Calculate the force exerted by point mass m on rod of uniformly distributed mass `M` and length l (Placed as shown in figure)
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Explanation

`because` Direction of force is changing at every element. We have to make components of force and then integrate.
Net vertical force `=0`
`dF` = force element `= (G.dM.m)/((x^(2)+a^(2))`
`dF_(h) = dF cos theta` = Force on element in horizontal direction `= (G.dM.m)/((x^(2)+a^(2)) cos theta`
`:.F_(h)=int(G.M.mcosthetadx)/(l(x^2+a^2))=(G.M.m)/l underset(-l//2)overset(l//2)int(costheta.dx)/((x^(2)+a^(2)))=(GMm)/(la^(2))underset(-l//2)overset(l//2)int(costheta.dx)/(sec^(2)theta)`
where `x =a` than `theta` then `dx = a sec^(2) theta d theta`
`=(GMm)/(la)[sintheta]_(l//2)^(l//2)tantheta=(x)/(a),thesintheta=(x)/(sqrt(x^(2)+a^(2)))`
=`(GMm)/(la)[(x)/(sqrt(x^(2)+a^(2)))]_(-l//2)^(l//2) =sqrt(GMml)/(lasqrt((l^(2))/(4)+a^(2)))=(GMm)/(asqrt(((l^(2))/(4)+a^(2))))`
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