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, \(b = g\tan \theta \)
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