Missing something (Newtons law)

[pinsky]

Hello there!

I'm trying to solve this in two ways, and i keep getting different solutions. I need to find the force by which M should be pushed for m1 and m2 stand still (compared tom M). There is no friction between anything.

First solution:
$$F_{rp} = m_2 \cdot g \: \: \: F_{rp} = m_1 \cdot a \: \: \Rightarrow a = \frac{m_2}{m_1} g$$

$$F = (M + m_1 + m_2) \cdot a = (M + m_1 + m_2) \cdot \frac{m_2}{m_1} g$$

So in this solution, I'v observed all the objects as a system, so that's how i got the F=m_total * a.


Second solution:

$$F_{rp} = m_2 \cdot g \\\\\\\ F_{rp} = m_1 \cdot a \: \: \Rightarrow a = \frac{m_2}{m_1} g$$

$$m_2 \cdot a = F_p$$

$$F - F_p = a \cdot M \: \: \Rightarrow F = a \cdot (M+m_2)$$

I've done a step by step decomposition of all forces, but I'm missing the influence of m1.

What am I missing here?

 

[weesiang_loke]

for second solution:
F - Fp = a . ( M + m1)
Then the final answer will appear as F = (M + m1 + m2) . a

 

[pinsky]

Why so?

If I'm observing the forces diagram for the mass M. There is not a force m1*a having an influence on it.

Could you explain please?

 

[tiny-tim]

hi pinsky!

essentially, you're asking is F = (M + m₁ + m₂)a, or is it only F = (M + m₂)a ?

you're thinking that there's no horizontal force between m₁ (the top block) and M-plus-m₂, the force F is pushing M and m₂ but isn't pushing m₁, and so why should m₁ be included in the F = ma for M-plus-m₂ ?

but you're ignoring the forces at the pulley … if the pulley was mounted on a spring, the rope would pull it diagonally down … basically, there's both a horizontal force and a vertical force at the pulley, each of strength F_rp, and that horizontal force is an external force on M-plus-m₂, making the M equation (F - F_rp - F_p) = (M + m₂)a

 

[pinsky]

YES!

Thank you!

I've failed to see the pulley can be considered as a part of M, and that is how m1 influences M.

[SOLVED]