Elevator forces

We have an elevator with mass M with a person with mass m inside moving downwards with an acceleration $$-a$$.

The person is accelerated downwards by the force $$-g\cdot m$$ then by Newton's third law the ground will exert a force equal in magnitude $$g\cdot m$$. As the ground is now moving down with $$-a$$ we get another force on the ground, but why dont we take for that the mass of the accelerator M? Then the total force on the accelerator is $$m\cdot g - M \cdot a$$ and the total force on the person $$-g\cdot m + a\cdot M$$.

Why is that wrong?

• The problem is that you have taken the acceleration of $Lift$ , but you're talking bout the acceleration of the Man
– user303440
Commented Jun 8, 2021 at 11:06
• @Keshav Singh But the acceleration of the lift influcenes the acceleration of the Man? Commented Jun 8, 2021 at 17:26

Since the person is moving with acceleration $$-a$$ (where we take upwards to be positive, and we assume $$a <= g$$) then the net force on them must be $$-ma$$. Since gravity exerts a force $$-mg$$ on the person, the force $$F_1$$ exerted on them by the lift must satisfy
$$-ma = -mg + F_1 \\ \Rightarrow F_1 = m(g-a)$$
By Newton's third law, the person exerts an equal and opposite force $$-F_1$$ on the lift. The lift is also accelerating with acceleration $$-a$$ so the net force on the lift must be $$-Ma$$. Taking into account the force of gravity on the lift which is $$-Mg$$, then there must be a further force $$F_2$$ exerted on the lift by its mechanism, which satisfies
$$-Ma = -Mg -F_1 + F_2 \\ \Rightarrow F_2 = M(g-a) + F_1 = (M+m)(g-a)$$
Of course, you could reach the same conclusion more directly by treating the person and the lift as a single object with mass $$M+m$$.
There must be a Pseudo-Force Upward , (Opposite to the direction of $$a$$) . $$ma^*=mg-ma$$ Hence , $$a^*=g-a$$