Newton's Second Law of Motion -- Three masses, an inclined plane and a pulley

[mustafamistik]

As you correctly stated in post #28, the forces that act on M are gravity and the tension. Those are the only forces that should appear in the equation. Yes, it may turn out that the tension is equal to $(m_1+m_2)*g*\sin(16)$, but take it one step at a time.
And think about the way each force acts on M.

T= $(m_1+m_2)*g*\sin(16)$
Gravity = M*g
Tension and Gravity are opposite forces. So we should consider the Gravity negative?

 

[PeroK]

T= $(m_1+m_2)*g*\sin(16)$
Gravity = M*g
Tension and Gravity are opposite forces. So we should consider the Gravity negative?

It doesn't matter which direction you take as positive as long as you are consistent. The important point is that tension and gravity are acting in opposite directions

 

[mustafamistik]

It doesn't matter which direction you take as positive as long as you are consistent. The important point is that tension and gravity are acting in opposite directions

I get it thanks sir. But i don't know how to calculate max M still.

 

[PeroK]

I get it thanks sir. But i don't know how to calculate max M still.

Let's take a step back. If there is no slipping between $m_1$ and $m_2$, then we have a simple pulley system with a mass $M$ pulling a mass $m_1 + m_2$ up a slope.

Can you calculate the acceleration of the system?

 

[mustafamistik]

Using $F=m*a$;
$M*g-(m_1+m_2)*sin(16)*g/(M+m_1+m_2)=a$

 

[PeroK]

Using $F=m*a$;
$M*g-(m_1+m_2)*sin(16)*g/(M+m_1+m_2)=a$

You might need to tidy that up a bit. And check you aren't missing some brackets.

What does that tell you about the acceleration of $m_1$?

 

[mustafamistik]

You might need to tidy that up a bit. And check you aren't missing some brackets.

What does that tell you about the acceleration of $m_1$?

You are right.
$(M*g-(m_1+m_2)*sin(16)*g)/(M+m_1+m_2)=a$
=> $(g*(M-(m_1+m_2)*sin(16)))$
a is acceleration of the all system.

 

[PeroK]

You are right.
$(M*g-(m_1+m_2)*sin(16)*g)/(M+m_1+m_2)=a$
=> $(g*(M-(m_1+m_2)*sin(16)))$
a is acceleration of the all system.

And there's an upper limit on $a$, correct? Depending on $\mu$?

 

[mustafamistik]

And there's an upper limit on $a$, correct? Depending on $\mu$?

This is the point where I stuck.

 

[PeroK]

This is the point where I stuck.

I don't know what you mean. You did that already:

This equation to find maximum acceleration. (m2*g*cos(16)* μ) -(m2*g*sin(16))=m2*a (F=ma)

Maybe it was so long ago!

 

[mustafamistik]

I don't know what you mean. You did that already:
Maybe it was so long ago!

Is that equation true ?

 

[PeroK]

Is that equation true ?

I think so. You just have to put everything together now.

 

[mustafamistik]

I think so. You just have to put everything together now.

Thank you and @haruspex for helping solve and understand this question. Also thanks for your interest.