I am new to Newtonian mechanics, and was wondering regarding the following:

The system pictured below is moving at acceleration $a$ towards the left. The mass of the ball is $m$, and there is no friction. If I take this as an non-inertial framework, I would analyse the situation as follows:

there is a force pressing the ball leftwards and a fictitious force "pushing backwards" at the wall. Of course gravity is pulling the ball downwards.

However, I was wondering what will happen to the normal force in such a scenario. Its y-component must of course be identical to $mg$, but what about its horizontal component, in light of the real force and the fictitious force to the right of it?

Thank you

enter image description here

  • $\begingroup$ Just apply Newton's second law. Based on your post it seems like you understand how to do this. $\endgroup$ – BioPhysicist Jul 18 '19 at 12:26
  • $\begingroup$ so the fictitious force lowers the normal force, because the two of them need to be equal to the leftward force? But it still remains? $\endgroup$ – Pregunto Jul 18 '19 at 12:29
  • 2
    $\begingroup$ The normal force doesn't depend on what frame you choose to be in. $\endgroup$ – BioPhysicist Jul 18 '19 at 12:31
  • $\begingroup$ But within this non-inertial frame the ball is at rest, so the x-component of the normal force + the fictitious force need to be equal to the leftward force. Correct or not? $\endgroup$ – Pregunto Jul 18 '19 at 12:33
  • $\begingroup$ Yes, that is correct $\endgroup$ – BioPhysicist Jul 18 '19 at 13:09

In this case all you need to do is apply Newton's second law to the ball of mass $m$. The interpretation of a fictitious force in this case really just depends on which side of the equation you put the $ma$ term.

Let's first work in an inertial frame. As you said, in the horizontal direction we have a leftward force $F_w$ acting on the ball to the left due to the wall, and we the horizontal component of the normal force $N_x$ acting to the right. If we define left to be positive, then with an acceleration $a$ to the left we have using Newton's second law: $$F_w-N_x=ma$$

What if we want to work in an accelerating frame moving to the left with the system? Well then this is equivalent to just moving the $ma$ term to the left side of the equation: $$F_w-N_x-ma=0$$

We can say that $ma$ is a fictitious force that acts to the right so that there is no net force in our frame of reference.

Notice how changing reference frames does nothing to influence the normal force. The fictitious force does not lower the value of the normal force.

| cite | improve this answer | |

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.