Elevator physics with pseudo force
When an elevator is moving with an acceleration, it becomes a non-nertial frame of reference. We find it difficult to apply Newton’s laws of motion for a non-nertial frame of reference, when the observer is a part of that reference. Here we will discuss the contribution of the concepts of pseudo force and discuss elevator physics with pseudo force. Once we start applying this concept, we can solve elevator physics problems with ease! Now you may read it with a cup of tea!
Observers in inertial and noninertial frame of references
Observer B is in an elevator. The elevator is moving upwards with an acceleration a. A box of mass M is resting on the elevator floor.
Observer A is stationary and standing on a static surface (say ground) and watching the box on the elevator floor.(the elevator has glass walls 🙂 )
Experience of Observer A – Elevator physics
Both observer B and the box are moving upwards with an acceleration a.
So the observer A (who is on inertial frame of reference) will draw the Free Body Diagram (FBD) of the box on the elevator floor in this way:
In the FBD, N is the Normal reaction exerted by the elevator floor on the box. The gravitational pull on the box is mg. (m is the mass of the box and g is the acceleration due to gravity)
Observer A will observe the box in the elevator to move upwards with an acceleration a.
So he will write the net force equation following Newton’s Second Law of motion in the following way:
Net force = N – mg
we know, Net force = mass X acceleration = m a
hence, ma = N – mg
or, N = m (g + a) ………………………………… (1)
Experience of observer B – Elevator physics
Now we take observer B, who is travelling with the box in the same elevator.
As both observer B and the box are within the same elevator frame of reference, observer B will find the box to be static. So for observer B, acceleration of the box equals zero.
Now observer B will draw the FBD of the box in the following way:
Observer B also will write motion equations following Newton’s 2nd law.
N – mg = Net force
Net force = mass X acceleration = m x 0= 0 ( as the box appears to be static to observer B)
hence, 0 = N – mg
or, N = m g ………………………………… (2)
Now what did you see? Elevator physics
2 observers are giving different observations and interpretations of Newton’s law. We know why. Observer A watches the motion of the box from an inertial frame of reference. And we have read in one of our earlier posts that Newton’s laws are well applicable in inertial frame of reference.
On the contrary observer B watches the box being a part of an accelerating or non-inertial frame of reference. And we know this type of frame of reference doesn’t directly support the Newton’s laws of motion, when the observer is also a part of the system.
Then what to do for non-inertial frame of reference while solving numerical? Yes, you are right.
We have to incorporate the concept of Pseudo force.
Let’s see how we can do the same here in this case.
Now following the concept of pseudo force we can get the magnitude and direction of the pseudo force on the box. Note that here the observer B is having an acceleration a upwards.(as the elevator he is in is moving with that acceleration)
Pseudo force on the box – Elevator physics
Pseudo force on the box = mass of the box X acceleration of the observer B = m a
Direction of this force = opposite to that of the reference frame = downwards
want a quick revision? read this: pseudo force
Now with the inclusion of pseudo force if we draw the FBD of the box for observer B then see what we find.
So now for observer B we write equation:
Net force = N – mg – ma
As observer B finds the box static so a = 0
So Net force on box = 0
As a result, N – mg – ma = 0
hence, N = m ( g + a) ………………………….. (3)
So we have just seen how just by including the concept of pseudo force we can successfully apply Newton’s 2nd law of motion even in an accelerating non-inertial frame of reference.