A) 100 N
B) 80 N
C) 120 N
D) 150 N
Correct Answer: C
Solution :
[c] F.B.D. of block A and block B for block A: If block A is at equilibrium, then \[{{f}_{1}}={{W}_{A}}=20N\] If friction is static, \[{{f}_{1}}\le {{\mu }_{1}}{{N}_{1}}\] Or \[20\le 0.1\times F\Rightarrow F\ge \frac{20}{0.1}=200N\] Hence, limiting value of friction between the blocks. \[{{f}_{1}}=0.1\times 200=20N\] From F.B.D, of block B: \[{{f}_{2}}={{f}_{1}}+{{W}_{b}}=20+100=120N\] Alternative method: If we take both blocks together as system, them For equilibrium of the system I vertical direction. \[{{f}_{2}}={{W}_{A}}+{{W}_{B}}=20+100=120N\]You need to login to perform this action.
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