A wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force F applied perpendicular to the angle of inclination is `30 ^(@)` and the box and the inclined plane is 0.2 , the minimum magnitude of F is
( Use ` g= 10 m//s^2`)

Options

A. 0 N as `30^(@)` is less than angle of repose
B. `le 1N`
C. `le 3.3 N`
D. `le 16.3 N`

Answer

D

Explanation

According to question we can draw the following diagram .
` (##ARH_5Y_SP_17_E01_011_S01.png" width="80%">
` mg sin theta = mu ( F + mg cos theta )`
` F = ( mg sin theta ) /( mu ) - mg cos theta `
` = mg [ ( sin theta )/ (mu)- cos theta ]`
Here ,
` m=1Kg , g=10 m//s ^2 , theta = 30 ^@ = mu = 0.2`
` F=1xx10 [ (sin 30 )/(02 ) - cos 30 ^@]`
` = 10 [ (1) /( 2xx0.2 ) -( sqrt(3))/(2)]`
` =10 [(5)/(2) - (sqrt(3))/(2)]= 5[5=- sqrt(3)]`
`=5[5-1.732 ]=16.34 N`