A uniform bar of mass M and length L is horizontally suspended from the ceiling by two vertical light cables as shown. Cable A is connected `(1)/(4)`th distance from the left end of the bar. Cable B is attached at the far right end of the bar. What is the tension in cable A?

Options

A. `1//4` Mg
B. `1//3` Mg
C. `2//3` Mg
D. `3//4` Mg

Answer

C

Explanation

`sum tau_(B) = T_(A)(3L//4) - Mg(1//2L)=0`
Therefore,
`T_(A) = (MgL//2)//(3L//4) = (MgL//2) (4//3L) = 2Mg//3`