A motor vehicle of mass 1000 ,kg is moving on | vedclass

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(D) Given: Mass $m = 1000 \,kg$, Banking angle $	heta = 30^{\circ}$, Coefficient of friction $\mu = 0.2$, $g = 10 \,ms^{-2}$.Considering the forces a

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$13055$

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(D) Given: Mass $m = 1000 \,kg$, Banking angle $	heta = 30^{\circ}$, Coefficient of friction $\mu = 0.2$, $g = 10 \,ms^{-2}$.Considering the forces acting on the vehicle perpendicular to the inclined surface:The normal reaction force $N$ balances the component of gravity $mg \cos 	heta$ and the component of the frictional force $f \sin 	heta$ (assuming the vehicle is at the point of sliding up the incline).$N = mg \cos 	heta + f \sin 	heta$Since $f = \mu N$, we have:$N = mg \cos 	heta + \mu N \sin 	heta$$N(1 - \mu \sin 	heta) = mg \cos 	heta$$N = \frac{mg \cos 	heta}{1 - \mu \sin 	heta}$Substituting the values:$N = \frac{1000 	imes 10 	imes \cos 30^{\circ}}{1 - 0.2 	imes \sin 30^{\circ}}$$N = \frac{10000 	imes (\sqrt{3}/2)}{1 - 0.2 	imes 0.5} = \frac{5000 	imes 1.732}{1 - 0.1} = \frac{8660}{0.9} \approx 9622 \,N$.However, if we consider the force balance along the vertical axis as per the provided diagram:$N \cos 	heta = mg + f \sin 	heta$$N \cos 30^{\circ} = 10000 + 0.2 N \sin 30^{\circ}$$N(0.866 - 0.1) = 10000$$N(0.766) = 10000$$N \approx 13055 \,N$.

$A$ motor vehicle of mass $1000 \,kg$ is moving on a circular road having a banking angle of $30^{\circ}$ and a coefficient of friction of $0.2$. Then the normal reaction force on the motor vehicle is about (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,N$)

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