You need to learn all this in India too .well if preparing for competitive examinations just before going to university .I SUGGEST YOU TO CHECK THE LEVEL OF PHYSICS IN JEE ADVANCED IN INDIA .

Is the lowest number 3123237195 divided by 3. To 375

Now I would like to see this an event horizon.

As my annotation a little earlier points out, this is a simple explanation and things in practice are inevitably more complicated. But for A Level, I suspect this is all you'll need.

DrPhysicsA Ok cool. Thanks :)

on a flat, horizontal surface, the direction (vector) of gravity on a mass is downwards, or think of it as inwards, towards the center of the earth. When there's no incline, there's no horizontal component, or..as you could say, the horizontal component is zero. One natural way of viewing such a system would be on the unit circle; at what angle is the sin negative and the cosine zero? The sin(3π/2) = -1 and the cos(3π/2) = 0, it also just so happens that the angle sum trig identities tell us that cos(3π/2+θ) = sinθ and sin(3π/2+θ) = -cosθ from this we can find that ||mg||*cos(3π/2+θ)*i + ||mg||*sin(3π/2+θ)*j = ||mg||*sinθ*i - ||mg||*cosθ*j notice what happens, on both sides of the equality sign, when you set the angle θ = zero; ||mg||*cos(3π/2+0)*i + ||mg||*sin(3π/2+0)*j = ||mg||*sin(0)*i - ||mg||*cos(0)*j ||mg||*cos(3π/2)*i + ||mg||*sin(3π/2)*j = ||mg||*(sin0)*i - ||mg||*(cos0)*j ||mg||*0*i + ||mg||*(-1)*j = ||mg||*0*i - ||mg||*(1)*j :. ||mg||*(-1)*j = -||mg||*j hope that gives you enough to mess around with it a bit! :)

No, because you first must derive why m is equal to all of that.

If the velocity is zero, then it will simply depend on whether the frictional forces are sufficient to stop the car from sliding down the bank.

DrPhysicsA Ok great!

He used the word curvature, rather than radius. Curvature is by definition equal to the reciprocal of the radius of curvature. A tight curve with a small radius has a lot of curvature. A large circle has a large radius and very little curvature.

This is all geometry. Look at the right angles triangles. If one angle is theta the other is 90- theta (and the third is by definition a right angle).

DrPhysicsA Yes , I know that the sum of all the the angles of a triangle is 180 . The thing is that which angle should I take theta and which one 90-theta?

The forward motion would've been caused by the car's existing speed, that it developed in a part of the road where there wasn't sheet ice. The vehicle would move tangent to the circle, if all of a sudden it encountered sheet ice while attempting to travel around an unbanked curve. The banked curve doesn't only supply the centripetal force, it also supplies the vertical component of the normal force that keeps it from vertically accelerating.

+lolzomgz1337 You use the formulae in this video which balances the force which would cause the car to slide down the slope and the force which would cause it to go over the top.

DrPhysicsA Yeah, thanks. What was messing me up is that we don't consider a component of the weight down the slope for some reason...

im confused. the way i would do it is tilting the plane theta(@) degrees then i would say mg.cos@ = N and mg.sin@ = mv2/r the reason why i would do that is because i thought the centripetal force had to come on an angle @ not simply to the right. if you are oscillating at an angle wouldnt make sense that the force would also be at that angle?

Newtons Law of equal and opposite forces

The normal force will be AS LARGE AS NECESSARY to prevent the car from sinking through the asphalt. It is a constraint force. Its purpose is to repel two objects apart once they are in contact, and stop them from passing through each other.

Just resolving the vertical and horizontal components of N.

How can we Resolve N

That's right. The forces won't balance if the vehicle slides. I am talking about stability where the vehicle drives round the ramp without either shooting off the top or sliding down.

Oh now I see, thanks a lot these videos are really helpful :)

Coefficient of friction.

Use the formulae at 11:25 and solve for theta.

v^2=g*r*tan(θ).. you have the velocity and the radius and g=9.81, plug it in and find the angle. If I'm not wrong, and if what I understood from the video is correct, that question should be a frictionless banked turn.

Kenny Chong the frictional force provides the centripetal force...

Max Hollander At what time in video does this arise?

10:10

Max Hollander Basic geometry. Try drawing a line parallel to the ground through the point at which the Normal meets the incline.

You fire an engine which pushes out exhaust gas in one direction and you go in the opposite direction (conservation of momentum).

Forces shouldn't add up to zero, if an object is accelerating. They should add up to m*a. In the vertical direction, the forces should add up to zero, because all acceleration is horizontal in this example. In the horizontal direction, forces should add up to mass*acceleration, which in this case is centripetal acceleration, equal to v^2/r. Vertical force addition gives us: N*cos(theta) - F*sin(theta) - m*g = 0 N*cos(theta) is the vertical component of the normal force. Cosine applies here, because when theta = 0, the normal force is vertical, and the trig term would equal 1. F*sin(theta) is the traction force holdiing the car on the road, and the vertical component of it. Sine applies here, because there would be zero vertical component of this force when theta = 0, and the trig term should also be zero. Horizontal force addition gives us: N*sin(theta) + F*cos(theta) = m*v^2/r The opposite trig terms apply in this case. When theta = 0, normal force doesn't act horizontally at all, and traction is the only force enabling the car to steer. Two equations, two unknowns (F & N). Now it's just algebra to isolate one of the unknowns, and combine it with the other equation.

