Answer
(a) The force $F$ needs to be 104 N.
(b) The component $F_y$ perpendicular to the ramp will be 52.0 N.
Work Step by Step
(a) $\frac{F_x}{F} = cos(\theta)$
$F_x = F~cos(30.0^{\circ}) = 90.0~N$
$F = \frac{90.0~N}{cos(30.0^{\circ})} = 104~N$
The force $F$ needs to be 104 N.
(b) $\frac{F_y}{F} = sin(\theta)$
$F_y = F~sin(30.0^{\circ})$
$F_y = (104~N)~sin(30.0^{\circ}) = 52.0 ~N$
The component $F_y$ perpendicular to the ramp will be 52.0 N.
