 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? May 26, 2013 at 8:24
• Actually I collected the picture from website and tried to aggregate my idea with the picture. 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.