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So previously we looked at how can we solve kinetics problems using Newton's second law, which states

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that the force is equal to the mass, multiplied by the acceleration in the direction of the force.

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That is why we've broken the force up into its components.

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And when we've used rectangular components in two dimensions, that would be an X component and a Y

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component or an eye and a J component.

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And we could calculate the force in each one of those components.

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If we have the mass of the particle that accelerates and we have the acceleration of the particle using

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it in second law.

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So we looked at an example like that.

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Now we can also apply Newton's second law within in T coordinates that'll say normal and tangential

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coordinates.

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We looked at rectangular coordinates, but we can also do it in any coordinates and that becomes handy

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when we are faced with problems of circular motion later vehicle going around a curved road, something

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like a roller coaster going around a hoop.

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Those kind of problems are easier to solve with empty coordinates.

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So.

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What I've drawn here is a curved path and there is a particle with a certain mass moving on the spot.

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We could say it's a vehicle.

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And as it moves the direction of the tangential component of the velocity changes and the normal component

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points into the radius of curvature.

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So if you look at this point, that tangential direction is tangent to the curvature and the normal

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direction points to the center of curvature.

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Now, what we want to do is we want to apply Newton's second law within this coordinate system.

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So what we do is we say that there's a certain force in the tangential direction and a certain force

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in the normal direction, and we apply Newton's second law to that so we can say the sum of the forces

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is equal to the mass, multiplied by the acceleration with the forces, the acceleration, those are

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vectors.

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And we can break it up in the tangential and the normal direction.

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So the force in a tangential direction is equal to mass, multiplied by the acceleration in the tangential

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direction, or we can say that is multiplied by V Dot.

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Now, Vidocq, as you'll know by this time, is the time derivative of velocity.

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So that gives us acceleration.

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And more specifically, that is how we calculate acceleration in the tangential direction.

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In the next line you'll see when we calculate the force in the normal direction, we say it's mass multiplied

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by the normal acceleration in the normal direction.

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And as we know, that is V squared of zero.

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So acceleration in the normal direction towards the center of curvature that is equal to V squared overwrote

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as we've seen in a previous video.

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So some of the forces, the net forces in the normal direction is in the squared overall row row is

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our radius of curvature.

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And this force is also called the centripetal force.

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So that is the force when you swing a tennis ball on a string.

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It's the tension in the string.

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So the moment you cut, that's that string, the ball will go out and it'll move out radially with that

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force or due to that force.

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So we'll now look at an example that we have looked at earlier.

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But I'm going to change the example a little bit to try to demonstrate Newton's second law in and coordinates.

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So a vehicle travels along a curved road, the speed of the vehicle is uniformly increased from a value

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of 20 meters per second to 30 meters per second during a period of two seconds.

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So from that information, we can already calculate the tangential acceleration.

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The moment the vehicle travels at a speed of 25 meters per second, the radius of curvature of the road

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is 250 meters.

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OK, now I'm going to change the question.

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I'm going to say if the vehicle weighs a thousand kilograms, determine the efficiency of friction in

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the normal direction.

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So if this vehicle goes around a curve with a curvature of 250 metres or a radius of curvature 250 metres,

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then it wants to slide out.

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And the friction of the of the course of the tires of the wheels of the car on the road keeps that vehicle

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from slipping out.

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And those tires have a certain coefficient of friction.

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And so the vehicle doesn't slip out.

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And we want to know what is that coefficient of friction that keeps the vehicle on the road when it

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goes around the bend.

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So we use Newton's second law to solve this problem.

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We know that the normal force is what is resisted by the friction.

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The normal force is calculated by if N is equal to M multiplied by eight and we have the mass that's

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a thousand kilograms multiplied by.

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How do we calculate and how do we calculate the normal acceleration that.

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Well that's V squared.

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Overall, we've been given V, that's 25 meters per second.

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So that would be squared over our radius of curvature and that gives us a force of 2500 Newtons.

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So this is the force that wants to push the car to the outside, and that is the force that is resisted

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by the friction of the vehicle's tires.

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So now we know how do we calculate the frictional force?

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Well, it is the efficient of friction multiplied by the normal force, the normal force that is the

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mass of the vehicle multiplied by nine point eight one.

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So if we manipulate this equation, we can say mu like afficionado friction is two thousand five hundred

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over the normal force.

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Why 2014 2500?

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Well, that is the force that needs to be resisted by the friction.

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So that is the frictional force of the tires and of the road on the tires, because that is the normal

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force that we have just calculated.

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And so we substitute in two thousand five hundred, divided by a thousand, multiplied by nine point

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eight one, and we get a frictional coefficient of zero point two five.

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It's a unit just to go fishing.

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So this demonstrates the practicalities of calculating the normal acceleration because then you can

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get the normal force and then you can calculate what is the force that needs to work in against the

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normal force of the car.

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Doesn't slip out when going around this bend.