Take the Centre of Mass, I Dare You!

In this problem, taking centre of mass of the rod and finding its attraction with the point is wrong. This method will yield

$F=\frac{4GMm}{9L^2}$ , thus $a+b=13$ .

This method is incorrect as the force distribution for each point on the rod is different, and non-contradictory.

Thus, let us use calculus to solve this problem.

Consider a small part of the rod of length $dx$ , with mass $dM$ , and let this small part's distance from the point-mass be $x$

Thus, $dF=\frac{G(dM)m}{x^2}$ . However, we cannot integrate this, as the variables do not match. Thus, we must find $dM$ in terms of $dx$ , or $x$ in terms of $M$ .

Let us do the former option, as it would eliminate the variable.

Now, consider linear density $\lambda$ . Since the rod is homogenous, and the small part of the rod has the same linear density, we get

$\lambda=\frac{dM}{dx}=\frac{M}{L}$ , which means that $dM=\frac{M}{L}dx$

Plugging into the force equation, we get $dF=(\frac{GMm}{L})\frac{1}{x^2}dx$

Now, the limits of this integration must be from $L$ to $2L$ , as the minimum distance of any point on the rod from the point mass is $L$ , and the maximum distance is $2L$

Thus, integrating, we get $F=\int_L^{2L}(\frac{GMm}{L})(\frac{1}{x^2})dx=\frac{GMm}{L}[\frac{-1}{x^2}]_L^{2L}=\frac{-GMm}{L}[\frac{1}{2L}-\frac{1}{L}]=\boxed{\frac{GMm}{2L^2}}$

Thus, $a=1,b=2$ and $a+b=3$

No thanks , i would use integration instead