Why is the $y$-component of acceleration twice the $x$-component?

Question

We were given this question. the answer said that when $m$ released, the $y$ component of acceleration of $m$ should be $2$ times the $x$ component of acceleration of $m$. I can get that the $x$ components of acceleration of $M$ and $m$ are equal but I can’t understand why $y$ and $x$ components aren’t equal. if $m$ moves $x$ meters $M$ has to move $x$ meters as well at the same time. and thus shouldn’t the $x$ and $y$ components of acceleration be equal?

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AccidentalTaylorExpansion · Answer

The pulleys make it trickey. You can see how this works by looking at the lengths of the individual rope pieces. When the cart moves 1 meter to the right the bottom part gets 1 meter shorter. The top part also gets 1 meter shorter. Since the rope can't change length the left part (that is attached to $m$) gets longer by 2 meters. So for every meter that the cart moves to the right the mass drops by 2 meter.

Answered by AccidentalTaylorExpansion on January 3, 2021

Young Kindaichi · Answer

Consider the following figure

The length of string is constant so
$$l_1+l_2+l_3+l_4=l$$
So that time derivative : $$l''_1+l''_2+l''_3+l''_4=0$$
but $l''_3=0$ as string between $BC$ can not increase.
$$l''_1+l''_2+l''_4=0$$
The acceleration of $M$ and $m $ respectively
$$a_M=a_{Mx}$$
and
$$a_m=a_{mx}hat{i}+a_{my}hat{j}$$
Now from figure it's clear that : $a_{Mx}=l''_4=l''_2=a_{mx}$ and $a_{my}=l''_1$. Putting everything we get
$$2a_{Mx}=a_{my}$$
QED :)

Why is the $y$-component of acceleration twice the $x$-component?

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