# Fictitious forces in relation to normal force

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 • Just apply Newton's second law. Based on your post it seems like you understand how to do this. – BioPhysicist Jul 18 '19 at 12:26
• 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? – Pregunto Jul 18 '19 at 12:29
• The normal force doesn't depend on what frame you choose to be in. – BioPhysicist Jul 18 '19 at 12:31
• 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? – Pregunto Jul 18 '19 at 12:33
• Yes, that is correct – 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.