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...
						

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							MgBecause all components depend on