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One end of a string 0.5 m long is fixed ...

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 force `F`.

A

6N

B

12N

C

18N

D

18N

Text Solution

Verified by Experts

The correct Answer is:
A
`AC = 0.5m, BC = 0.3m`
:. `AB =0.4m`
and it `/_BAC = theta.`
Then `cos theta=(AB)/(AC)=(0.4)/(0.5)=(4)/(5)`
and `sin theta=(BC)/(AC)=(0.3)/(0.5)=(3)/(5)`
Here , the object is in equilibrium under three concurrent forces. So we can apply Lami's theorem.
or `(F)/(sin (180^(@)-theta))=(8)/(sin (90^(@)+theta))=(T)/(sin 90^(@))`
or `(F)/(sin theta)=(8)/(cos theta)=T`
and `F=(8 sin theta)/(cos theta)=((8)(3//5))/((4//5))=6N`

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