Answer
The weight of the sand and the box is $~~3300~N$
Work Step by Step
We can consider the horizontal forces on the box to find an expression for $m$:
$F~cos~\theta-[(mg-F~sin~\theta)~\mu] = ma$
$F~cos~\theta+F~\mu~sin~\theta = ma+mg~\mu$
$m = \frac{F~cos~\theta+F~\mu~sin~\theta}{a+g~\mu}$
To find the angle $\theta$ which maximizes the mass, we can find an expression for $\frac{dm}{d\theta}$:
$\frac{dm}{d\theta} = \frac{(F~\mu~cos~\theta-F~sin~\theta)(a+g~\mu)}{(a+g~\mu)^2} = 0$
$F~\mu~cos~\theta-F~sin~\theta = 0$
$F~\mu~cos~\theta = F~sin~\theta$
$\mu~cos~\theta = sin~\theta$
$\mu = tan~\theta$
$\theta = tan^{-1}~(\mu)$
$\theta = tan^{-1}~(0.35)$
$\theta = 19^{\circ}$
We can assume that the box will just barely move so we can let $a = 0$. We can find the weight $mg$:
$m = \frac{F~cos~\theta+F~\mu~sin~\theta}{a+g~\mu}$
$mg = \frac{F~cos~\theta+F~\mu~sin~\theta}{\mu}$
$mg = \frac{(1100~N)~cos~19^{\circ}+(1100~N)~(0.35)~sin~19^{\circ}}{0.35}$
$mg = 3300~N$
The weight of the sand and the box is $~~3300~N$
