Free Body Diagram

Free body Diagram | Motion of a block on inclined plane | Part: 2

A block placed on a frictionless inclined plane with angle of inclination $\theta$. Here the inclined plane is at rest and the body slides down the pnclined plane.

From the free body diagram we get,
$R = mg\cos \theta$ and
the net effective force on the block is $F = mg\sin \theta$
Therefore, $F = mg\sin \theta$
or, $ma = mg\sin \theta$
or, $a = g\sin \theta$

A block placed on a inclined plane with angle of inclination $\theta$ and whose coefficient of friction between the surface of the block and the inclined plane is $\mu$. Here the inclined plane is at rest and the body slides down the inclined plane.

From free body diagram we get,
$R = mg\cos \theta$ and
friction force $f = \mu R = \mu mg\cos \theta$ acts upward opposes the motion.
And net effective force on the block is
$F = mg\sin \theta - \mu R$
or, $F = mg\sin \theta - \mu mg\cos \theta$
Therefore, $F = ma = mg\sin \theta - \mu mg\cos \theta$
or, $a = g\sin \theta - \mu g\cos \theta$
or, $a = g\left( {\sin \theta - \mu cos\theta } \right)$

Retardation of a block up through inclined plane.

From free body diagram we get, $R = mg\cos \theta$

Here friction force $f = \mu R = \mu mg\cos \theta$acts downward opposes the upward motion of the block.
And net effective force on the block is $F = mg\sin \theta + \mu mg\cos \theta$
From Newton's second law of motion we get, $F = mg\sin \theta + \mu mg\cos \theta$
Therefore, $ma = mg\sin \theta + \mu mg\cos \theta$
or, $a = g\sin \theta + \mu g\cos \theta$
or, $a = g\left( {\sin \theta + \mu \cos \theta } \right)$

A block placed on a frictionless inclined plane with angle of inclination $\theta$. Here the inclined plane is in an acceleration $b$ and the body slides down the inclined plane.

Since the body lies in ARF (Accelerated Reference Frame), an inertial force $mb$ acts on it in the opposite direction.
From free body diagram we get,
$R = mg\cos \theta + mb\sin \theta$

And net effective force on the block is
$F = mg\sin \theta - mb\cos \theta$

From newton's second law of motion we get, $F = ma$
or, $ma = mg\sin \theta - mb\cos \theta$
or, $a = g\sin \theta - b\cos \theta$

Special Notes:
The condition for the body to be rest relative to the inclined plane if $a = 0$
Thus, $g\sin \theta - b\cos \theta = 0$
or, $b = g\tan \theta$