Answer
$F=2.91\times 10^{-12}N$
$\alpha=83.4^{\circ}$
Work Step by Step
We know that
$F_1=\frac{(6.67\times 10^{-11}Nm^2/Kg^2)(1.00Kg)(3.00Kg)}{(10.0m)^2}$
$\implies F_1=0.2001\times 10^{-11}N$
and $F_2=\frac{(6.67\times 10^{-11}Nm^2/Kg^2)(1.00Kg)(2.00Kg)}{(10.00m)^2}$
$\implies F_2=0.1334\times 10^{-11}N$
Now $F_{x}=F_{1x}+F_{2x}$
$\implies F_{x}=-(0.2001\times 10^{-11}N)cos60+(0.1334\times 10^{-11}N)cos60$
$\implies F_{x}=-0.03335\times 10^{-11}N$
and $F_{y}=-(0.2001\times 10^{-11}N)sin60-(0.1334\times 10^{-11})sin60$
$\implies F_{y}=-0.2888\times 10^{-11}N$
The magnitude of the force is
$F=\sqrt{F_x^2+F_y^2}$
$F=\sqrt{(-0.03335\times 10^{-11}N)^2+(-0.2888\times 10^{-11}N)^2}$
$\implies F=2.91\times 10^{-12}N$
Now the direction can be determined as:
$\alpha=tan^{-1}(\frac{F_y}{F_x})$
We plug in the known values to obtain:
$\alpha=tan^{-1}(\frac{-0.2888\times 10^{-11}N}{-0.03335\times 10^{-11}N})$
$\implies \alpha=180^{\circ}+tan^{-1}(\frac{-0.2888\times 10^{-11}N}{-0.03335\times 10^{-11}N})$
$\implies \alpha=83.4^{\circ}$ below the horizontal
