Let us consider +Y as upward and -Y axis as downward. So the acceleration that works these objects is always $-g$.

1. For the first picture, if the lift is going upward with acceleration $a$, then will the weight of the man of mass $m$ be $W=m(a-g)$? (but I found in textbook $g-a$) $a$ is positive here because the man gets acceleration due to lift's motion and which is +Y axis.
2. For the 2nd picture, if the lift is going upward with acceleration $g$, then will the weight of the man of mass $m$ be, $W=m(g-g)=0$?(but I found in textbook $g+g$)

These results contradicts with weightlessness. Can someone clarify my mistakes?

• Your text is not consisten with the figure. Can you please correct this? Can you also explain what in your opinion the contraction with weightlessness is? Commented May 26, 2013 at 8:24
• Actually I collected the picture from website and tried to aggregate my idea with the picture. Commented May 26, 2013 at 9:04

The feeling of weightlessness or the feeling of weighing heavier is due to the force that acts on you from the floor.

Now, if a lift is going upward with an acceleration $a$, then the weight of a man inside it will experience a weight of:

$$F = ma$$

where $F$ is the net force acting on the man and $a$ is the net acceleration of the man (and lift of course).

The force that gravity exerts on the man is given by $F_g = mg$

The net force, $F$, is obtained from the difference of the force acting on the man from the floor of the lift (which we are interested in), and the force of gravity:

$$F_{floor} - Fg = F$$

so that:

$$F_{floor} = ma + mg = m(a+g)$$

Therefore, when the lift is going up, you feel as though there is an acceleration of $(a+g)$ acting on you (I think that the $(a-g)$ in your question was a typo and should be $(a+g)$, since you feel heavier when a lift is going down and lighter when the lift is going down).

If the lift is going up at an acceleration $g$, then you feel an acceleration of $(g+g)$.

Now, to feel weightlessness, the lift has to have a certain acceleration value downwards. If this downward acceleration is equal to $g$, we get (we substitute $a$ with $-g$):

$$F_{floor} = m(-g+g) = 0$$

So that you don't feel the force the floor is acting on you and are said to be free falling.