Acceleration and tension of a block on an inclined plane

[Like Tony Stark]

Homework Statement

The coefficient of static and dynamic friction are 0.45 and 0.3 respectively. Find the maximun value that $m_2$ could have so that the system is in equilibrium. Then, if $m_2$ is the double of the mass calculated previously, determine what's the acceleration of the system.

Relevant Equations

Newton's equations

I considered the downwards direction and left direction as negative. For $m_1$, Newton's equations are:
$x) Fr + W_x - T=0$
$y) N - W_y =0$

For $m_2$:
$y) T - W =0$

Then, if I replace the data, I get $T=22.2 N$ and then $m_2=2.2 kg$.
With that, for the second question $m_2=4.4 kg$, and then solving Newton's equations I get for $m_1$
$x) 151.96 - T=20.a_x$
And for $m_2$ I get
$y) T -44=4,4a_y$.
Then $a_x=a_y$ and if I solve the system I get $T=63.92 N$ and $a=4,39 m/s^2$

But I have some doubts with the sign of the acceleration and tension. The acceleration must be negative, because $m_2$ will be moving downwards and $m_1$ will be moving to the left.

 

[haruspex]

For $m_2$:
$y) T - W =0$

Did you draw an FBD for that?

 

[Like Tony Stark]

Did you draw an FBD for that?

Yes, but I didn't upload a photo because I thought that the equations were enough

 

[haruspex]

Yes, but I didn't upload a photo because I thought that the equations were enough

Well, the equation is wrong. Take another look at the diagram. Can you describe it or upload it?

 

[Like Tony Stark]

Well, the equation is wrong. Take another look at the diagram. Can you describe it or upload it?

 

[haruspex]

It's a pulley with a rope running around under it. You only show one straight length of rope.

 

[Like Tony Stark]

It's a pulley with a rope running around under it. You only show one straight length of rope.

Oh, I forgot the other tension acting on $m_2$, didn't I? Well, apart from that my calculations are ok? Don't I have to change anything in the signs of the acceleration?

 

[haruspex]

Oh, I forgot the other tension acting on $m_2$, didn't I? Well, apart from that my calculations are ok? Don't I have to change anything in the signs of the acceleration?

Yes, you have a sign error.

Please develop the habit of working entirely algebraically. Refrain from plugging in numbers until the end. Where the question supplies only a number (like 30 degrees) replace it with a symbol, like θ.
This has many advantages, one being that it is much easier for others to follow and check your reasoning.

 

[Like Tony Stark]

Yes, you have a sign error.

Please develop the habit of working entirely algebraically. Refrain from plugging in numbers until the end. Where the question supplies only a number (like 30 degrees) replace it with a symbol, like θ.
This has many advantages, one being that it is much easier for others to follow and check your reasoning.

Ok. I'll take your advice! Thanks

 

[Delta2]

Well, the equation is wrong. Take another look at the diagram. Can you describe it or upload it?

Is the correct equation $2T-W_{m_2}=0$?

 

[haruspex]

Is the correct equation $2T-W_{m_2}=0$?

Yes.