I'm trying to solve this problem from the book "Fundamentals of Applied Dynamics" by R.A.Tenenbaum.
I start it by trying to find the resultant gravitational force by the means of integration.
$$d\overrightarrow{F}=\frac{GmdM}{d^2}\overrightarrow{n}=\frac{GmMdy}{2a\sqrt{(r^2sin^2(\psi)+y^2)^2}}*\overrightarrow{n}= \frac{GmMdy}{2a(r^2sin^2(\psi)+y^2)^{3/2}}*(rsin(\psi)\overrightarrow{n_1}-y\overrightarrow{n_2})$$
$$\overrightarrow{F}=\int_{rcos(\psi)-a}^{rcos(\psi)+a}\frac{GmMdy}{2a(r^2sin^2(\psi)+y^2)^{3/2}}(rsin(\psi)\overrightarrow{n_1}-y\overrightarrow{n_2})$$
$$ \overrightarrow{F}=(\frac{GmM(rcos(\psi)+a)}{2arsin(\psi)\sqrt{r^2sin^2(\psi)+(rcos+a)^2}}-\frac{GmM(rcos(\psi)-a)}{2arsin(\psi)\sqrt{r^2sin^2(\psi)+(rcos-a)^2}})\overrightarrow{n_1}+(\frac{GmM}{2a\sqrt{r^2sin^2(\psi)+(rcos+a)^2}}-\frac{GmM}{2a\sqrt{r^2sin^2(\psi)+(rcos-a)^2}})\overrightarrow{n_2}$$$$ \overrightarrow{F}=(\frac{GmM(rcos(\psi)+a)}{2arsin(\psi)\sqrt{r^2sin^2(\psi)+(rcos(\psi)+a)^2}}-\frac{GmM(rcos(\psi)-a)}{2arsin(\psi)\sqrt{r^2sin^2(\psi)+(rcos(\psi)-a)^2}})\overrightarrow{n_1}+(\frac{GmM}{2a\sqrt{r^2sin^2(\psi)+(rcos(\psi)+a)^2}}-\frac{GmM}{2a\sqrt{r^2sin^2(\psi)+(rcos(\psi)-a)^2}})\overrightarrow{n_2}$$
When I let a/r go to zero, I get $$F=GmM/r^2*\overrightarrow{n_1}$$ (I'm not sure that this step is correct, it probably isn't). Now the common sense tells me, that if a/r goes to zero, the bar is reduced to a point mass and I would expect that the resultant should be of the form: $$GmM/r^2 (sin(\psi)\overrightarrow{n_1} - cos(\psi)\overrightarrow{n_2})$$ without any gravitational torque. However the answer according to the author is: $GmM/r^2$ and $GmMa^2/2r^3*sin(2\psi)$ (the force and the torque). I would appreciate some guidance on this.
