A body of mass 2 ,kg is on an inclined plane | vedclass

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(C) Given:Mass of the body,$m = 2 \,kg$Angle of inclination,$	heta = 30^{\circ}$Coefficient of friction,$\mu = \frac{1}{\sqrt{3}}$Acceleration due to

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$20$

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(C) Given:Mass of the body,$m = 2 \,kg$Angle of inclination,$	heta = 30^{\circ}$Coefficient of friction,$\mu = \frac{1}{\sqrt{3}}$Acceleration due to gravity,$g = 10 \,ms^{-2}$To move the body up the inclined plane,the applied force $F$ must overcome both the component of gravitational force acting down the plane $(mg \sin 	heta)$ and the limiting frictional force $(f_r)$ acting down the plane.The normal force $N$ acting on the body is $N = mg \cos 	heta$.The limiting frictional force is $f_r = \mu N = \mu mg \cos 	heta$.The minimum force $F$ required is:$F = mg \sin 	heta + f_r$$F = mg \sin 	heta + \mu mg \cos 	heta$$F = mg (\sin 	heta + \mu \cos 	heta)$Substituting the values:$F = 2 	imes 10 	imes (\sin 30^{\circ} + \frac{1}{\sqrt{3}} \cos 30^{\circ})$$F = 20 	imes (\frac{1}{2} + \frac{1}{\sqrt{3}} 	imes \frac{\sqrt{3}}{2})$$F = 20 	imes (\frac{1}{2} + \frac{1}{2})$$F = 20 	imes 1 = 20 \,N$

$A$ body of mass $2 \,kg$ is on an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is $\frac{1}{\sqrt{3}}$. The minimum force required to move the body up the inclined plane is (Acceleration due to gravity $g = 10 \,ms^{-2}$) (in $\,N$)

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