Any plane surface which makes an angle q with the horizontal surface is called inclined plane such that0<q <90. . Inclined plane is an example of simple machine which is used to lift heavy bodies without applying very huge force.
MOTION OF A BODY ON AN INCLINED PLANE
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.
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
MECHANICAL ADVANTAGE OF INCLINED PLANE
MECHANICAL ADVANTAGE OF INCLINED PLANE IS GIVEN BY M.A = 1/sinq
(i) When q = 0 a = g sin0 a = g x 0 a = o
(ii) When q = 30 a = g sin30 a = g x 0.5 a = g/2 a = 4.9 m/s2
(iii) When q = 60 a = g sin60 a = g x 0.866 a = 4.9 m/s2
(iv) When q = 90 a = g sin90 a = g x 1 a = g a = 9.8 m/s2
Above analysis shows that the accelerationof a body on an inclined plane depends upon the angle of inclination.