Pseudo force always acts in a non-inertial frame and it is always opposite to the direction of the body's acceleration. The backward push is of course inertia, but can we consider it in this manner (pseudo force) also?
If it is a pseudo force, then why is it visible? Pseudo forces are imaginary/fake forces that we can't see but we imagine it to be there.
$\begingroup$
$\endgroup$
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
8
Yes... that can also be considered as a pseudo force only when you are analyzing the dynamic of the motion in the accelerating frame (of the train in this case) The pseudo force acts through the centre of mass of the person and pushes him backwards in the frame of the train.
-
$\begingroup$ If it is a psuedo force , then why is it visible ? Pseudo forces are imaginary/fake forces that we can't see but we imagine it to be there. $\endgroup$ Commented Sep 17, 2020 at 12:46
-
1$\begingroup$ Pseudo forces are just a mathematical trick. It is not present there. But if you assume it to be present then you can explain the motion without violation of basic laws.The main analysis is done in the inertial frame. So that we have the same equations in other frames too, we need to consider that force $\endgroup$ Commented Sep 17, 2020 at 12:52
-
$\begingroup$ @Soumyadwipchandra , you said that "It is not present there" .But in this case ,it is not only present but also visible .So can we consider it as psuedo force? $\endgroup$ Commented Sep 17, 2020 at 12:56
-
$\begingroup$ Even in this case there is no such force. We must assume the force to explain the motion inside the non-inertial frame. In that sense, it is visible without being present. Take it the other way round. $\endgroup$ Commented Sep 17, 2020 at 12:58
-
$\begingroup$ So during calculation, we draw a force in backward direction 'F'(due to inertia) ,should we draw another force(psuedo force) opposite to the direction of acceleration of the body (for calculation)if we consider non-inertial frame of reference? $\endgroup$ Commented Sep 17, 2020 at 13:09
$\begingroup$
$\endgroup$
Indeed, in this case the train can be treated as a non-inertial reference frame. If we were to describe the motion in such a reference frame, we would need to introduce a fictitious force acting on all the objects.
This becomes by far more interesting when dealing with rotating reference frames, which requires introducing a bunch of additional fictitious forces (Centrifugal, Coriolis and Euler forces). You may want to look up the Newton laws in rotating reference frame and Foucault pendulum.
$\begingroup$
$\endgroup$
To clear up a misunderstanding about pseudo-forces:
Pseudo-forces are not real forces - because they have no physical cause and disappear in an inertial frame of reference
...but they have real effects - they appear as terms in the equations of motion when co-ordinates are measured relative to a non-inertial frame of reference
An analogy is as follows. Latitude is a true geographic co-ordinate - there is nothing arbitrary about where we place the equator. On the other hand, longitude is a pseudo-co-ordinate because it depends on the arbitrary placement of a prime meridian (which is why it is more difficult to measure than latitude). However, even though longitude is a pseudo-co-ordinate, getting it wrong has real effects.
answered Sep 17, 2020 at 13:25
$\begingroup$
$\endgroup$
2
If it is a psuedo force , then why is it visible ?
What is visible is not a force, but the effect of the inertia of the person's upper body.
Let's say the person is standing in a train going at constant speed in straight line. The train then suddenly accelerates. In the inertial frame of the train tracks Newton's first law says a body in uniform motion tends to stay in uniform motion and a body at rest tends to stay at rest unless acted upon by an net external force. The person's feet do not move relative to the floor of the train due to the external static friction force. But the upper part of the person's body is not so restrained by the static friction force, so it wants to continue at constant speed rather than accelerate. Consequently it appears to fall back during the acceleration. It is the lack of an external force to change its speed that causes it to fall back, not the application of an external force.
In the non-inertial frame of the train one observes the upper part of the body accelerate backwards. In order to explain this acceleration using Newton's second law, $a=\frac{F}{M}$, one has to assume a force $F$ acts on the upper body. Since there is no actual contact force causing the upper body to accelerate backwards in the frame of the train, the "force" $F$ is called a pseudo or fictitious force.
While Newton's laws of motion are the same in all inertial frames, in non-inertial frames they vary from frame to frame depending on the acceleration. Generally, in non-inertial frames pseudo forces are needed to explain the effects of the acceleration of the non-inertial frame.
Pseudo forces are imaginary/fake forces that we can't see but we imagine it to be there.
Yes, but we only need to imagine such forces in the non-inertial frame of reference in order to apply Newton's laws of motion. We don't need to imagine them in inertial frames of reference.
So the effect of inertia is not considered as force ?
The inertia of an object is a property of its mass and the effect of the inertia is for the object to resist a force. Newton says $F=ma$ or $a=F/m$. The greater the mass of an object the greater the inertia of the object and the greater the force needed to cause the object to accelerate.
If we are given a question like this and said to calculate acceleration of the train, we need not consider the effect of inertia?
The effect of the inertia of the train is its mass. The acceleration of train in the inertial frame of the tracks and is $a=F/m$ where $F$ is the net force applied to the train and $m$ is its mass. The greater the mass of the train the greater its inertia and its resistance to acceleration.
Hope this helps.
-
$\begingroup$ So the effect of inertia is not considered as force ? If we are given a quesion like this and said to calculate acceleration of the train, we need not consider the effect of inertia? $\endgroup$ Commented Sep 17, 2020 at 13:39
-
1$\begingroup$ @Angelinevarghese I have updated my answer to respond to your follow up questions. Hope it helps. $\endgroup$– Bob DCommented Sep 17, 2020 at 14:25