If I am travelling in a train and the train accelerates forward then I experience a pseudo force backward.
But if someone is free falling under gravity where does he experience pseudo-force because pseudo is experienced in an accelerated frame.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ FWIW, in general relativity free falling means being in an inertial frame, thus no pseudo force. But since you are apparently asking about newtonian mechanics, I am not sure about this. $\endgroup$– jng224Feb 12, 2021 at 8:31
-
6$\begingroup$ Does this answer your question? Why does a free-falling body experience no force despite accelerating? $\endgroup$– jng224Feb 12, 2021 at 8:32
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
2
In a Newtonian model we would say that objects in a room are pulled down by a gravitational force. They accelerate downward until they reach some other object or the floor.
If we take the room and accelerate it downward at 1g, then we can say a pseudo force pointing upwards appears in that frame. This force exactly balances gravity. Now objects in the room have no net force at all on them. They don't accelerate with respect to the room and can remain at rest away from other objects. This matches our observations of objects "floating" in freefall.
answered Feb 12, 2021 at 8:50
-
$\begingroup$ So let us assume that a huge hand is pulling someone with a acceleration greater than 'g'. Now where is the pseudo force applied? $\endgroup$– ShivFeb 18, 2021 at 15:53
-
$\begingroup$ Psuedo forces appear based on your choice of frame. You can pick a frame accelerating faster than $g$ with or without a force on it being present. The psuedo force will appear for objects in that frame on their center of mass pointed in the opposite direction of the frame's acceleration. $\endgroup$ Feb 18, 2021 at 18:12
$\begingroup$
$\endgroup$
2
Pseudo force is fictitious force or we can say a mathematical correction term that we add in order for Newton's 2nd Law to work properly.
For example, a car is moving with acceleration $a$ on a road, and on side of the road, there is a man standing stationary. Driver will see that, first man approaches him, and then recedes in distance as he passes by the man in his car.
If driver will make Free Body Diagram of man, he will only include normal reaction, ang gravity, in $y$ direction, which is balanced. But he is still moving in $x$ direction as according to driver, in absence of any force. So Newton's 2nd law is broken. In order to prevent this, driver includes a pseduo force in direction of motion of man.
$$F_{pseudo}=M_{man}×a_{car}$$
In your case of train, you don't actually feel pseudo, since it's fictitious, but it can be explained by inertia of rest. (Pseudo force is a frame dependant quantity.)
In your case of free fall, you will apply pseudo to other objects around you especially the ground, which will seem to come to you with $g$ acceleration.
-
$\begingroup$ So let us assume that a huge hand is pulling someone with a acceleration greater than 'g'. Now where is the pseudo force applied? $\endgroup$– ShivFeb 20, 2021 at 7:36
-
$\begingroup$ @Ayush Bora, if let's say that the person is falling, with acceleration $a$, $a>g$, then the pseudo force will be applied to all other objects around person, if we are taking perspective of the person in free fall. . $\endgroup$ Jun 30, 2021 at 7:37
$\begingroup$
$\endgroup$
2
In free fall there is a pseudo force that, locally, exactly cancels gravity.
-
$\begingroup$ So let us assume that a huge hand is pulling someone with a acceleration greater than 'g'. Now where is the pseudo force applied? $\endgroup$– ShivFeb 18, 2021 at 15:54
-
$\begingroup$ @Ayush Bora, like i said that pseudo force is applied frame wise, to keep Newton second law from breaking. Let us say that the object is falling with acceleration $a$ such that $a>g$, then the obserever wrt object will see that somehow some 'pseudo force' is pulling up the ground with force $M_{earth}a$. This is because the observer will think that he is at rest, while others are moving. $\endgroup$ Feb 20, 2021 at 7:40
$\begingroup$
$\endgroup$
The opposite force exerted is drag. Once drag is equal to weight, acceleration become zero and velocity is constant. Once this happens, $F = ma$ tells us that there is no force.
