Two masses A and B of 10kg and 5kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table (as shown in figure). The coefficient of friction between the table and the block is 0.2. The minimum mass (in kg ) of C that may be placed on A to prevent it from moving is equal to
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answer is 15.00.
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Detailed Solution
Let T be the tension in the string ;f= frictional force between block A and table; m'= minimum mass of C . For just motion of block A on table T=f=μR=μ(m+m')g=0.2(10+m')g For the just motion of block B,T=5 g∴5 g=0.2(10+m')g⇒m'=5−20.2=15 kg