Block B lying on a table weighs W. The coefficient of static friction between the block and the table is μ. 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. `(Wtantheta)/(mu)`
B. `muWtantheta`
C. `muWsqrt(1+tan^(2)theta)`
D. `muWsintheta`

Answer

B

Explanation

Weight of block B = W
Weight of block A = W.
T = tension in the string
Normal on block B, N= W
Friction force on block B,f= μN= μW

System will be in equilibrium, if
`Tcostheta=f=muW` ...(i)
`Tsintheta=W.` ...(ii)
Divide eqn (ii) by eqn (i), we get
`(sintheta)/(costheta)=(W.)/(muW),thereforeW.=muWtantheta`