Kinematics- Laws Of Motion- Apparent Weight In A Lift

When a body of mass m is put on a weighing machine and it is in a lift, the body’s actual weight is mg. This affects a weighing machine, which produces a reaction R based on the weighing machine’s reading. The apparent weight in the lift is the reaction produced by the area of contact on the body. The vector sum of an object’s actual weight and the negative of its acceleration forces equals the apparent weight of the object that is accelerating.

Everything is attracted towards its centre by the earth. The force of attraction exerted by the earth on a body is called the ‘gravity- force’. If m is the mass of a body, then the gravity- force on it will be mg. It is due to this gravity force that a body presses down the plane on which it is placed. The force exerted by the body on the plane is called the weight (W) of the body. The weight of a body equals the gravity force (W = mg). But, if the body is on a plane that is accelerated up or down, then the force exerted by the body on the plane, that is, the weight W of the body changes, while the gravity- force on the body is still mg. Generally, there is no difference between the weight of the body and the gravity force exerted on it. But, under special conditions, like an apparent weight in a lift, there is a difference.

How do we feel the weight of our bodies?

When we stood on the earth’s surface, we pressed the earth downwards due to the gravity- force. This pressing force is our weight. Earth also exerts an upward- force on our feet. It is due to this reactionary- force that we experience our weight. Generally, we do not take any note of it. But whenever there is a change in the reactionary- force exerted by the plane (on which we are standing) on our feet, then we experience a change in our weight. For example, our weight appears to be increased in a lift accelerated upward and decreased in a lift accelerated downward.

 Let us now consider the following cases to easily understand the concept of apparent weight in a lift:

 Case 1:

Suppose a person with mass m  is standing on a weighing machine. The gravity- force acts on him vertically downward. The person exerts on the machine a force vertically downward which is the weight of the person.  

Let it be W (given by the machine). The machine also exerts a reactionary-force R on the person in the upward direction, where R=W according to  Newton’s third law. 

Thus, two forces are acting on the person, they are:

1.     Gravity force mg; and
2.     Reactionary force R.

Since these two forces are in opposite directions, the net force on the person is given by the following formula:

                               F = m g – R (downward)

But the person is stationary with no acceleration in him, hence the net force on him should be zero, that is:

                             F = mg – R = 0; or

                             R = mg, but R = W

Therefore, W = mg

Thus, the weight of the person shown on the machine equals the gravity- force.

Case 2:

 Suppose the person is standing on a weighing machine placed in a lift which is going up with acceleration a. Again, the net force on the person is

                 F = mg – R (downward).

Now the person has an acceleration (upward). Hence, according to Newton’s second law, now the net force on him is given by F = ma (upward) or F = ma (downward). Thus: 

                               -ma = mg – R; or

                               R = mg + ma but R = W

Therefore, W = mg + ma.

Thus, under this condition, the apparent weight W of the person is greater than the gravity–force mg which means the person will feel his weight to be increased than in the normal state.

 Case 3:

 Suppose that the lift is coming down with acceleration a. Again, the net force on the person is

               F = mg – R (downward).

Now, the acceleration a is directed downward. Hence, the net force F = ma should also be directed downward. That is, 

                           ma = mg – R; or

                          R = mg – ma but R = W

Therefore, W = m g – m a

Under this condition, the person will feel his weight to be decreased.

If the string of the descending lift is broken, then it will fall like a free body. Under this condition, a = g, and so we get from the above equation:

                           W = m g – m g = 0

Now the machine will read zero, which means that the person will feel his weight zero.

 We may now say that if an observer himself is in a state of accelerated motion, then Newton’s law of motion will appear to be violated.

If the acceleration of the descending lift is greater than the acceleration due to gravity g, then the weight of the person  W = m g – ma (a > g ), will be negative. 

Under this condition, the person will rise from the floor of the lift and stick to the ceiling of the lift.

If the lift is moving up or down with a constant velocity ( a= 0), then

W = m g, that is, the weight of the person will remain equal to the gravity force.

 Conclusion

The apparent weight of an object is a characteristic that describes its perceived weight. One example is your apparent weight in a lift. Due to a downward push, when the list goes up, the perceived weight changes. When you step on a scale in an elevator that is speeding upward, you will feel heavier since the elevator’s floor pushes more on your feet. When the elevator goes downward, on the other side, you feel lighter.