Answer
The Magnitude Of The Force On The Mass $m_1$ Due To The Rod Is :
$3.01\times10^{-10}$$N$.
Work Step by Step
Consider a small segment of the rod with length $dr$ at a distance of $r$ from the point mass $m_1$.
Let the linear mass density of the rod is $\lambda$.
The the mass of the segment $dm$ is
$\lambda$$dr$
Gravitational force on the mass $m_1$ due to the small segment acts along $x-axis$ and is given by :
$$dF = \frac{Gm_1dm}{r^2}$$
$$dF = \frac{Gm_1\lambda dr}{r^2}$$
Total force on the point mass $m_1$ is :
$$F= \int_{d}^{d+L}\frac{Gm_1\lambda}{r^2}dr$$
Where $d$ is the distance between the end of rod closer to the point mass $m_1$ and $L$ is the length of the rod.
$$F= Gm_1\lambda\int_{d}^{d+L}\frac{1}{r^2}dr$$
$$F=-Gm_1\lambda\frac{1}{r}\left. \right|_{d} ^{d+L}$$
$$F=-Gm_1\lambda(\frac{1}{d+L}-\frac{1}{d})$$
$$F=Gm_1\lambda(\frac{1}{d}-\frac{1}{d+L})$$
$$F=Gm_1\lambda L(\frac{1}{d\times(d+L)})$$
$$ \lambda = \frac{M}{L}$$
Where $M$ is the mass of the rod.
Therefore,
$$F=Gm_1M (\frac{1}{d\times(d+L)})$$
Substituting for the values, we get :
$$F= 6.67\times10^{-11}\times0.67\times5(\frac{1}{0.23\times(0.23+3)})$$
$$F= 22.3\times10^{-11}\times(\frac{1}{0.23\times(3.23)})$$
$$F= 22.3\times10^{-11}\times(\frac{1}{0.74})$$
$$F= 30.1\times10^{-11}$$
Therefore magnitude of force on the point mass $m_1$ is :
$$F= 3.01\times10^{-10}$$
