IIT JEE , AIEEE Forum - Physics , Mechanics - Lift-pulley problem

Prakriteesh

A man of mass 60kg is standing on a light weighing machine kept in a box of mass 30kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine? What force should he exert to get his correct weight on the machine? [Ans: 15 kg, 1800 N].

This problem is the 39th problem of the EXERCISES in chapter 5 of HCverma.

INDIAN_ARMY19890

i have already solved it in ur previous posts

govindsuku

hey i am gettin the first part right.......but the 2nd ans is comin out to be 900 not 1800......u sure its the ans?

INDIAN_ARMY19890

yes the answer is 1800N

Prakriteesh

Thanks, INDIAN ARMY. I could not solve it because by mistake I have considered the box and man together as the system. That way T comes to be 90g or 900N upward. Solving, the normal reaction on the man comes in negative. However the previous post in which U had solved was not mine.

INDIAN_ARMY19890

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

INDIAN_ARMY19890

THIS IS SOLUTION OF SECOND PART BUDDY_GUY

INDIAN_ARMY19890

HAVE U GOT IT?

joyfrancis

The man manages to keep the block at rest , so why have you used acceleration in your equations/??

Prakriteesh

See Joyfrancis, in the 1st case the man manages to keep the block at rest and then the weight shown by the machine is 15 kg. But in the 2nd case, as the weight shown by the w-machine is 60 kg so the force applied by the man has to be different from (and more than) that applied in the 1st case. So, the block is no longer at rest but is accelerating upward in the 2nd case. Hence the acceleration term is included by INDIAN ARMY in his solution. And he is correct.

INDIAN_ARMY19890

THANKS PRAKRITEESH U R ABSOLUTELY RIGHT THE MAN MANAGED TO KEEP SYSTEM AT REST ONLY IN FIRST CASE IT NOT IN SECOND CASE IN FACT IT IS NOT POSSIBLE TO DO SO

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