Is the Calculation of Spring Deformation in an Accelerating Room Correct?

[Like Tony Stark]

Homework Statement

Two bodies of equal mass are connected by an ideal rope and are inside a room which can accelerate vertically. The second body is attached to a spring of constant $200 \frac{N}{m}$. Find the deformation of the spring when
A) the room moves with constant velocity
B) the room is accelerating upwards with $2 \frac{m}{s^2}$.
C) the room is accelerating downwards with $2 \frac{m}{s^2}$.

Relevant Equations

Newton's equations

I've solved all the cases in the non inertial system.

A) For $m_1$ we have
$x) P_{1x} -T=m.a_x$
$y) N_1 -P_{1y}=m.a_y$

For $m_2$ we have
$y) T+F_e -P_2=m.a_y$

As it moves with constant velocity I solve it setting $a_x=0$. So for $m_1$ $mgsin(\alpha)=T$, then I replace it in the tension of $m_2$ and find $\Delta x$

B) For $m_1$
$x) mgsin(\alpha)-f*.sin(\alpha)-T=0$

For $m_2$
$y) Fe + T -mg-f*=0$

Where $f*=2m$

Then, I replaced the tension from $m_1$ in $m_2$ and found the deformation. (As you see, I considered the elastic force pointing upwards in this case)

C) the same for $B$ just changing the sign of $f*$ and the elastic force.Is this right? Because I had some doubts about the sign of the elastic force.

 

[tnich]

I don't understand your equations. Assume that $T$ is the tension on the rope, that $x$ is the horizontal axis, and $y$ is the vertical axis. Why does $T$ act in the $x$ direction on mass 1? Could you post your free body diagrams, please?

 

[Like Tony Stark]

I don't understand your equations. Assume that $T$ is the tension on the rope, that $x$ is the horizontal axis, and $y$ is the vertical axis. Why does $T$ act in the $x$ direction on mass 1? Could you post your free body diagrams, please?

For $m_1$ I considered inclined axis and for $m_2$ I considered the typical vertical and horizontal axis.
I've attached the F.B.D.

($P$ is weight)

 

[kuruman]

For $m_1$ I considered inclined axis and for $m_2$ I considered the typical vertical and horizontal axis.
I've attached the F.B.D.

($P$ is weight)

Why inclined? The room accelerates along the vertical in one-dimensional motion.

 

[Like Tony Stark]

Why inclined? The room accelerates along the vertical in one-dimensional motion.

Yes, but as I suppose it doesn't move, the acceleration (with respect to the room) is 0, so it doesn't matter which axis I take.

 

[kuruman]

Yes, but as I suppose it doesn't move, the acceleration (with respect to the room) is 0, so it doesn't matter which axis I take.

But the fictitious force in the non-inertial frame is along the same axis as the real force of gravity. There is no angle involved, the problem is one-dimensional.

 

[Like Tony Stark]

But the fictitious force in the non-inertial frame is along the same axis as the real force of gravity. There is no angle involved, the problem is one-dimensional.

I've corrected it in the first post. Is it ok now?

 

[SammyS]

Homework Statement: Two bodies of equal mass are connected by an ideal rope and are inside a room which can accelerate vertically. The second body is attached to a spring of constant $200 \frac{N}{m}$. Find the deformation of the spring when
A) the room moves with constant velocity

Homework Equations: Newton's equations

A) For $m_1$ we have
$x) P_{1x} -T=m.a_x$
$y) N_1 -P_{1y}=m.a_y$

For $m_2$ we have
$y) T+F_e -P_2=m.a_y$

As it moves with constant velocity I solve it setting $a_x=0$. So for $m_1$ $mg\sin(\alpha)=T$, then I replace it in the tension of $m_2$ and find $\Delta x$
...

For $m_1$ I considered inclined axis and for $m_2$ I considered the typical vertical and horizontal axis.
I've attached the F.B.D. ( $P$ is weight )

You need more clarity in your choice of variable. You need to distinguish those variables used for Body #1 (on the incline) as opposed to those for Body #2 (the body hanging from the rope while being supported from below by the spring).
As both have the same mass, it is fine to use $m$ for the mass.
However, the acceleration of Body #1 is not the necessarily the same as the acceleration of Body #2. Of course I suppose that the Bodies are at rest in the reference frame attached to the room, although that is not stated. Therefore, the accelerations with respect to the room are both zero.
At any rate, it is recommended that you be clearer as regards defining you variables.

As for your solution, you state:

As it moves with constant velocity I solve it setting $a_x=0$. So for $m_1$ $mg\sin(\alpha)=T$, then I replace it in the tension of $m_2$ and find $\Delta x$

What are the "it"s you refer to?

What have you done with $a_y$ ?

Where does $\Delta x$ appear in any of your expressions?

 

[Like Tony Stark]

You need more clarity in your choice of variable. You need to distinguish those variables used for Body #1 (on the incline) as opposed to those for Body #2 (the body hanging from the rope while being supported from below by the spring).
As both have the same mass, it is fine to use $m$ for the mass.
However, the acceleration of Body #1 is not the necessarily the same as the acceleration of Body #2. Of course I suppose that the Bodies are at rest in the reference frame attached to the room, although that is not stated. Therefore, the accelerations with respect to the room are both zero.
At any rate, it is recommended that you be clearer as regards defining you variables.

As for your solution, you state:

What are the "it"s you refer to?

What have you done with $a_y$ ?

Where does $\Delta x$ appear in any of your expressions?

Sorry
Using "It" in that sentence I refer to the room
I didn't write $a_y$ since they are stationary with respect to the room, so I wrote 0
$\Delta x$ is "inside" the elastic force ($Fe=-k \Delta x$)I didn't write all the steps because I thought that the ones that I'd written were enough to understand my ideas. Excuse me

 

[tnich]

As @karuman says, the acceleration is in the same direction as gravity. So in the accelerating frame you can treat it as an increase (if positive) or decrease (if negative) of the acceleration of gravity. That means you can solve the equations once with the pseudo-gravity equal to $g+a$ and then plug in the different values of $a$ for the three cases.

 

[kuruman]

As @karuman says, the acceleration is in the same direction as gravity. So in the accelerating frame you can treat it as an increase (if positive) or decrease (if negative) of the acceleration of gravity. That means you can solve the equations once with the pseudo-gravity equal to $g+a$ and then plug in the different values of $a$ for the three cases.

Amen to that. BTW, it's ku[/color]ruman.

 

[tnich]

You would think I would know that by now.