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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 |
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Answer» `(Wtantheta)/(mu)` 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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