A block of mass 10 ,kg is kept on a fixed rou | vedclass

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(A) Given:Mass of the block,$m = 10 \,kg$Angle of inclination,$	heta = 30^{\circ}$Coefficient of static friction,$\mu_s = 0.8$Acceleration due to gra

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$50$

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(A) Given:Mass of the block,$m = 10 \,kg$Angle of inclination,$	heta = 30^{\circ}$Coefficient of static friction,$\mu_s = 0.8$Acceleration due to gravity,$g = 10 \,m/s^2$ (taking $g=10$ for simplicity)Step $1$: Calculate the force component tending to slide the block down the plane.$F_{down} = mg \sin 	heta = 10 	imes 10 	imes \sin 30^{\circ} = 100 	imes 0.5 = 50 \,N$Step $2$: Calculate the maximum static frictional force available.$f_{max} = \mu_s N = \mu_s mg \cos 	heta = 0.8 	imes 10 	imes 10 	imes \cos 30^{\circ}$$f_{max} = 80 	imes \frac{\sqrt{3}}{2} = 40 \sqrt{3} \approx 40 	imes 1.732 = 69.28 \,N$Step $3$: Compare the forces.Since the force tending to slide the block $(50 \,N)$ is less than the maximum static frictional force $(69.28 \,N)$,the block remains at rest.Therefore,the static frictional force acting on the block is equal to the force tending to slide it down.$f = F_{down} = 50 \,N$.

$A$ block of mass $10 \,kg$ is kept on a fixed rough $(\mu=0.8)$ inclined plane of angle of inclination $30^{\circ}$. The frictional force acting on the block is ........... $N$.

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