Question

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 magnitude of the tension `T` in the string and force `F`.

Answer 1

AC=0.5 m, BC=0.3 m,
`therefore AB=0.4 m`
and if `angleBAC=theta`
Then , `costheta=(AB)/(AC)=(0.4)/(0.5)=(4)/(5)`
and `sintheta=(BC)/(AC)=(0.3)/(0.5)=(3)/(5)`

Here, the object is in equilibrium under three concurrent forces. So, we can apply Lami, therorem.
or `(F)/(sin(180^(@)-theta))=(8)/(sin(90^(@)+theta))=(T)/(sin90^(@))`
or `(F)/(sintheta)=(8)/(cos theta)=T`
`therefore T=(8)/(costheta)=(8)/(4//5)=10N " and " F=(8 sin theta)/(costheta)=((8)(3//5))/((4//5))=6N`