that is multiplied by V Dot.
Now, Vidocq, as you'll know by this time, is the time derivative of velocity.
So that gives us acceleration.
And more specifically, that is how we calculate acceleration in the tangential direction.
In the next line you'll see when we calculate the force in the normal direction, we say it's mass multiplied
by the normal acceleration in the normal direction.
And as we know, that is V squared of zero.
So acceleration in the normal direction towards the center of curvature that is equal to V squared overwrote
as we've seen in a previous video.
So some of the forces, the net forces in the normal direction is in the squared overall row row is
our radius of curvature.
And this force is also called the centripetal force.
So that is the force when you swing a tennis ball on a string.
It's the tension in the string.
So the moment you cut, that's that string, the ball will go out and it'll move out radially with that
force or due to that force.
So we'll now look at an example that we have looked at earlier.
But I'm going to change the example a little bit to try to demonstrate Newton's second law in and coordinates.
So a vehicle travels along a curved road, the speed of the vehicle is uniformly increased from a value
of 20 meters per second to 30 meters per second during a period of two seconds.
So from that information, we can already calculate the tangential acceleration.
The moment the vehicle travels at a speed of 25 meters per second, the radius of curvature of the road
is 250 meters.
OK, now I'm going to change the question.
I'm going to say if the vehicle weighs a thousand kilograms, determine the efficiency of friction in
the normal direction.
So if this vehicle goes around a curve with a curvature of 250 metres or a radius of curvature 250 metres,
then it wants to slide out.
And the friction of the of the course of the tires of the wheels of the car on the road keeps that vehicle
from slipping out.
And those tires have a certain coefficient of friction.
And so the vehicle doesn't slip out.
And we want to know what is that coefficient of friction that keeps the vehicle on the road when it
goes around the bend.
So we use Newton's second law to solve this problem.
We know that the normal force is what is resisted by the friction.
The normal force is calculated by if N is equal to M multiplied by eight and we have the mass that's
a thousand kilograms multiplied by.
How do we calculate and how do we calculate the normal acceleration that.
Well that's V squared.
Overall, we've been given V, that's 25 meters per second.
So that would be squared over our radius of curvature and that gives us a force of 2500 Newtons.
So this is the force that wants to push the car to the outside, and that is the force that is resisted
by the friction of the vehicle's tires.
So now we know how do we calculate the frictional force?
Well, it is the efficient of friction multiplied by the normal force, the normal force that is the
mass of the vehicle multiplied by nine point eight one.
So if we manipulate this equation, we can say mu like afficionado friction is two thousand five hundred
over the normal force.
Why 2014 2500?
Well, that is the force that needs to be resisted by the friction.
So that is the frictional force of the tires and of the road on the tires, because that is the normal
force that we have just calculated.
And so we substitute in two thousand five hundred, divided by a thousand, multiplied by nine point
eight one, and we get a frictional coefficient of zero point two five.
It's a unit just to go fishing.
So this demonstrates the practicalities of calculating the normal acceleration because then you can
get the normal force and then you can calculate what is the force that needs to work in against the
normal force of the car.
Doesn't slip out when going around this bend.