Question from Laws of motion

[Prashant_Gupta]

This is a question from laws of motion(H.C.Verma) :

there is a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. 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. Question is : What force should the man exert on the rope to get his correct weight on the machine?

PLease explain just with equations.

 

[polymath]

Hi Prashant_Gupta

The man cannot attain a steady-state mass reading of his weight on the machine.

The total mass suspended is 60+30 kgf = 90kgf

The light pulley will rotate until the rope force is equal on both sides.

The man:
body force = 60kgf down
applied force from arm = rope force = 90/2 kgf = 45kgf up
applied force from feet (w/mch) = 60-45kgf = 15kgf up
Man is in static equilibrium.

If man pulls additionally on rope he will rise, pulley will rotate, box will rise until equilibrium is again achieved.

Only a transient reading of his weight may be obtained - depending upon drag on system (bandwidth). If there is no drag then the system has infinite response and no change will occur no matter how hard the man pulls.

regards ... Polymath

 

[safwatonline]

hi polymath,
i get the 90(45) thing, but i dont get the 15 what does it represent ?isnt the answer simply 90 or what?
thnx

 

[aryajur]

Prashant,
Polymath is right that if the box and man are in equilibrium then the man is pulling the rope with a force of 45g and this does not show the true weight of the man on the machine. So the only possibility is to be in a dynamic equilibrium, so the man and the box will be rising with accesleration a.
Let us analze both cases separately. You can separate out the Man and the box as separate bodies.
On the man the forces are T(Tension) upwards by the rope that he is puliling down. A normal reaction N upwards and of course his weigth 60g downwards.
On the box the forces are T(Tension) upwards by the rope connected to the box to the pulley, the normal reaction N downwards and its weight 30g downwards.
For case 1 if the box is in static equilibrium then the forces should balance and cancel out hence we get the equations:

On Man: T+N=60g
On Box: N+30g=T

Solving we get N=15g, T=45g
since N=15g (weight shown on weighing machine is=15kg) it is not the true weight of the man.

Now suppose the man is pulling the rope with a continuous force irrespective of his motion, with a force T (not equal to 45g since now we assume the man is moving with the box upwards). Assuming the man and box are moving with an acceleration 'a' upwards, then the motion equations become:

On Man: T+N-60g=60a
On Box: T-N-30g=30a

Since we want N to be 60g (to show the true weight of the man), we can solve for T and 'a' to get:
T = 60a and a=3g
Thus T=180g
So if the man pulls the rope with a force 180g continuously then the machine will show his true weight.

 

[polymath]

post removed

 

[aryajur]

Hello Polymath,
I still think my analysis is correct. The monkey on the rope which I know of was that a monkey on one side and the mirror on the other side of the pulley of equal mass and the monkey cannot escape his image. If this is the one you are talking then this is quite different from the problem at hand.
If the man can maintain a constant pull then from the equations of motion of Newtonian mechanics he will pull himself and the box up, the equations I wrote will apply. Maybe I do not understand your reasoning to why it is wrong.

 

[polymath]

Hi aryajur

Yes, your analysis is correct.
I have removed my last posting to avoid confusing others and hide my embarrassment.:|

best regards ... Polymath

 

[Shweta_S]

hey
can you post ecopy of H.C.Verma
or atleast the problems in mechanics in H.C.Verma

Thanks
Shweta

 

[newelltech]

The answer is 0.
If he exerts no force on the rope he will be on the ground and the scale will show the correct weight.