Homework Answers
a) consider an element of length dz and mass dm at a distance of z from mass m on the rod. Force due to this element will be Gmdm/z2. dm = Mdz/L.
Now for total mass integrate it from x = z to x = z+L.
F = GMm/z(z+L). Towards - z direction.
b). Consider an element of length dz at a distance of z from the upper end of lower rod having mass dm.
F on that dm will be GMdm/z(z+L).
DM will be again Mdz/L.
Now integrate it from z= L to z= 2L.
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