Question

A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the mass 60 is grams and the force F is applied at the opposite end of the stick and it makes an angle of 60° with the stick.

Answer 1

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

$F=\frac{2\times0.588}{\sqrt{3} }$

F = 0.679 N