Auto-Closing Gate (Relative Acceleration problem homework)

[Zer0]

1. Homework Statement
It's an auto closing gate. Two parts of the gate are identical. I had given to find the time taken to close the gate. The distance to close point (P) of the gate from closed event of the gate is given as "a". Sorry for my bad English.
What i need to know is how the two masses (which are m connected by a wire through two pulleys which are on the two gate parts move in relative to the gate parts. Masses of the gate parts are M and magnitude of the inclined plane of the gates are α. again sorry for my bad english. I have attached the image of the problem.

what i need to know is how the masses (m) move.
Then i can mange to work on the acceleration of the gate part and use constant acceleration equation which is s=ut+1/2(at)^2
Oh and no friction is involved and the wire is nonelastic.

This is my first thread on this forum. Sorry for my bad English. Thank You!

Homework Equations

The Attempt at a Solution

What i think is that when two gates are getting closer the masses (m) are going downwards on the inclined plane on the gates.

 

[haruspex]

By symmetry, the middle of the wire is fixed, so only need to analyse one side.
With this type of problem, it can be easier to work in the reference frame of one of the accelerating objects, the gate in this case. That will involve use of pseudo forces. Are you comfortable with that, or do you prefer to work always in the ground frame?

 

[Zer0]

Thank you so much for replying. But I am supposed to work on ground frame. When the gates are fetting closer two masses should move downwards on the plane right? Or since they are attached to one another are they suppose to move on a same direction (not exactly same direction, like when one goes down other should go up) ? When that also happenes the gates should close right? Beacuse the forces applied on the gate parts don't change with the direction masses go? Hope you understood that. Thank you for advice that i should woek on one gate :)

 

[Zer0]

I think you meant that middle poiny is fixed that they are moving downwards when gates getting closer am i right?

 

[haruspex]

When the gates are fetting closer two masses should move downwards on the plane right?

Yes. There is mirror symmetry, so the two masses will move in mirror symmetry.
The tricky part is the direction of acceleration of a mass.

 

[Zer0]

This is what I am currently working

on is this correct to mark acceleration dirrection like that? I mean it'll look like they are trying to move opposite directions right? Thank you so much sir!
Also forced involved to move a gate are one of the perpendixular forces from "m" mass and (T - Tcosa) right?

 

[Zer0]

This is what I am currently working on

http://i44.photobucket.com/albums/f13/madumalindrajith/Mobile%20Uploads/IMG_20170913_110938_zpssn5rvo8d_edit_1505281688474_zpssqzprpx6.jpg on is this correct to mark acceleration dirrection like that? I mean it'll look like they are trying to move opposite directions right? Thank you so much sir!
Also forced involved to move a gate are one of the perpendixular forces from "m" mass and (T - Tcosa) right?[/QUOTE]

 

[haruspex]

forced involved to move a gate are one of the perpendixular forces from "m" mass and (T - Tcosa) right?

No, it's trickier than that because of the mass's own acceleration. Let the normal force be N and derive the accelerations in terms of that.

 

[Zer0]

If i marked the acceleration like this and calculated acceleration in relative to Earth would that be correct?
http://i44.photobucket.com/albums/f13/madumalindrajith/Mobile%20Uploads/IMG_20170913_110938_zpssn5rvo8d_edit_1505281688474_zpssqzprpx6.jpg

 

[haruspex]

If i marked the acceleration like this and calculated acceleration in relative to Earth would that be correct?

The link to your image does not work for me.

 

[Zer0]

Im sorry for that. :(

 

[Zer0]

Check the link

 

[haruspex]

Check the link

It just says "please update your account to enable third party hosting". Same in post #8.

 

[Zer0]

I'll check on that and upload my photo. Thanks :) sorry for your time.

 

[Zer0]

If i marked the acceleration like this and calculated acceleration in relative to Earth would that be correct?

 

[haruspex]

If i marked the acceleration like this and calculated acceleration in relative to Earth would that be correct?
View attachment 211026

Could be... it's a bit hard to tell from that. Please post some equations, not as an image. And please define your variables.

 

[Zer0]

Okay Thanks!
I have chosen the left side gate. I have calculated the acceleration of mass m like this (e-earth)
I took,
F as the acceleration of left side mass M in relative to Earth and, {a(M,e) = F→}
f as the acceleration of left side mass m in relative to mass M {a(m,M) = f in parallel to inclined plane}

\begin{equation}
a(m,e) = a(m,M) + a(M,e) \\
= f \swarrow α + F\rightarrow \\
\end{equation}

by using F=ma
\begin{equation}
To "M" f=ma→; \quad T-Tcosα+S.Sinα=MF \quad\quad\quad\quad -1 \\
To "m" f=ma \swarrow α ; \quad mgsinα-T=m(f-Fcosα) \quad\quad\quad\quad -2\\
To "whole" f=ma → \quad ; T=MF+m(F-fcosα) \quad\quad\quad\quad -3\\
To "m" f=ma \nwarrow (perpendicular\quad to\quad plane) \quad ; \quad S-mgcos∝=m(-Fsin∝) \quad\quad\quad\quad -4\\
\end{equation}

Please check these are correct with the diagram i sent. If those are correct then i got 4 equations to find "F→" which is the acceleration of Left Gate part in relative to Earth. Also I'm not sure with equation #4 :/
Right?Information given :
two equal masses "m"
Masses of two equal gates "M"
Magnitude of inclined plane of the gates "α"
Distance "a" from opened point of a gate to it's closed point.

