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mr_godi [17]
3 years ago
9

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 rul

er, 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.
Physics
1 answer:
kow [346]3 years ago
6 0

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

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