Answer
(a) The magnitude of ${{\bf{F}}_1}$ in terms of the magnitude of ${{\bf{F}}_2}$:
${F_1} = \frac{2}{{\sqrt 3 }}{F_2}$
${F_1} = 980$ N, ${\ \ }$ ${F_2} = 490\sqrt 3 $ N
(b) the maximal magnitude of ${{\bf{F}}_1}$ is ${F_1} = 980$ N
Work Step by Step
(a) Since the wagon does not move, by Newton's first law, the net forces that act on the wagon is zero, that is,
$\sum {\bf{F}} = {\bf{W}} + {{\bf{F}}_1} + {{\bf{F}}_2} = {\bf{0}}$
where ${\bf{W}} = - 50g{\bf{j}}$ is the gravitational force that acts on the wagon and $g$ is the gravitational acceleration.
From the figure we see that
${{\bf{F}}_1} = {F_1}\cos 30^\circ {\bf{i}} + {F_1}\sin 30^\circ {\bf{j}}$
${{\bf{F}}_2} = - {F_2}{\bf{i}}$,
where ${F_1} = ||{{\bf{F}}_1}||$ and ${F_2} = ||{{\bf{F}}_2}||$.
So, the components of the total net forces are
${F_1}\cos 30^\circ - {F_2} = \frac{1}{2}\sqrt 3 {F_1} - {F_2} = 0$
$ - 50g + {F_1}\sin 30^\circ = - 50g + \frac{1}{2}{F_1} = 0$
From the first equation we obtain the magnitude of ${{\bf{F}}_1}$ in terms of the magnitude of ${{\bf{F}}_2}$:
${F_1} = \frac{2}{{\sqrt 3 }}{F_2}$
From the second equation we obtain ${F_1} = 100g$ N.
So, ${F_2} = \frac{1}{2}\sqrt 3 {F_1} = 50\sqrt 3 g{\rm{N}}$
Let $g=9.8$ $m/{s^2}$. So,
${F_1} = 100g = 980$ N, ${\ \ }$ ${F_2} = 490\sqrt 3 $ N
(b) From part (a) we obtain the results ${F_1} = 980$ N and ${F_2} = 490\sqrt 3 $ N
Since the direction vector of ${{\bf{F}}_2}$ is horizontal, it does not contribute to the upward force. Thus, we only consider ${{\bf{F}}_1}$. From part (a) we obtain ${F_1} = 980$ N. The system is in equilibrium with this magnitude that balances the weight of the wagon. Therefore, this is the maximal magnitude of ${{\bf{F}}_1}$ that can be applied to the wagon without lifting it.