There is no x component of the force of gravity. Gravity is wholly in the y (down) direction.

Thank you, I figured that out after doing a few of these. Thanks for posting these, they are very lucid!

MrAdy0207 No. I've still got loads. Just wanted to see whether a white board presentation would be any better.

Sometimes a white board. Sometimes A3 paper and pen.

DrPhysicsA Sir, your videos have been extremely helpful this semester for me! Thank you so much!

This is why: kzfaq.info/get/bejne/irSIYL1nvbiXl3U.html

Not Legato I think I can maybe help? My understanding of this topic is quite broken but I believe that the normal force is actually greater than the force of its weight, this is because the vertical component of the normal force MUST be equal to the weight so that the car stays at the same vertical height, but then the horizontal component of the normal force is the reason that the car is pulled towards the centre of the circle (the centripetal force)

The surface was horizontal. So cos(0)=1 so n=(mg)(1)=mg

N isn't always equal to the same function of m, g, and theta. It depends on the specifics of the problem. N is a constraint force that will be as large as necessary to prevent an object from penetrating the surface.

Indeed it's N=mgcostheta vertically woth respect to inclined plane. But taking forces vertically in y-axis irrespecitive of the inclined plane, we see that Ncostheta = mg.

*with *irrespective Never mind.

but if that were true, the y component of the mg with respect to the slope would have to equal mg/cos(theta) to balance out the normal force(they must be balanced or else the object on the slope would rise off the slope). This cant be true because a component of the gravitational force can't be greater than the gravitational force itself. mg

We must remember that this is a banked turn not just an object on an inclined plane. So the object is moving and the normal force is therefore a combination of the gravitational force and the centripetal force. Consider for example a motorcyclist who rides round the so called wall of death at a funfair. The walls are almost vertical and the motorcyclist rides round at sufficient speed that he doesn't fall off. The gravitational component of the normal in this case is almost zero. So what normal force is acting on the wall. It is the component of the centripetal force.

DrPhysicsA But how is this possible in a frictionless banked curve scenario where the centripetal force is due to the normal force, which is also due to gravity. If gravity is the force responsible for everything, it doesn't make sense for the normal force component to be greater than gravities component because the centripetal force is caused by Fn due to gravity. I could understand where the extra leverage would be coming from if there was an external force such as friction or tension acting upon the object towards the center.

Can you tell me where on the video this arises so I can check it?

DrPhysicsA it is not on this video , symply when we are deriving Ff (uN) we are saying that Normal force is the component of mg perpendicular to the inclain and i was trying to substitute the N by that and i got confused

This may be a late reply but I think you're resolving perpendicular and parallel to the plane whereas as he is resolving vertically ad horizontally

+Ali Moazzam I see your point. I think you have missed that mg cos theta does not equal N. We say that N cos theta = mg in the vertical direction because there is no vertical movement. therefore the forces are equal. But if you resolve mg along the N direction the forces are not equal because a part of the N force is providing the centripetal acceleration.

+DrPhysicsA But if N cos theta = mg then mg /cos theta = N which means N is greater than mg. And as theta increases N remains constant while mg shrivels to practically zero? On the other hand when we measure static friction using a slope we always use Fn (or N) = mg cos theta meaning mg is always greater than Fn (or N) which makes more sense than N greater than mg. So I have the same problem as Ali Moazzam.

+Ali Moazzam I agree. If N cos theta = mg then mg /cos theta = N which means N is greater than mg. And as theta increases N remains constant while mg shrivels to practically zero? On the other hand when we measure static friction using a slope we always use Fn (or N) = mg cos theta meaning mg is always greater than Fn (or N) which makes more sense than N greater than mg. So I have the same problem as you are having. There has to be an explanation but it is eluding me at this time. I don't like N being greater than mg. It should always be the reverse.

+Ali Moazzam Like you I was initially puzzled myself but I think I know the answer. When measuring Us with object (mass) on slope it should be noted that the mass is at rest so the Fn is derived from mg = mg cos theta i.e Fn is LESS than mg. However when finding V on Banked turns the object (mass) is not at rest but has a velocity = V. Now the Fn is not derived from just mg. Here Fn = V^2/r sin theta and is greater than mg. So the two Fns are quite DIFFERENT. One when mass on slope is at rest and the other where it has a velocity = V. It has nothing to do with different coordinate systems or axii as is often claimed.

+qualquan Thank you for the help

N = mg costheta is only true if there is no acceleration. The normal force is determined by the method described in the video when there is acceleration.

In this case it isn't because the object is on an inclined plane. There is a normal force which is perpendicular to the plane but it does not equal the weight of the block.

DrPhysicsA Right. Thanks. I need to review normal forces

DrPhysicsA Am I right in saying that it's because, in the initial situation (where you state N = mg), there's no incline, so θ = 0. But cos 0 = 1. Therefore, 'N.cos θ = mg' is actually always applicable, as cos 0 = 1, so N.cos θ = N, when θ = 0. Is that correct?

Alex Khouri If θ = 0, then the ground is flat and so the normal force will be perpendicular to the surface (as usual) but in this case it will be wholly in the vertical direction. So N =mg as you say.

Basically, we are saying that the horizontal component of the normal force is providing the centripetal force. since the mass term m appears on both sides of the equation we can cancel it.

Please delete this comment,People will believe what you are saying. Jk you're not clear on the concepts.Both ways are correct. I know this is 7 months old but leaving this here for other people.