One end of a string $0.5 m$ long is fixed to a point A and the other end is fastened to a small object of weight $8 N$ . The object is pulled aside by a horizontal force F, until it is $0.3 m$ from the vertical through A. Find the magnitudes of the tension T in the string and the force F.

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10N, 6N

7N, 6N

6N, 8N

9N, 10N

answer is A.

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Detailed Solution

$A C = 0.5 m$ , $B C = 0.3 m$

AB = 0.4 m

and if

Then

$∠ B A C = θ$

and $\cos θ = \frac{A B}{A C} = \frac{0.4}{0.5} = \frac{4}{5} \sin θ = \frac{B C}{A C} = \frac{0.3}{0.5} = \frac{3}{5}$

Here, the object is in equilibrium under three concurrent forces. So, we can apply Lami's theorem.

$\frac{F}{\sin (180 ° - θ)} = \frac{8}{\sin (90 ° + θ)} = \frac{T}{\sin 90 °}$

$\frac{F}{\sin θ} = \frac{8}{\cos θ} = T T = \frac{8}{\cos θ} = \frac{8}{4 / 5} = 10 N F = \frac{8 \sin θ}{\cos θ} = \frac{(8) (3 / 5)}{(4 / 5)} = 6 N$

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