Auto-Closing Gate (Relative Acceleration problem homework)

[Zer0]

So i can't use the displacement to find the acceleration?

 

[haruspex]

So i can't use the displacement to find the acceleration?

You can use distances, not displacements.

 

[Zer0]

And the quantity of the accelerations are same? Right? I mean if the acceleration if mass m is 'f' and the mass M is 'a' then |f|=|a| right?

 

[Zer0]

Is there a name for that concept? Using distances to find acceleration in relative acceleration?

 

[haruspex]

Is there a name for that concept? Using distances to find acceleration in relative acceleration?

Kinematics, as distinct from kinetics. https://en.m.wikipedia.org/wiki/Kinematics

 

[Zer0]

The quantities of acceleration of two masses are same right?

 

[haruspex]

The quantities of acceleration of two masses are same right?

The distance from the central point to the pulley plus the distance from the pulley to the sliding mass is constant. Thus the second derivatives of those distances are equal and opposite. The first of those two is indeed the scalar acceleration of the gate, but the second is not the scalar acceleration of the sliding mass. Rather, it is the acceleration of the mass relative to the gate.

 

[Zer0]

So if i take " x +(x+y) " woild that be a constant?

 

[haruspex]

So if i take " x +(x+y) " woild that be a constant?

No, you were right to say that |x|+|y| is constant, but $\ddot {|y|}$ is not the magnitude of the acceleration of the sliding mass in the ground frame. It is the magnitude of its acceleration relative to the gate.

 

[Zer0]

Yeah i just relaized that is there a way to get displacement from middlepoint to the mass and get x (x as mention before) both a constant?

 

[Zer0]

So'll have the mass's acceleration in relative to ground

 

[haruspex]

So'll have the mass's acceleration in relative to ground

Just add the accelerations vectorially.
a_{mass relative to ground}=a_{gate relative to ground}+a_{mass relative to gate}

 

[Zer0]

Thank you so much. I just managed to understand relative acceleration much more than i knew earlier. Thank you so much. I've found the same problem which has given the answer and i got it. Thank you.