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Question

Two blocks A and B of mass mA=1 kg and mB=3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F (in newtons) that can be applied on B horizontally, so that the block A does not slide over the block B is [Take , g=10 m/s2]

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Solution

Step 1Acceleration a of system of blocks A and B isa=Net forceTotal mass=F−f11mA+mBwhere, fl1 = limiting friction between B and the surface=μ(mA+mB)gSo,a=F−μ(mA+mB)g(mA+mB)...........(i)Here, μ=0.2,mA=1 kg,mB=3 kg,g=10 ms−2Substituting the above values in Eq. (i), we have a=F−0.2(1+3)×101+3a=F−84.............(ii)Due to acceleration of block B, a pseudo force F' acts on AStep 2This force F' is given byF′=mAa where, a is acceleration of A and B caused by net force acting on B.For A to slide over B; pseudo force on A, i.e. F' must be greater than limiting frictionbetween A and B.⇒mAa≥fl2We consider limiting case,mAa=fl2⇒mAa=μ(mA)g ⇒a=μg=0.2×10=2 ms−2......(iii)Putting the value of a from Eq. (iii) into Eq. (ii). we getF−84=2∴F=16 N

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