Two blocks each of mass m are placed on a rough horizontal surface and connected by a massless inelastic string as shown. The coefficient of friction between each block and horizontal surface is μ. The string connecting both the blocks initially has zero tension. The minimum force to be applied on block A to just move the two block system horizontally (without the string getting slack) is
2μmg
2μmg√μ2+1
2μmg√4μ2+1
μmg√μ2+1
A
2μmg
B
2μmg√μ2+1
C
2μmg√4μ2+1
D
μmg√μ2+1
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Solution
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From FBD of block B, T=μN1=μmg .... (i) From FBD of block A, Fsinθ+N2=mg .... (ii) Fcosθ=T+μN2 .... (iii) From equations (i), (ii) and (iii) Fcosθ=μmg+μ(mg−Fsinθ) F(cosθ+μsinθ)=2μmg F=2μmgcosθ+μsinθ hence, Fmin=2μmg√1+μ2
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