# Motion on a rough slope - Textbook question [closed]

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 $S$.

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 $2a$.

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

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• 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. – John Rennie Aug 19 '18 at 15:29
• 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. – mmesser314 Aug 19 '18 at 15:58

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