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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
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The correct answer is:
$45^{\circ}$
$$
\mu=0.5, \frac{F}{R}=\frac{1}{2}
$$
Net force acting downward $=m g \sin \theta-f$ where, $f=$ frictional force $=\mu R(R=$ Normal reaction $)$ $=\mu m g \cos \theta$
$$
\frac{F}{R}=\frac{1}{2}
$$
$\begin{aligned} & \frac{m g \sin \theta-\mu m g \cos \theta}{m g \cos \theta}=\frac{1}{2} \\ & \tan \theta-\mu=\frac{1}{2} \Rightarrow \tan \theta=\frac{1}{2}+\mu \\ &=\frac{1}{2}+0.5=1 \\ & \tan \theta=\tan 45\\ & \therefore \quad \theta=45^{\circ}\end{aligned}$
\mu=0.5, \frac{F}{R}=\frac{1}{2}
$$
Net force acting downward $=m g \sin \theta-f$ where, $f=$ frictional force $=\mu R(R=$ Normal reaction $)$ $=\mu m g \cos \theta$
$$
\frac{F}{R}=\frac{1}{2}
$$
$\begin{aligned} & \frac{m g \sin \theta-\mu m g \cos \theta}{m g \cos \theta}=\frac{1}{2} \\ & \tan \theta-\mu=\frac{1}{2} \Rightarrow \tan \theta=\frac{1}{2}+\mu \\ &=\frac{1}{2}+0.5=1 \\ & \tan \theta=\tan 45\\ & \therefore \quad \theta=45^{\circ}\end{aligned}$
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