Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown in fig. The coefficient of friction of A with the table is 0.2 The minimum mass of C that may be placed on A to prevent it from moving is equal to

Let T = tension in the string, f = frictional force between block A and table, x = minimum mass of C. For the system to be in equilibriumf = T (for block A) ...(1)where f=μmg=μ(10+x)g=0⋅2(10+x)g∴T=0⋅2(10+x)g      ...(2)For the equilibrium of B,t = 5g   …(3)From eqs. (2) and (3), we get0⋅2(10+x)g=5g or 2+0⋅2x=5 or 0⋅2x=3 or  X=30⋅2=15kg.