A body is sliding down a rough inclined plane | vedclass

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(C) Given: Coefficient of friction $\mu = 0.5$. The ratio of net force $F$ to normal reaction $R$ is $\frac{F}{R} = \frac{1}{2}$.The net force $F$ act

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) Given: Coefficient of friction $\mu = 0.5$. The ratio of net force $F$ to normal reaction $R$ is $\frac{F}{R} = \frac{1}{2}$.The net force $F$ acting down the plane is given by $F = mg \sin 	heta - f$,where $f$ is the frictional force.The normal reaction $R$ is $mg \cos 	heta$.The frictional force $f$ is given by $f = \mu R = \mu mg \cos 	heta$.Substituting these into the ratio:$\frac{mg \sin 	heta - \mu mg \cos 	heta}{mg \cos 	heta} = \frac{1}{2}$Dividing the numerator by the denominator:$	an 	heta - \mu = \frac{1}{2}$Given $\mu = 0.5$,we have:$	an 	heta - 0.5 = 0.5$$	an 	heta = 1$Therefore,$	heta = 45^{\circ}$.

$A$ body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is $0.5$. The ratio of the net force required for the body to slide down and the normal reaction on the body is $1:2$. Then the angle of the inclined plane is: (in $^{\circ}$)

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