A block of mass m is placed on a smooth inclined plane of inclination q with the horizontal. The force

exerted by the plane on the block has a magnitude.

(a) mg (b) mg/cos q (c) mg cos q (d) mg tan q

From FBD of the situations we get R= mg cos A

Dea Jeet,

Consider a block of mass "m" placed on an inclined plane, which makes an angle q with the horizontal plane. The weight "W" of the block is acting vertically downward. The weight of the block can be resolved into two rectangular components: wcosq and wsinq other forces acting on the block are: (i) Surface reaction (R) which is perpendicular to the plane (ii) Force of friction (f) acting opposite to the direction of motion of block. Let us take x-axis perpendicular to the inclined plane. If the block is at rest, then wsinq acting down the plane balances the opposing frictional force. Applying 1st condition of equilibrium.
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Fx = 0 f – wsinq = 0 -------(1)
Fy = 0 R– wcosq = 0 -------(2)
Since there is no motion in the direction perpendicular to the inclined plane, therefore wcosq is balanced by R i.e. R = wcosq
If block slides down with an acceleration equal to ''a'', then the resultant force is equal to ''ma'' and the force on block will be:
wsinq – f
According to 2nd law of motion
wsinq – f = ma
If the force of friction is negligible, then
wsinq = ma
but w = mg
This expression shows that if friction is negligible the acceleration of a body on an inclined plane is independent of mass but is directly proportional to sinq.
WHEN FRICTION IS NOT NEGLIGIBLE
(i) If block moves upward
f - wsinq = ma
(ii) If block moves downward
wsinq – f = ma

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                            mg cos q - because the plane is horizontal...