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I'm not sure in which blocks is $\vec{F}$ acting on, I know it is clearly acting on the block of mass $M$ and clearly acting on the block of mass $m_2$. I'm not certain about the effects of $\vec{F}$ on the block of mass $m_1$.

Two blocks, on hanging and the the other laying, on top of a third one.

This is a problem with no friction which consists on finding $F$ with respect to the masses such that the mass $m_2$ has no acceleration on it's $y$ axis using $F=ma$.

At first I though $\vec{F}$ was not affecting $m_1$ since there is no friction. So $m_1$ is basically "free-falling" to the right.

Then I saw the text book answer which was $F=(M+m_1+m_2)\cdot g\frac{m_2}{m_1}$ (my answer was $F=(M+m_2)\cdot g\frac{m_2}{m_1}$) which suggest that the author believes $F$ acts on $m_1$. We clearly got the same value for acceleration even though I didn't took into account the effects of $F$ on $m_1$.

Is this a mistake by the author or is it something that I'm not getting? If $F$ acts on $m_1$, how so?

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  1. Taking $M$ and $m_{2}$ as system, then if a force is applied surely both of this masses would accelerate with same acceleration

  2. Now the question states there shouldn't be any acceleration of $m_{2}$ in $y-$ axis then the string of pulley should not accelerate as well , which means there shouldn't be any acceleration of $m_{1}$ with respect to system of $M$ and $m_{2}$

Reason :

If mass $m_{1}$ doesn't accelerate with same magnitude and in same direction as that of $M$ and $m_{2}$ then there would be a relative acceleration of $m_{1}$ with respect to $M$ which would cause string to move upwards for $m_{2}$ hence accelerating it up which we don't want in question

So force $F$ will affect all masses as all of them would accelerate with same acceleration vector

So $$a = \cfrac{F}{M +m_{2} +m_{1}}\tag 1$$

Now looking from a inertial frame of reference, we could see there should only be tension of string (acting in right direction for $m_1$) that would cause acceleration of $m_1$ in horizontal direction as there is no friction between all masses so

For $m_1$:

$$T = m_1 a \tag 2$$

Where $a$ is given in equation $1$

Now $T$ for $m_2$ is $m_2g$ upwards since it would not accelerate in $y$ direction

So putting $T = m_2 g$ in equation $2$ :

$$F = \cfrac{(M +m_2 +m_1)gm_2}{m_1}$$

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Look at the forces:

•) $m_2$ has no acceleration along the $y-$direction, so $\vec F_{net}$ along the $y-$direction is $0$. Therefore, the force of tension in the string is: $$T = g m_2$$

•) Since $m_2$ is not accelerating along $y-$direction, the string is not rising or falling with acceleration; so the horizontal length of the string is not changing with acceleration. Therefore, $m_1$ shares the same acceleration $a$ (in the $x-$direction) as the pulley, which is firmly attached to $M$, so the string tension is :

$$a\ m_1 = T$$ Hence, $a \ m_1 = T = g \ m_2$ : $$a = g\frac {\ m_2}{m_1}$$

But

•) Remembering that ALL the masses have that same acceleration $a$, and that the only external force along the $x-$direction is $F$ : $$F = a\ (M + m_1 + m_2)$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left (g \frac {\ m_2}{m_1}\right )(M + m_1 + m_2)$$

The author is correct. What you missed is that the string is connecting $m_1$ to $M$, just as the wheels are connecting $m_2$ to $M$.

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It seems to me that $F$ will apply a force to the string through the pulley and thus could act on $m_1$.

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  • $\begingroup$ Do you know what the magnitude of the force acting through the string due to $\vec{F}$ could be? $\endgroup$ – Random User Jun 4 at 20:11

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