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I am trying to understand this problem intuitively. The wedge accelerates to the left with $a$. According to the solution, the rope on the left must accelerate down the slope with $a\cos\alpha$? And the rope on the right must accelerate up the slope with $a\cos\beta$? Then the ropes will not get displaced relative to the wedge? Is this what the solution says?

Also, if we observe the system from ground’s frame of reference, then of course the wedge is accelerating to the left with $a$. What about the rope on the left? Does it have two accelerations from ground’s reference frame, i.e $a\cos\alpha$ down the wedge and $a$ (because of the wedge)? Same goes for the rope on the right?

EDIT : It looks like (as @Bob D has explained) acceleration of the rope when observed from Earth’s reference frame would be same as the acceleration of the wedge (to the left along the horizontal plane). Because as the wedge is accelerating to the left, the rope is not getting displaced relative to it.

Since $a_{r/w}$ = $0$

$=>$ $a_r$ $-$ $a_w$ $=$ $0$

$=>$ $a_r$ $=$ $a_w$

My question is, what ‘net force’ is making the rope accelerate to the left (along the horizontal) when observed from Earth’s reference frame? If I analyse the FBD of the section of rope on the right, I can say that there are two forces, tension and $\frac{m}{2}g\sin\alpha$ acting on it that are parallel to the surface of the wedge. And there’s the normal force $N_1$ and $\frac{m}{2}g\sin\alpha$ acting on the rope that are perpendicular to the surface of the wedge it is on. I can’t seem to figure out how these forces add up to accelerate the rope to the left along the horizontal (when observed from Earth’s frame)?

Any help is much appreciated.

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    $\begingroup$ Note that $sin(x)$ is the product of $s$ times the imaginary unit times some function of $x$. If you intended on the sine of $x$, then please use \sin in your MathJax $\endgroup$ – Kyle Kanos Dec 9 '19 at 13:08
  • $\begingroup$ @Kyle Kanos Just did. Thanks for pointing out $\endgroup$ – π times e Dec 9 '19 at 13:39
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Intuitively it might help if you think about it this way. If the angle to the left is greater than the the angle to right and the wedge is stationary the rope will slide down on the left side. When the wedge accelerates to the left if the component of that acceleration down the incline on the left exactly equals the downward acceleration of the rope that would have occurred with the wedge stationary, then there will be no RELATIVE motion between the rope and the wedge.

Keep in mind that the wedge doesn’t exert any force on the rope parallel to the surface of the wedge. That would require friction between the rope and the wedge.

Hope this helps.

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  • $\begingroup$ Thanks a lot. "If the component of that acceleration down the incline on the left exactly equals the downward acceleration of the rope that would have occurred with the wedge stationary, then there will be no RELATIVE motion between the rope and the wedge." It's another way of saying that the speed of separation, or speed of approach between the two is zero? What I am confused about is, I am struggling to visualise it from ground's reference frame. What would the motion of the rope on the left look like to an observer in Earth's frame? Would the block have two accelerations in Earth's frame? $\endgroup$ – π times e Dec 8 '19 at 15:52
  • $\begingroup$ Although mathematically it is obvious that if there is no relative motion between the rope and the block, the rope would undergo zero displacement relative to the wedge. But it's so counter intuitive to think that even though the rope is sliding, the wedge is accelerating such that the rope does not get displaced relative to it $\endgroup$ – π times e Dec 8 '19 at 15:55
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    $\begingroup$ @πtimese Yes it means the ropes acceleration relative to the wedge is zero, be it in the reference frame of the wedge or ground. I admit it is counter-intuitive. But consider another example. Suppose you release a marble on the front windshield of a stationary car. It will roll down the window. Now suppose the marble were released while the car accelerates. Would the marble roll down with the same acceleration, even with no air drag? What if you really hit the gas. Is it not possible that the acceleration could be great enough that the marble would roll up the windshield? $\endgroup$ – Bob D Dec 8 '19 at 18:18
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    $\begingroup$ @πtimese Actually rain drops are probably not good examples. You probably notice that rain drops, if small enough, stay on the vertical surface of the windshield. This is not due to friction but adhesion- an electrostatic force due to the polar nature of both water and glass molecules. See this link for explanation: quora.com/Why-is-glass-polar The adhesion force is greater than the downward force of gravity until the size (mass) of the water drop is large enough for gravity to win over. Also air drag readily breaks up water droplets. $\endgroup$ – Bob D Dec 9 '19 at 13:07
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    $\begingroup$ @πtimese The electrostatic force is the result of the silicon dioxide glass molecules and the water molecules being polar. You will learn that. Regarding the horizontal acceleration of the rope, although the wedge does not exert a force on the rope parallel to its surface it does exert a force normal (perpendicular) to the rope. That component of the wedge force causes the rope to accelerate horizontally along with the wedge. Hope that helps as the site is now complaining about extended discussions in comments. If you want to continue, click on the link to move our discussion to chat. $\endgroup$ – Bob D Dec 9 '19 at 13:53

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