# Should I include this fictitious force for the upper block? [closed]

The question is what will be the highest value of $$\theta$$ for which the two blocks will move together?

For the block of mass $$m_1$$, I got the equation: $$m_1 g (\sin \theta - \mu _k \cos \theta) = m_1 a$$ where $$a$$ is its acceleration.

But I am really confused about the block of mass $$m_2$$. The upper block will have to stay still for both of the blocks to move together. As the lower block will accelerate, the upper block will experience a fictitious force of $$m_2 a$$ in the opposite direction. Should I include this force in my calculation for $$m_2$$?

You only need to include fictitious forces if you are working in a non-inertial reference frame. If you do your calculations from the perspective of an inertial frame, e.g. a frame at rest relative to the incline, then you do not need to include fictitious forces.

As a simple example, imagine a block accelerating horizontally due to some applied force $$f$$. One can either use Newton's law in an inertial frame: $$F_\text{net}=f=ma$$, i.e. the block's acceleration is $$a=f/m$$.

Or we can work in some frame with a constant horizontal acceleration $$a_\text{frame}$$ so that in this frame we have to include a fictitious force $$-ma_\text{frame}$$ acting on the block. Applying Newton's second law in this frame gives us $$F'_\text{net}=f-ma_\text{frame}=ma'$$. Note that if our frame moves with the block we have $$a_\text{frame}=a=f/m$$, and we correctly calculate no acceleration in the accelerating frame, as then $$a'=0$$.

Something to keep in mind is that we call them "fictitious forces" because they violate Newton's third law. In non-inertial frames Newton's laws do not hold. We can either choose to drop the second law and have accelerations arise in situations where forces are not present, or we can bring in fictitious forces and drop the third law, as now we have "forces" that are not arising due to interactions. However, it is not correct to think, "Ah, an object is accelerating, therefore we need to bring in fictitious forces." Fictitious forces arise from the reference frame, not the object. These often get conflated, however, because often we choose a frame that moves with the accelerating object, and so the distinction becomes irrelevant at that point.

• So I can just use $m_2 g sin \theta - m_2 g \mu _k cos \theta =0$? And also, is my equation for the lower block correct? Commented Aug 26, 2020 at 16:34
• @Theoretical Are you assuming that block m2 has no acceleration but is sliding relative to m1? Commented Aug 26, 2020 at 16:47
• No. I am assuming that m1 is accelerating and m2 is still on it. Commented Aug 26, 2020 at 16:53
• @Theoretical Ok, so then I am not sure why your equation is involving kinetic friction coefficient, or why you have set the right side to $0$. I am assuming the equation in your comment here is applying Newton's second law to m2? Remember that Newton's second law is $F_\text{net}=ma$. Commented Aug 26, 2020 at 16:55
• Yes. The equation in my comment is applying the second law to m2. I think I have made a mistake. Shouldn't the acceleration of m2 be a instead of 0? Commented Aug 26, 2020 at 17:39