All my teacher said was that this is an auto-closing gate. We open the gate then it closes automatically. Then he gave us the diagram with masses and without marking forces applied on each objects or how the masses (which are m) move. Then he gave us that distance from fully opened point of a gate to its closing point is "a". And he told us to find the time taken to close the gate itself.

 

[haruspex]

Okay Thanks!
I have chosen the left side gate. I have calculated the acceleration of mass m like this (e-earth)
I took,
F as the acceleration of left side mass M in relative to Earth and, {a(M,e) = F→}
f as the acceleration of left side mass m in relative to mass M {a(m,M) = f in parallel to inclined plane}

\begin{equation}
a(m,e) = a(m,M) + a(M,e) \\
= f \swarrow α + F\rightarrow \\
\end{equation}

by using F=ma
\begin{equation}
To "M" f=ma→; \quad T-Tcosα+S.Sinα=MF \quad\quad\quad\quad -1 \\
To "m" f=ma \swarrow α ; \quad mgsinα-T=m(f-Fcosα) \quad\quad\quad\quad -2\\
To "whole" f=ma → \quad ; T=MF+m(F-fcosα) \quad\quad\quad\quad -3\\
To "m" f=ma \nwarrow (perpendicular\quad to\quad plane) \quad ; \quad S-mgcos∝=m(-Fsin∝) \quad\quad\quad\quad -4\\
\end{equation}

Please check these are correct with the diagram i sent. If those are correct then i got 4 equations to find "F→" which is the acceleration of Left Gate part in relative to Earth. Also I'm not sure with equation #4 :/
Right?Information given :
two equal masses "m"
Masses of two equal gates "M"
Magnitude of inclined plane of the gates "α"
Distance "a" from opened point of a gate to it's closed point.

All my teacher said was that this is an auto-closing gate. We open the gate then it closes automatically. Then he gave us the diagram with masses and without marking forces applied on each objects or how the masses (which are m) move. Then he gave us that distance from fully opened point of a gate to its closing point is "a". And he told us to find the time taken to close the gate itself.

Your four equations look right to me.

 

[Zer0]

Ok Thanks!
Then i have to use \begin{equation}
s=ut+(1/2)at^2
\end{equation}

and it's initial velocity is 0 and calculate time taken by the gate to it's closing gate which is S=a and get positive value by the square root because time cannot get minus value right?

 

[Zer0]

Your four equations look right to me.

Ok Thanks!
Then i have to use \begin{equation}
s=ut+(1/2)at^2
\end{equation}

and it's initial velocity is 0 and calculate time taken by the gate to it's closing gate which is S=a and get positive value by the square root because time cannot get minus value right?

 

[haruspex]

Ok Thanks!
Then i have to use \begin{equation}
s=ut+(1/2)at^2
\end{equation}

and it's initial velocity is 0 and calculate time taken by the gate to it's closing gate which is S=a and get positive value by the square root because time cannot get minus value right?

Yes.

 

[Zer0]

Is there any chance that they would go one after another i meant they are attached by the wire so one goes down in the inclined plane so other one goes up and still gate will work fine?

 

[haruspex]

Is there any chance that they would go one after another i meant they are attached by the wire so one goes down in the inclined plane so other one goes up and still gate will work fine?

Consider the tension in the wire. What can you deduce?

 

[Zer0]

Consider the tension in the wire. What can you deduce?

All of the forced applied on the masses are equal including the tension and they should not move one after another right?

 

[haruspex]

All of the forced applied on the masses are equal including the tension and they should not move one after another right?

Yes, they are subject to exactly the same forces so should accelerate in the same way.

 

[Zer0]

Thank you so much! :)

 

[Zer0]

The masses has an acceleration in relative to Earth right? From The middle point of the wire to the pulley and from pulley to mass is a constant and if i take the second derivative of it then i have an acceleration of the mass in relative to Earth right? That means if the acceleration of the gate is |a| so the mass's is also |a| (i don't know how to tell thr quantity of acceleration in english) am i correct?

 

[Zer0]

I think previously i was wrong

 

[Zer0]

If the displacement of the fixed point if the wire to the pulley is x and from pulley to the mass is y and

(X+Y)=constant
X(double dot) = -Y(double dot)

Right?

I don't know how to type diuble dot

 

[haruspex]

That means if the acceleration of the gate is |a| so the mass's is also |a|

No. Consider e.g. if all were in a straight line. The gate could move but the mass stay put.

(X+Y)=constant

You defined X and Y as displacements, which makes them vectors. |X|+|Y| is constant.

how to type diuble dot

You can do it with LaTeX. Type a double # to start LaTeX, \ddot X for $\ddot X$, then another double # to terminate.

 

[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