Find the upward acceleration of the man

[Doc Al]

We can consider the force exerted by the man is same as the weight hanging from one end.right?

I don't know what you mean.

Here we are told how hard he pulls. So the tension is given.

In the case of a hanging mass, you are not told the tension you have to figure it out.

 

[BvU]

We can consider the force exerted by the man is same as the weight hanging from one end.right?

If I try to be a good listener (something like a lawyer trying to clear you), I could say: somewhat, given the very unfortunate wording of the original problem:

If he pulls each rope with a force to half its weight

[strike]That can not be in the law book of physics, because the dimensions are different. You may want to check on that wording. I already sneered at that in post #23. Doc backs me up. But for now I take your word that that is how it was mischievously (actually: wrongly) formulated. But[/strike] it means m/2 times g and that way we have a force [strike]again.[/strike]

And your the "weight hanging from one end" would then mean: the tension force at the lower end of e.g. the rightmost rope. Also a force. "Weight hanging from one end" [strike]times g[/strike] is only almost right. The m/2 times g trick doesn't work here, because we have not yet established that there is no acceleration ! the factor 2 can be admitted because of the symmetry. But the factor g has to be (g - a).

Do you think you now have enough to rephrase in a way that Doc doesn't have a chance to claim he doesn't know what you mean ? Your idea is right (I think), but the formulation leads to an answer like his (for your own good!).

 

[Doc Al]

That can not be in the law book of physics, because the dimensions are different. You may want to check on that wording.

I suspect it's a typo.
It should read:

If he pulls each rope with a force to half [STRIKE]its[/STRIKE] his weight

 

[BvU]

I stand corrected. So often seen weight expressed in kg that it deformed me. And then sneering at lb and lbf speaking nations. I should be ashamed and I am !

Feels great, though, because --being quite asine-- I won't make that mistake any more...

Also a bit obnoxious: his weight happens to be its weight (the platforms weight). Mightily confusing for our avi.

 

[avistein]

If he pulls each rope with a force to half its weight

It's a typo.It should be "half his weight".

But here in the man+platform problem,how can we consider that the force applied by the man has no acceleration? then only we can write T-mg/2=0=>T=mg/2

Suppose a pulley is there and one end has a mass and the other end has a force exerted by the man.Then here also will the tension equal to the force applied ? Then why not the tension=force when the force is replaced by a mass?

 

[avistein]

because we have not yet established that there is no acceleration

But here in the man+platform problem,how can we consider that the force applied by the man has no acceleration? then only we can write T-mg/2=0=>T=mg/2

 

[Doc Al]

But here in the man+platform problem,how can we consider that the force applied by the man has no acceleration? then only we can write T-mg/2=0=>T=mg/2

You do not assume no acceleration. You apply Newton's 2nd law and deduce it. (Here you are told that T = mg/2.)

 

[BvU]

Ah, now we are back to the original exercise somewhat. The idea is that you draw all the forces, including the gravitational forces (ahem, the weights). With the pulleys transferring the man's pulling force to the tensions in the ropes to the platform, you then have all that is needed to establish the acceleration of the platform. And then you will find ...

So: back to business and an answer to the question: what other forces play a role on the system in the dotted ellipse ? (post #21)

 

[BvU]

But here in the man+platform problem,how can we consider that the force applied by the man has no acceleration? then only we can write T-mg/2=0=>T=mg/2

Could it be that there is something in the concept of a tight rope -- with or without a pulley -- that hasn't completely settled in your mind? Just a hunch on my part, and believe me: it is so common and so basic that it is very, very often overseen.

So pardon me if I repeat the obvious (it will be of use many more times in PF, anyway):

In statics and in dynamics a tight, massless rope

• transfers force from one end to the other end.
  
  you pull the rope, the rope pulls whatever is attached at the other end​
• the forces on each end are equal and opposite
  
  Like action = - reaction at a distance
  the magnitudes of the forces must be exactly equal
  otherwise the rope itself would accelerate (with "infinite" acceleration, according to ΔF = ma)​
• the forces are aligned with the rope
  otherwise the rope would move sideways
  

  (Try it: push sideways against a dangling rope!)​

  If the rope is slung over a pulley, this is still true for all sections of the rope
  (a section is a straight piece of rope with a distinct direction
  -- it can be considered as a separate rope attached to something else
  -- the something else can also be a section of rope)
  ​

(open for clarifications, error correction, further questions)