TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given, mass of block A and B are (mA),(mB) be 3kg and 7kg, respectively.
Coefficient of friction between block A and B,B and floor (µA),(µB) be 0.4 and 0.55 respectively,
Applied force (F) on block B be 50N.
Let fA,fB be the friction forces between blocks A,B and B, floor respectively. Now, from block B diagram,
fB=µBNB . . . (i)
where, NB be the normal reaction on block B
‌‌=µB(mA+mB)g
‌‌=0.55(3+7)10=55N
Similarly, fA=µANA
where, NA is normal reaction on block A.
∴‌‌fA‌‌=µAmAg
fA‌‌=0.4×3×10
‌‌=12N
Now, since, fB>f
∴ the block B will not move.
Hence, fA=0N