# Block, pulley and an external force moving the whole system problem (classical physics)

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 $$m_1$$.

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?

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 $$2$$ :

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

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 $$M$$.

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

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