0
$\begingroup$

I am familiar with Pseudo forces and how we use them in accelerating reference frames. My question is a bit specific. Let’s say I am accelerating at $a\frac{m}{s^2}$ and holding a tennis ball of mass $m$. I understand it is the push force provided by my hand that is accelerating the ball with me. When I observe the ball in my own (accelerating) frame, I would say that it is at rest relative to me. I am asked to write its force equations, and I come up with this,

$F_{push}$ - $ma$ = $m$$a_{ball}$, where ‘$ma$‘ is a pseudo force acting on the ball.

Since $a_{ball}$ = $0$ (relative to my frame), I’d say, $F_{push}$ = $ma$, and that’s why the ball is at rest relative to me.

My question is, can I use a pseudo force for myself? Because I am at rest relative to my own non-inertial frame, but I am accelerating relative to an inertial frame. Now since I am accelerating relative to an inertial frame, there must be a force (say, $F_{applied}$) acting on me that is accelerating me. I can see (or feel) that force in my own (non-inertial frame) because individual forces are frame-independent.

Let’s say my mass is $M$. I am at rest relative to my frame, so if I write the force equation for myself, can I say, $F_{applied}$ - $Ma$ = $M$$a_M$, where $a_M$ is my acceleration relative to my own frame.

Since $a_M$ = $0$ $\Rightarrow$ $F_{applied} = $M$a_M$

Is it correct? My question is, can I use the pseudo force for myself in my own accelerating frame? I understand I can use it for the ball I am holding, or any other body that I am observing. Can I use it for myself if I want to explain my state of rest relative to my own frame?

$\endgroup$

2 Answers 2

3
$\begingroup$

The short answer is yes, you can say a pseudo force is acting on you in your non inertial frame. If you are stationary in your frame it is because there is a real restraining force acting on you as well preventing you from accelerating in the non inertial frame.

For example, when you accelerate your car a force appears to be pushing you back against your seat back. That’s the pseudo force. In reality it is because of your inertia. Your seat back exerts a real force forward on you equal and opposite to the pseudo force so that you don’t accelerate in the non inertial frame of the car. That force is responsible for your acceleration in the inertial frame of the road.

Another example of a pseudo force is the centrifugal that appears to push you to one side of the car when cornering. The restraining force preventing you from sliding on your seat is the real centripetal force, in this case the static friction force between you and the surface of the seat.

Hope this helps.

$\endgroup$
1
  • $\begingroup$ Very helpful, thanks a lot. The references that you made to real-life situations helped me understand it better. $\endgroup$
    – 4d_
    May 16, 2020 at 12:28
2
$\begingroup$

...can I use the pseudo force for myself in my own accelerating frame?

Yes you can, as long as you include all the forces acting on you and then compute their resultant. However considering any force on you in the reference frame attached to you is useless, because the result is bound to be that all the forces cancel out and your acceleration becomes zero, always. Why? It primarily relies on the way we define a non inertial frame of reference and subsequently apply pseudo forces. Mathematically, let your acceleration in the ground frame be $\mathbf a_{\text{inertial}}$ and your mass be $M$. Thus by Newton's second law,

$$\mathbf F_{\text{net,inertial}}=M\mathbf a_{\text{inertial}}$$

Now let's move to the non inertial frame attached to you. Writing the equations again, including relevant pseudo forces, we get

$$\mathbf F_{\text{net,non-inertial}}=\mathbf F_{\text{net,inertial}}+\mathbf F_{\text{pseudo}}=M\mathbf a_{\text{inertial}}+M(-\mathbf a_{\text{inertial}})=0$$

Thus by Newton's second law applied in the non-inertial frame

$$\mathbf F_{\text{net,non-inertial}}=M\mathbf a_{\text{non-inertial}}=0\Longrightarrow \mathbf a_{\text{non-inertial}}=0 \quad \text{(always)}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.