# Newton’s second law - a uniform massive rope on a wedge

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.

• 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 – Kyle Kanos Dec 9 '19 at 13:08
• @Kyle Kanos Just did. Thanks for pointing out – π times e Dec 9 '19 at 13:39