Calculate the force of friciton for shown situtation.

Explanation

`F_("lim")=muN=mu(Mg-F sin 30^(@))`
`=0.5(5xx9.8-20(1)/(2))`
`0.5(49.0-10)=0.5(39)=19.5N`
`F_("applied")=F cos 30^(@)=(20sqrt(3))/(2)=17.3N`
Since `F_("applied") lt F_("lim")`
`therefore` Force of friciton `=F_("applied")=17.3N`