Vectors!

Classical Mechanics Level 2

Two equal forces act on a body. It is found that the square of their resultant is twice the square of each force acting on it, then find the least angle between the two forces (in degrees).

180 ∘ {180}^\circ 53 ∘ {53}^\circ 37 ∘ {37}^\circ 90 ∘ {90}^\circ

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
May 4, 2016

Resultant = A 2 + B 2 + 2 A B cos ⁡ θ \sqrt{A^2 + B^2 + 2AB\cos\theta}
Let the two forces be x x ,
Then according to the given conditions:-
x 2 + x 2 + 2 × x × x cos ⁡ θ = 2 F 2 2 x 2 + 2 x 2 cos ⁡ θ = 2 F 2 2 x 2 ( 1 + cos ⁡ θ ) = 2 F 2 1 + cos ⁡ θ = 1 cos ⁡ θ = 0 ∴ θ = 90 ° ( or ) 270 ° x^2 + x^2 + 2×x×x\cos\theta = 2F^2\\ 2x^2 + 2x^2\cos\theta = 2F^2\\ 2x^2(1 + \cos\theta) = 2F^2\\ 1 + \cos\theta = 1\\ \cos\theta = 0\\ \therefore \theta = {90}° (\text{or}) {270}°

So, the least angle is 90 ° \boxed{{90}°} .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Did the same way :)

Abhay Tiwari - 5 years, 1 month ago

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Great :+1: :)

Ashish Menon - 5 years, 1 month ago

And 270 ain't there in the options lol. Vectors is quite interesting.

Abhiram Rao - 5 years, 1 month ago

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Heeh forgot that.

Ashish Menon - 5 years, 1 month ago

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No problem. Btw this is our first chapter this year.

Abhiram Rao - 5 years, 1 month ago

Are you in Narayana ?

A Former Brilliant Member - 4 years, 12 months ago

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Nope. I study in FIITJEE.

Abhiram Rao - 4 years, 12 months ago

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