Why does lift acceleration depend on reactionary force on the person in this question? Was working through a past paper and asked the question below to my teacher. She explains that there is a reactionary force upwards from the lift to the boy and that subtracting that by the weight will get me the resultant force upwards and by substituting 70 as m in F=ma, the answer is 3. This answer would be fine if I was able understand specifically 2 things:
 

*

*Why are we ignoring the force of the student on the lift in the expression? 

*And even in the case we are meant to ignore it, why is the reactionary force of the lift on the student being used to calculate the acceleration of the lift upwards? (Comes from my understanding that we treat them as separate entities)
 
What I do understand is that, a is the resultant acceleration upwards, the force of the boy on the lift is acting downwards and hence the reactionary force of the lift on the boy is acting equally upwards.


A lift moves upwards from rest with an acceleration a. A student of mass 70 kg standing in the lift exerts a force of 800N on the floor of the lift. Determine an expression for a.

*

*70g = 70a

*800 = 70a

*800 - 70g = 70a

*70g - 800 = 70a


 A: *

*We ignore the force of the boy on the lift because we are actually calculating the acceleration of the boy, not the acceleration of the lift (although both accelerations happen to have the same value $a$). The only use we make of the force of the boy on the lift is to use it to give us the force of the lift on the boy, via Newton's third law.

*We are not directly calculating the acceleration of the lift. Instead we are calculating the acceleration of the boy and using it to infer the acceleration of the lift. We cannot calculate the acceleration of the lift directly because we do not know its mass or the other forces acting on it. We can however infer the acceleration of the lift indirectly by assuming it is the same as the acceleration of the boy.

To clarify the difference between calculating the boy’s acceleration and the lift’s acceleration, suppose the boy was wearing jet boots which exerted a force of $800$ N on the boy and an equal and opposite force of $800$ N on the lift. Then the boy’s acceleration is still given by
$$800-70g=70a$$
but now the lift’s acceleration is completely unknown.
