Question

A cyclist with combined mass 80 kg goes
around a curved road with a uniform speed
20 m/s. He has to bend inward by an angle
0 = tan (0.50) with the vertical. The force of
friction acting at the point of contact of tyres
and road surface is
(g = 10 m/s?)​

Answer 1

400

..................ok

Answer 2

The force of  friction acting at the point of contact of tyres  and road surface is 400 N.

Explanation:

Mass of the cyclist, m = 80 kg

Speed of the cyclist, v = 20 m/s

He has to bend inward by an angle  $\tan^{-1}(0.5)$  with the vertical. Let F is the force of  friction acting at the point of contact of tyres  and road surface. In order to take a turn, the frictional force is balanced by the centripetal force as :

$\mu mg=\dfrac{mv^2}{r}\\\\\mu mg=\dfrac{mgv^2}{rg}$

Since, $\dfrac{v^2}{rg}=\tan\theta$

$mu mg=mg\tan\theta=F\\\\F=80\times 10\times 0.5\\\\F=400\ N$

So, the force of  friction acting at the point of contact of tyres  and road surface is 400 N.

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