1. 5 .

Let the forces on the three ropes be $\boldsymbol{F}_{1}, \boldsymbol{F}_{2}, \boldsymbol{F}_{3}$, then $\left|\boldsymbol{F}_{1}\right|=1,\left|\boldsymbol{F}_{2}\right|=2,\left|\boldsymbol{F}_{3}\right|=3$, and the angles between $\boldsymbol{F}_{1}, \boldsymbol{F}_{2}$, $\boldsymbol{F}_{3}$ are all $60^{\circ}$. Thus, $\boldsymbol{F}_{1} \cdot \boldsymbol{F}_{2}=\left|\boldsymbol{F}_{1}\right| \cdot\left|\boldsymbol{F}_{2}\right| \cos 60^{\circ}=1, \boldsymbol{F}_{2} \cdot \boldsymbol{F}_{3}=\left|\boldsymbol{F}_{2}\right| \cdot\left|\boldsymbol{F}_{3}\right| \cos 60^{\circ}=3$, $\boldsymbol{F}_{3} \cdot \boldsymbol{F}_{1}=\left|\boldsymbol{F}_{3}\right| \cdot\left|\boldsymbol{F}_{1}\right| \cos 60^{\circ}=\frac{3}{2}$. Let the gravitational force on the object be $\boldsymbol{F}$, then $\boldsymbol{F}+\boldsymbol{F}_{1}+\boldsymbol{F}_{2}+\boldsymbol{F}_{3}=\mathbf{0}$, hence
$$
\begin{aligned}
|\boldsymbol{F}| & =\left|\boldsymbol{F}_{1}+\boldsymbol{F}_{2}+\boldsymbol{F}_{3}\right|=\sqrt{\left(\boldsymbol{F}_{1}+\boldsymbol{F}_{2}+\boldsymbol{F}_{3}\right)^{2}} \\
& =\sqrt{\left|\boldsymbol{F}_{1}\right|^{2}+\left|\boldsymbol{F}_{2}\right|^{2}+\left|\boldsymbol{F}_{3}\right|^{2}+2\left(\boldsymbol{F}_{1} \cdot \boldsymbol{F}_{2}+\boldsymbol{F}_{2} \cdot \boldsymbol{F}_{3}+\boldsymbol{F}_{3} \cdot \boldsymbol{F}_{1}\right)}=5 .
\end{aligned}
$$
1. An object at rest in space is suspended by three ropes. It is known that the magnitudes of the forces in the three ropes are $1 \mathrm{~N}, 2 \mathrm{~N}, 3 \mathrm{~N}$, and the angle between any two ropes is $60^{\circ}$. The magnitude of the gravitational force acting on the object is $\qquad$ $\mathrm{N}$.
(Gold 供题)