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Suppose you wish to find the max speed at which a car can travel on a banked curve of radius R and angle theta to the horizontal and coeff. of static friction mu.

I don't understand the relevant free body diagram.

We should have $F_g$ downwards, $F_f$ pointing towards the center of the track parallel to the track, and $F_n$ perpindicular to the track. The net force in the direction parallel to the track points towards the center of the track, so the car should accelerate down the track right (towards the center), right? What is keeping the track up - its the centrifugal force, right? But why don't we draw this on the FBD; what is providing the centrifugal force?

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As you seem (correctly) to understand, your free body diagram for the car should be as in the left of the drawing below. The force summation does not close to a polygon.

Banked Curve FBD

You know neither $F_N$ nor $F_F$, the components of the force on the car from the road. But you know their directions ($\theta$ is the banking angle) and you know that all the forces must sum, as in the right of my drawing, to a net force (shown in red) pointing towards the centre of curvature, and whose magnitude is $m\,v^2/R$, where $v$ is the car's speed and $R$ the radius of curvature. So there is a net unbalanced force on the car. This is the centripetal force that accelerates the car to keep it moving along the curved track. You can work out $F_N$ and $F_F$ by trigonometry: from my force diagram it is readily seen that:

$$F_N\,\cos\theta - F_F\sin\theta = \frac{m\,v^2}{R}$$ $$F_N\,\sin\theta + F_F\,\cos\theta = m\,g$$

which is an easy set of equations to invert, given that the inverse of the rotation matrix is easily found

$$\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)^{-1}=\left(\begin{array}{cc}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{array}\right)$$

Note that $F_F$ can be positive or negative: at a critical speed given by $v^2\,\tan\theta/R = g$ it will be zero. Your last step is to check that $|F_F|\leq\mu\,F_N$; if this condition doesn't hold, then your car will slip (either upwards if its going too fast, or downwards if it's going too slow) and uniform circular motion around the track at the speed in question is thus not possible.

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    $\begingroup$ +1 - Good drawing. In particular it is helpful how you show the general case where the friction needs to be "greater or equal to x". The rotation matrix may be more confusing than it needs to be - just a decomposition of forces makes more sense (this isn't really a rotation, is it). $\endgroup$ – Floris Nov 11 '14 at 6:56
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When you are moving along a banked curve, three forces are at play in your frame of reference:

  1. Gravity: pulling you down; and because of the bank, pulling you inward
  2. Friction: preventing you from moving either in or out. Magnitude and direction depend on the coefficient of friction, the normal force, and the direction in which you are trying to move
  3. Centrifugal force: the force that appears in a rotating frame of reference because you try to stay in a circular path.

Now if we are looking at the "max speed", the friction should be preventing the car from moving outwards, so the friction is pointing inwards. This means the following forces come into play:

enter image description here

I hope it is clear what each of these forces (or their components) are. I put the friction and component of gravity head-to tail to show they need to cancel the component of the centrifugal force along the banked slope. It seems to me all these forces belong in a complete diagram; if some of them are left out, it may be in a misguided attempt to explain that some "forces" (e.g. centrifugal "force") are not real forces. But in a rotating frame of reference (the frame in which the car is stationary) they are every bit as real as other forces.

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Well, for starters, centrifugal force is just a pseudo force. Get it? Let me make it a bit simpler.. See, Centripetal force, which is mv^2/r is acting towards the center and centrifugal force away from the center. I thinking you're thinking why did we use the centripetal force in case of banking of road.. right? Well, we did that because, enter image description here

This is nothing but, just a half circle, andd the car is moving on a circle... so, we have centrifugal force.. ..

Part-2, for the cos components and all, well, the picture goes like this...enter image description here

Ahm, so, you see, it's a case of an incline plane, with a mass on it, moving in a circular direction. Hope you get it.

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    $\begingroup$ I don't think you meant to label centrifugal force on your free body diagram. At least, I hope you didn't, since as you correctly mentioned, it's a fictitious force! $\endgroup$ – Sean Nov 11 '14 at 13:13
  • $\begingroup$ Opps. I did a mistake. $\endgroup$ – Mrinal Gautam Nov 11 '14 at 15:02
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Centrifugal force is a 'fictional' force which only appears when the reference frame itself is rotating.

You would have centrifugal force if you described the situation from the frame of the car as it moves round the curve. In this case you are instead describing the situation frame the reference frame of the stationary track, so there is no centrifugal force.

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