At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
could you explain a bit how the FBD is drawn with Normal Reaction as mg/cos theta
you mean for case 2?
sure... there are only two forces on the block in that case... N and mg... the horizontal component of N must be F and the vertical component must be mg
Oh . yes i have not got the logic clear.
ah whoops, I see why you're asking..
I labelled the theta incorrectly...
there now it works...:)
N2 is mg/cos(theta)
But what about M at rest ? Is the Normal Reaction force mg/cos theta is sufficient to negate mg sin theta component. I am not so sure whether I have got the concept right. Could you please elaborate. I have got the values right.
After a second thought it struck me that the wedge is pitted against the Mass M and there is as such no component as mg sin theta...am i correct?
m is at rest with respect to the ramp when it's accelerating with the ramp...
yeah I mean, you could decompose mg into vectors along the ramp or into/out of the ramp, but it would be pretty pointless...