Question

A ruler of length 0.30m is pivoted at its centre. Equal and opposite forces of magnitude 2.0N are applied to the ends of the ruler, creating a couple which create an angle of 50o with the ruler. What is the magnitude of the torque of the couple on the ruler.

Answer 1

Answer:

0.3858 Nm

Explanation:

The torque of the couple is the dot product of the force vector and the couple vector from 1 end of the ruler to the center. This equals to the product of their magnitude times the cosine() of the angle made by their direction:

$T = \vec{F} \cdot \vec{s} = Fscos(50^0) = 2 * 0.3 * 0.643 = 0.3858 Nm$