1.

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

Answer»

`(Wtantheta)/(mu)`
`muWtantheta`
`muWsqrt(1+tan^(2)theta)`
`muWsintheta`

Solution :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`


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