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I've talked with 2 teachers about this situation:

mechanic system

one teacher said he was completely sure that B have twice the acceleration of A, the other said he was completely sure they have same acceleration. Can you have a better look on it? What do think? Consider it has no friction.

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4 Answers 4

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Let the initial length of the bottom segment of rope be $l_1$, the initial length of the middle segment be $l_2$, and the initial length of the top segment be $l_3$.

Since the total length of the rope is constant, we can write

$$l_1 + l_2 + l_3 = K$$

Now, displace the block $A$ by $\Delta x_A$ down the slope and then

$$(l_1 + \Delta x_A) + (l_2 + \Delta x_A) + (l_3 + \Delta l_3) = K$$

Thus

$$\Delta l_3 = -2 \Delta x_A$$

So, the length of the segment $l_3$ decreases twice as much as the displacement of block $A$.

However, this isn't the entire story. The top-most pulley moves down the slope with block $A$ and so the displacement of block $B$ is

$$\Delta x_B = \Delta l_3 + \Delta x_A = -\Delta x_A$$

And so we conclude that the blocks have accelerations of equal magnitude and opposite direction.

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The two blocks have the same acceleration:

enter image description here

The position of block B is determined by $l_1$ and $l_3$, that of A is determined by $l_3$ or $l_4$ (as they vary by the same amount).

Let $l_1$ increase by a distance $d$, then $(l_3+l_4)$ will decrease by the same $d$ (in case the string is inextensible and $l_2$ remains constant), and they will share this decrease equally, each one will decrease a $\dfrac{d}{2}$. Therefore, the net motion of B will be $d-\dfrac{d}{2}=\dfrac{d}{2}$ downwards, while that of A will be $\dfrac{d}{2}$ upwards. You can derive w.r.t time to get the accelerations.

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enter image description here

$$l=x_A+(x_A-k_1)+k_2+k_3+2\pi R$$ $$k_3=x_B-2R-(x_A-k_1)$$ $$l=x_A+x_B+constant$$ so, we have: $$0=v_A+v_B$$ and$$a_A=-a_B$$

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The accelerations are the same. They can't be different, since otherwise $B$ would move faster than $A$ and the rope wouldn't be tight anymore.

The reason your teacher thought that they are different is a common mistake: under the assumption $F_{AB} = -F_{BA}$ and $F=ma$ (force on A should be force from B and vice versa) you might think that same $F$ and different $m$ should result in different $a$, e.g. $a_{A}=\frac{F_{AB}}{m_{A}}=\frac{m_{A}-m_{B}}{m_{A}}g^{*}$ and $a_{B}=\frac{F_{BA}}{m_{B}}=-\frac{m_{A}-m_{B}}{m_{B}}g^{*}$ (where $g^{*} = g * sin(25°)$ in your example).

The reason that's wrong is the following: the resulting force, that is what's left if you substract the gravitational forces on both masses, $F=({m_{A}-m_{B}}) g^{*}$, has to accelerate BOTH masses. Image how the blocks would move without gravity, they are just two masses bound together behind each other that move in the same direction (= "the same direction as the rope"). Gravitational forces on $B$ are already "taken care of" by some part of the gravitional force on $A$ via the pulley. So if you (or the earth) pull on $A$ you have to move both $A$ and $B$.

So you get $a = \frac{F}{m_{A}+m_{B}} = \frac{{m_{A}-m_{B}}}{m_{A}+m_{B}}g^{*}$ (for both blocks).

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  • $\begingroup$ Hmm sorry, but I'm not sure if I followed you :/ tell me the accelerations of these blocks, the relation between them: [url=http://postimg.org/image/6we70mosn/][img]http://s31.postimg.org/6we70mosn/image.jpg[/img][/url] $\endgroup$ Apr 22, 2016 at 21:53
  • $\begingroup$ That's a different setup. The idea is the same: while $B$ moves $s$, $A$ moves $\frac{s}{2}$ (because of the movable pulley). So you get as the relation for the acceleration: $a_{A} = \frac{a_{B}}{2}$, since $s = \frac{1}{2}a t^2$. $\endgroup$
    – Solarflare
    Apr 23, 2016 at 1:30

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