Answer:
[tex]F = 56.75 N[/tex]
Explanation:
As per the free body diagram of the box we can say that the applied force must be greater than or equal to the sum of friction force and component of the weight of the box along the hill
So we can say that
[tex]F_{net} = F cos35 - (mgsin20 + F_f)[/tex]
on the other side the force perpendicular to the plane must be balanced so that it remains in equilibrium in that direction
so we can say that
[tex]F_n + Fsin35 = mgcos20[/tex]
now we will have
[tex]F_f = \mu F_n[/tex]
[tex]F_f = (0.20)(mg cos20 - Fsin35)[/tex]
now we have
[tex]0 = Fcos35 - mg sin20 - (0.20)(mg cos20 - Fsin35)[/tex]
[tex]F(cos35 + 0.20 sin35) = mg sin20 + 0.20 mgcos20[/tex]
[tex]F = \frac{mg sin20 + 0.20 mgcos20}{(cos35 + 0.20 sin35)}[/tex]
[tex]F = \frac{100(sin20 + 0.20 cos20)}{(cos35 + 0.20 sin35)}[/tex]
[tex]F = 56.75 N[/tex]
