Block B lying on a table weighs w. The coefficient of static friction between the block and the table is mu. Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is

Options

A. `(w tan theta)/(mu)`
B. `muw tan theta`
C. `mu w sqrt(1+tan^(2)theta)`
D. `mu w sin theta`

Answer

B

Explanation

(b) Let weight of A is w'. From the free body diagram, for equilibrium of the system,

`T cos theta=mu N=mu w`……….(i)
`T sin theta=w'`…….(ii)
where, T=tension in the thread lying between knot and the support.
On dividing Eq.(ii) by Eq. (i), we get
`(T sin theta)/(T cos theta)=(w')/(mu w) implies tan theta=(w')/(mu w)`
`implies " " w'=muw tan theta`