With two hands you hold a cone motionless upside doen , as shown in figure The mass of the cone is `(m = 1 kg)` and the coefficient of static friction between you fingers and the cone is `(mu = 0.5)` what is the minimum normal force you must applywith each hand in order to hold up the cone? Consider only translational equilibruim

Answer

`65 N`

Explanation

For equilibruim in vertical direction

` 2f sin theta - 2N cos theta - mg = 0` (1)
`2 muN sin theta - 2N cos theta = mg`
, `tan theta = ((pi)/(2) - theta ) = (5)/(12)`
`cos theta = (5)/(12) =sin theta = (12)/(13)` and `cos theta = (3)/(12)`
` 2(-5) ((12)/(13)) N - 2N(3)/(13) = (1) (10)`
`N((12)/(13) - (10)/(13)) = 100 rArr N = ((10 xx 13)/(2)) = 65 N`