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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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Answer:

Explanation:

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N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2

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N_{bond} = \dfrac{L}{da} \\ \\  = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}

\text{Finally; the stiffness of a single interatomic spring is:}

k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s

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s = ut + at²/2

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