Question 118: Work, Energy and Power & Circular Motion

Question

The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30$$^\circ$$ banked road at maximum possible speed without skidding is ____________ $$	imes$$ 10
3
kg m/s
2
. [Given cos30$$^\circ$$ = 0.87, $$\mu$$
s
= 0.2]

Accepted Answer

The given situation can be represented as
Equating forces perpendicular to the inclined plane,
$$N = mg\cos 30^\circ  + {{m{v^2}} \over R}\sin 30^\circ $$
$$ \Rightarrow N - mg\cos 30^\circ  = {{m{v^2}} \over R}\sin 30^\circ $$ .... (i)
Equating forces along the inclined plane,
$$mg\sin 30^\circ  + {\mu _s}N = {{m{v^2}} \over R}\cos 30^\circ $$ .... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$${{N - mg\cos 30^\circ } \over {mg\sin 30^\circ  + {\mu _s}N}} = 	an 30^\circ $$ [$$\because$$ $$\cos 30^\circ  = {{\sqrt 3 } \over 2}$$ and $$\sin 30^\circ  = {1 \over 2}$$]
$$\Rightarrow$$ $${{N - mg(\sqrt3/2) } \over {mg(1/2) + (0.2)N}} = {1 \over {\sqrt 3 }}$$
$$N\sqrt 3  - mg{{\sqrt 3 } \over 2}.\sqrt 3  = {{mg} \over 2} + 0.2N$$
$$ \Rightarrow (\sqrt 3  - 0.2)N = mg{{(1 + 3)} \over 2} = 2mg$$
$$ \Rightarrow N = {{2mg} \over {\sqrt 3  - 0.2}} = {{2 	imes 800 	imes 10} \over {1.532}}$$
$$ = 10.44 	imes {10^3}V$$
Therefore, $$N = 10.2 	imes {10^3}$$ kg-m/s
2

Answer

12.4

Answer

7.2

Answer

6.96

Answer

10.2

Work, Energy and Power & Circular Motion

The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30$^\circ banked road at maximum possible speed without skidding is ____________ \times 10 3 kg m/s 2 . [Given cos30^\circ = 0.87, \mu$ s = 0.2]

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