An object is in equilibrium under four concurrent forces in the direction shown in figure. Find the magnitudes of `F_(1)` and `F_(2)`

Options

A. ` F_(2)=(10)/(sqrt(3))N` , `F_(1)=(4)/(sqrt3)N`
B. ` F_(2)=(20)/(sqrt(3))N` , `F_(1)=(4)/(sqrt2)N`
C. ` F_(2)=(20)/(sqrt(2))N` , `F_(1)=(4)/(sqrt3)N`
D. ` F_(2)=(20)/(sqrt(3))N` , `F_(1)=(4)/(sqrt3)N`

Answer

D

Explanation

The object is in equilibrium. Hence,
(i) `sumF_(x)=0`
`therefore 8+4 cos 60^(@)-F_(2) cos 30^(@)=0`
or `8+2-F_(2)(sqrt3)/(2)=0 " or " F_(2)=(20)/(sqrt(3))N`
(ii) `sumF_(y)=0 therefore F_(1)+4sin 60^(@)-F_(2) sin 30^(@)=0`
or `F_(1)(4sqrt(3))/(2)-(F_(2))/(2)=0" or "F_(1)=(F_(2))/(2)-2 sqrt(3) =(10)/(sqrt3)-2sqrt(3)`
or `F_(1)=(4)/(sqrt3)N`