# Explain what happens to object in lift [closed]

In the following image,three cases have been mentioned. $N$ is the normal force acting on the object inside the lift and $mg$ is the force of attraction due to gravity.
In case 1, $N = mg$.
In case 2, $N = m(g+a)$ and
in case 3, $N = m(g-a)$. Why is it so in 2nd and 3rd case?

• why has my question been put on hold? Oct 31 '13 at 14:01
• see my updated answer. Nov 19 '13 at 14:27

You need to consider the problem in two steps:

1) What is the object doing? In all three cases, the object remains in contact with the floor of the elevator, so it shares exactly the acceleration of the elevator. Newton's Second Law then gives us the total of all forces acting on the object.

2) What real forces are acting, and how large must they be to give the result required by 1)?

So:

Case #1

Elevator acceleration $=0$, so Object Acceleration $=0$, and Total Force $=0$

Gravity exerts $mg$ downward, so floor must exert a normal force $mg$ upwards, to exactly cancel gravity.

Case #2

Elevator acceleration $=a$ upward, so Object Acceleration $=a$ upward, and Total Force $=ma$ upward

Gravity exerts $mg$ downward, so floor must exert a normal force $mg+ma=m(g+a)$ upwards, to cancel gravity and have enough left over to accelerate the object upward.

Case #3

Elevator acceleration $=a$ downward, so Object Acceleration $=a$ downward, and Total Force $=ma$ downward.

Gravity exerts $mg$ downward, so floor exerts a normal force $mg-ma=m(g-a)$ upwards, to cancel some of gravity and leave the right amount to accelerate the object downward.

• oops, I didn't think of it that way.. Oct 31 '13 at 14:03

In physics weight is defined as the net gravitational force acting on an object. But what the measuring scales measures is the amount of $N$ required to make the object at rest in the frame of reference of the measuring instrument. If we are measuring everything from lift's frame then you should consider the gravitational force carefully because the gravitational force is not $mg$ in lift's frame. Gravitational force is $mg$ if we measure it w.r.t to earth's frame of reference. If we measure the gravitational force from the frame of reference of lift then we have to modify the gravitational force.
In the lift's frame there exists a pseudo force $ma$ on the object, $a$ is the acceleration of the lift as measured from the earth's frame of reference. The reading of the measuring scale(which we usually assumes to be the weight of an object) is $N$, so if $N$ changes we say the weight of the object changes. Let's Say the gravitational force in lift's frame is $F^{'}$ and in earth's frame $F$. Now let us see how the weight is different in the 3 different cases that you depict in your question.

CASE:1
Lift is at rest w.r.t earth, $a$ is $0$, so in lift's frame $F^{'}$ is same as $F$.
since this object is at rest in lift's frame so the net force on it should be $0$
$\vec F^{'}+ \vec N = 0$ but $F^{'}=F=mg \$ so $N =mg$
$N$ is what the measuring scale measure. So the weight of the object remains $mg$ if the lift is stationary w.r.t. earth.

CASE:2
Lift moves up w.r.t earth with an acceleration $a$. Now $F^{'} \neq F.$ By gelilean transformations we have $F^{'}=m(g+a)$. Since net force on the object should be $0$
we will have $N=m(g+a)$, that is the weight of the object increase if the lift accelerates upwards w.r.t earth.

CASE:3
Lift moves down w.r.t earth with an acceleration $a$. Now too $F{'} \neq F.$ By gelilean transformations we have $F^{'}=m(g-a)$. Since net force on it should be $0$
we have $N=m(g-a)$, that is if the lift accelerates downwards w.r.t earth the weight of the object decreases.

• The formatting, spelling and grammar issues of this answer are so severe that they render it unreadable. Mar 14 '14 at 12:38