Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F
[tex]F=\frac{2\times0.588}{\sqrt{3} }[/tex]
F = 0.679 N
