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A particle with mass $4.5kg$ lies on a rough plane inclined at $30°$ to the horizontal. A light, inextensible string connects to $P$, runs parallel with the line of greatest slope of the plane to a smooth peg, then vertically downwards through a smooth, free ring $R$, with mass $2 kg$, and then vertically upwards to a fixed point $S$.

Model

The coefficient of friction between $P$ and the plane is $0.15$.

Let $a$ be the acceleration of the ring when the system is released from rest. By considering the distance moved by each object, explain why the acceleration of $P$ is $2a$.

I've been stuck on this for ages. How do I "consider the distance"? I know that $2g - 2T = 2a$, and $R = 4.5gcos(30)$, but that's about it. There is a 'worked solution', but it doesn't make sense to me:

Resolving at $R$:

$R(↓): 2g - 2T = 2a_{R}$

For the ring, vertical acceleration is given by $a = g - T$

Hence $a + T = g$

I don't understand where they considered the "distance travelled", or how they explained why the acceleration of $P$ is $2a$.

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closed as off-topic by John Rennie, stafusa, Kyle Kanos, Jon Custer, Emilio Pisanty Aug 21 '18 at 13:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

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    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Aug 19 '18 at 15:29
  • $\begingroup$ For "distance travelled", you are being asked to consider how far up the ring gets pulled if P moves down the slope a given distance like say 1 cm. From this, you should be able to figure out how the accelerations of P and R are related to each other. To get you started thinking in the right way, you might consider how the velocities of P and R are related to each other. $\endgroup$ – mmesser314 Aug 19 '18 at 15:58
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I don't understand where they considered the "distance travelled", or how they explained why the acceleration of $P$ is $2a$.

Imagine that the particle $P$ is moved $2$ metres up the slope.
You now have $2$ metres of "slack" string.
Move the ring down one metre and that slack is taken up adding one metre of string to each side of the ring.

So you now know that the distance travelled by particle $P$ is twice the distance travelled by the ring.

Differentiate the expression $2 \,x_{\rm ring} = x_{\rm particle}$ twice with respect to time to get the relationship between the accelerations.

$2 \,x_{\rm ring} = x_{\rm particle} \Rightarrow 2 \,\dot x_{\rm ring} = \dot x_{\rm particle} \Rightarrow 2 \,\ddot x_{\rm ring} = \ddot x_{\rm particle} \Rightarrow 2\,a_{\rm ring} = a_{\rm particle} $

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