# Finding friction on a banked road [closed]

Problem: A turn of radius $$20m$$ is banked for the vehicle of mass $$200kg$$ going at a speed of $$10ms^{-1}$$. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed of $$15ms^{-1}$$.

My attempt:

For the optimum velocity of banking, friction will not act as horizontal component of normal reaction provides the necessary centripetal force. Therefore, $$\frac{mv^2}{R} = F_N\sin\theta, \ F_N\cos\theta = mg\tag{1}\label{1}$$ $$\therefore \frac{mv^2}{R} = mg\tan\theta\\ \therefore \tan\theta = \frac{v^2}{Rg} =\frac{100}{20\times10} =0.5\\ F_N \stackrel{\eqref{1}}= 1000\sqrt{5} \ N$$

Now, when we drive too fast, the car will tend to slip radially outwards. Hence, friction will act radially inwards. Therefore, to find the magnitude of frictional force, we have two ways:

Either, solve the horizontal components or solve the vertical components
Note: magnitude of Normal reaction will remain unchanged.

Method 1: $$F_N\sin\theta + F_f\cos\theta = \frac{mv^2}{R}\\ \therefore F_f = \frac{\frac{200 \times 225}{20} - 1000\sqrt{5} \times \frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = 625\sqrt{5} \ N$$

Method 2: $$F_f\sin\theta = F_N\cos\theta - mg\\ \therefore F_f = \frac{1000\sqrt{5} \times \frac{2}{\sqrt{5}} - 200 \times 10}{\frac{1}{\sqrt{5}}} = 0 \ N$$

As we can see, the answers from the two methods are different. So, possibly I'm missing something. Can I get some hint?

## closed as off-topic by John Rennie, Jon Custer, ACuriousMind♦Jul 2 at 17:26

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• Note: magnitude of Normal reaction will remain unchanged. This is your error. The normal force will be different from the value you calculate initially. Take your two equations at the end, which are both valid, and solve for $F_f$ by eliminating $F_N$ – Aaron Stevens Jul 1 at 13:08
• @AaronStevens: Okay, I understood what you've said. But, if I resolve the forces parallel to the normal reaction. Then, I get $F_N = mg\cos\theta$ in both the situation (i.e., when v = 10 and v = 15). Isn't it? – rv7 Jul 1 at 13:19
• No, $F_N\neq mg\cos\theta$ There is an acceleration component perpendicular to the incline here – Aaron Stevens Jul 1 at 13:25
• Oh! It seems I missed out the centrifugal force in the second scenario. Thank you very much! :) – rv7 Jul 1 at 13:28
• Yes, the acceleration is centripetal, hence you have centripetal force components – Aaron Stevens Jul 1 at 13:29