Answer:
a.69.3 N
b.34.7 N
Step-by-step explanation:
We are given that
Angle of inclination of the ramp=\(20^{\circ}\)
Force F makes an angle with ramp=\(30^{\circ}\)
The component of F parallel to the ramp =\(60 N\)
a.We have to find the value of when its horizontal component is 60 N.
We know that
\(F_x=F cos 30^{\circ}\)
\(F=\frac{F_x}{cos 30^{\circ}}\)
\(F=\frac{60}{\frac{\sqrt3}{2}}\)
\(F=40\sqrt3=69.3N\)
b.We have to find \(F_y\) perpendicular to the ramp.
\(F_y=Fsin 30^{\circ}\)
\(F_y=69.3\times \frac{1}{2}=34.7 N\)
