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.