Aarti Gupta
Last Activity: 10 Years ago
Let us suppose a body which is sliding down the plane shown below.Now all force components should be taken along and perpendicular to the surface of the block.
Now,
1).Balancing components along the perpendicular of the inclined plane-
W cos theta = R
2).Balancing components along the surface of the inclined plane-
W sin theta – f = ma [ where, a is the acceleration of the block along the inclined surface]
mg sin theta –
R = ma
mg sin theta –
mg cos theta = ma
Therefore, a = g(sin theta –
cos theta)
The above equation gives the acceleration of a body sliding down a rough inclined plane.