Newtons second law perspective problems

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pinsky
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I'm having some problems in observing the total force which.
The original situation is numbered "1."

There is a resultant force [tex]F_r = m_2 \cdot g - k \cdot m_1 \cdot g[/tex]

and because of that force m1 has the acceleration a.

What I can't seem to figure out is is that resultant force equal.

[tex]F_r = m_1 \cdot a[/tex]

or

[tex]F_r = (m_1 + m_2) \cdot a[/tex]


In defense of the first argument, I've drawn the equivalent picture (2.) in which the second mass is replaced just by the force it exerts to m1.

In that case, it is clear that the first equation for Fr is the correct one.


But, I've also drawn a second equivalence for which the second equation would make more sense.

I know that one of them is wrong, i just can't seem to figure out which :)
 

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pinsky said:
I'm having some problems in observing the total force which.
The original situation is numbered "1."

There is a resultant force [tex]F_r = m_2 \cdot g - k \cdot m_1 \cdot g[/tex]
Careful! That's an equivalent 'resultant force' on the entire system along its direction of motion.

and because of that force m1 has the acceleration a.
The entire system has acceleration a along the direction of motion.

What I can't seem to figure out is is that resultant force equal.

[tex]F_r = m_1 \cdot a[/tex]

or

[tex]F_r = (m_1 + m_2) \cdot a[/tex]


In defense of the first argument, I've drawn the equivalent picture (2.) in which the second mass is replaced just by the force it exerts to m1.

In that case, it is clear that the first equation for Fr is the correct one.
Diagram 2 is incorrect. m2*g is not a force acting on m1. It's the cord tension that acts on m1.
 
Doc Al said:
Diagram 2 is incorrect. m2*g is not a force acting on m1. It's the cord tension that acts on m1.

But isn't Fstring=m2*g ?

Why can't I just ignore the string force, and just observe the string as a force carrier?
 
pinsky said:
But isn't Fstring=m2*g ?
No! Think about it. If the string force equaled m2*g, what would be the net force on and acceleration of m2?

Why can't I just ignore the string force, and just observe the string as a force carrier?
If you treat the system as a whole you can ignore the string force--it's just an internal force. But if you then choose to look at m1 alone, you must consider forces acting on m1.

I would advise against taking shortcuts until you are more practiced. It's generally easier to understand things if you treat each mass separately, apply Newton's 2nd law, and then combine the two equations to solve for the acceleration of the system. Of course, once you've done this kind of problem a few times, it's perfectly OK to jump to the solution immediately.
 
Thanks for the reply. I took me a while to SEE it right.

[SOLVED]