Why it's not apt to use $F=ma$ over here directly

Question

According to the Newton's second law $$ F_{net}=ma$$ (I am considering the scalar notation since my question only confines to motion in a single line.)

A question asked me about the acceleration of the wedge of mass $M$ when the mass of $m$ is moving over it. All the surfaces are perfectly smooth and as shown in the figure a force $F$ is being applied on the mass $M$.Initially, I thought that I can write the acceleration of the system just by writing $$a= \frac{F}{M+m}$$ but after making the whole free body diagrams and considering the dynamical equations of motion, the answer comes out to be different. Can anybody please help me in understanding, why we can't apply $F_{net}=(Mass)_{net} \cdot (acceleration)$ over here.

Answer 1

You are very smart! Yes, this is a way to do the problem. However the question that you have to ask is, "the acceleration of what?" What is accelerating here?

One step back: what are our real equations?

Well, we actually have two equations, here I will say that object "1" is the bigger wedge $M$ and object "2" is the smaller wedge $m$ and I will locate them at vector positions $\vec r_1 = [x_1(t), y_1(t)]$ and $\vec r_2 = [x_2(t), y_2(t)].$

There is some complicated constraint force $\vec U_{21} = -\vec U_{12}$ (by Newton's third law!) that keeps the wedge $m$ from passing through the wedge $M$. The rest of the puzzle is these two force equations, the first of which is$$m_1 ~\vec a_1 = \vec F + \vec U_{21} + m_1\vec g + \vec F_N,$$where $\vec F_N = [0, F_N]$ is the normal force that keeps $M$ on the ground and $\vec g = [0, -9.81\text m/\text s^2]$ is the gravitational acceleration. Basically $F_N$ will be whatever it has to be to constrain the y-component $(\vec a_1)_y = 0.$ So actually when we do out the problem the y-component equation is just trivial and we have $$ m_1 (\vec a_1)_x = |\vec F| - |\vec U_{21}|~\sin\theta,$$where $\theta$ is the angle of inclination of the wedge $M$.

What about the other wedge? It has the equation: $$m_2 ~\vec a_2 = m_2 \vec g + \vec U_{12},$$ and we might actually really care how it moves up/down in space as well as left/right. So you don't get any such simplification. However, and this will be important in a second: since it has to be on this line $y_2 = (x_2 - x_1 - L) ~\tan \theta,$ for some fixed $L$ and $\theta$, you can calculate $y_2$ from $x_1$ and $x_2.$

However remember now Newton's third law, and add these two equations together!$$m_1 ~\vec a_1 + m_2 ~\vec a_2 = \vec F + (m_1 + m_2) \vec g + \vec F_N.$$The complicated constraint force is gone! (I mean, it's not gone gone, it is being balanced on $M$ by $\vec F_N$ so it still has an effect here: but it's an indirect effect, not an explicit effect. The explicit terms are gone.)

Now the question

So now you want to define that there is some acceleration $$(m_1 + m_2) ~\vec a = m_1~\vec a_1 + m_2~\vec a_2$$ and my question is, what is this the acceleration of? If you are writing $\vec a$ then that is $d^2\vec r/dt^2$ and what is the $\vec r$?

And I will answer this question for you: the operator $d/dt$ is what we call a "linear" operator meaning that by standard calculus rules you can derive, $$ \vec r = \frac{m_1 ~\vec r_1 + m_2 ~\vec r_2}{m_1 + m_2}.$$ In other words it is the acceleration of the center of mass which obeys this equation.

If we similarly to the case for wedge $M$ ignore the $y$-motion of the center of mass, we get simply:$$(m_1 + m_2)~(\vec a)_x = |\vec F|,$$ which is precisely what you intuited. The $x$-component of the center of mass behaves in exactly this way.

Now you just need to figure out either the motion of $x_1,$ or the motion of $x_2$, or the motion of $y_2$, to figure out the rest of the problem.

Answer 2

Because, like mentioned in the problem, $m$ is moving over $M$. If they both started at rest relative to each other, but start moving relative to each other you know that $m$ and $M$ don't have the same acceleration.

In terms of physics, this means that there is either no friction between or that the coefficient of friction isn't large enough to stop them from moving relative to each other.

You need to write a force equation of the total force $m$ feels, and the total force equation for $M$. They aren't the same.

Answer 3

The formula $F = m_{tot} a$ is only valid if the system $\textit{as a whole}$ is a solid. It is not the case if $m$ is moving with respect to $M$ as the distance between an arbitrary point of $m$ and another point of $M$ is not constant in time.