A) \[210\,kg\]
B) \[190\,kg\]
C) \[185\,kg\]
D) \[162\,kg\]
Correct Answer: D
Solution :
[d] Maximum friction that can be obtained between A and B is \[{{f}_{1}}=\mu {{m}_{A}}g=(0.3)(100)(10)=300N\] and maximum friction between B and ground is \[{{f}_{2}}=\mu ({{m}_{A}}+{{m}_{B}})g=(0.3)(100+140)(10)=720N\]Drawing free body diagrams of A, B and C in limiting case, we get Equilibrium of A gives \[{{T}_{1}}={{f}_{1}}=300N\] ...(1) Equilibrium of B gives \[2{{T}_{1}}+{{f}_{1}}+{{f}_{2}}={{T}_{2}}\] or \[{{T}_{2}}=2(300)+300+720=1620N\] ?..(2) and equilibrium of C gives \[{{m}_{C}}g={{T}_{2}}\] or \[10{{m}_{C}}=1620\] or \[{{m}_{C}}=162kg\]You need to login to perform this action.
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