HERE IS THE SOLUTION OF PART 2 I ALREADY SOLVED PART-1 making FBD of the man u will find following forces on him 1-weight downward=Mg 2-nomal reaction upward let it beN 3-tension upward now the equation will be like this let acceleration of man=a let it be in upward direction hence T-+N-Mg=Ma..............1 here M is mass of MAN now making FBD of the box u will find following forces on him 1-weight downward=mg 2-normal reaction=N downward 3-tension upward=T if the acceleration of man is a then box will also have same acceleration a also in upward direction we get this eq T-N-mg=ma..........2 putting M=60kg m=30kg g=10m/s2 and N=Mg according to quection we get T+600-600=60a.............1.1 T-600-300=30a.............2.2 solving 1.1 and 2.2 u will get a=30m/s2 putting in 1.1 T=60a=60*30=1800N ans
Moderation Team
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