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I'm studying physics but in this question I can't figure out the free body diagram.

pic

link: http://prntscr.com/6j0t2u

For $m_3$:

$$T-m_3 g = m_3a $$

For $m_2$ along the y-axis:

$$N - (m_2 + m_1)g = 0 $$

But I can't figure out the x-axis forces on $m_2$.

Sorry about this basic question. I have no choice to ask anybody else.

Edit:

Now, I'm trying to find T equation on $m_3$

@Steeven said $$T = m_2a$$ so when I try to solve it $a=20m/s^2$

It is seems not right.

Final:

At the top first equation should be like this. $$m_3g - T = m_3a $$ After that it seems correct.

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  • $\begingroup$ There is no friction in the system, so only contact forces are normal reactions. As far as m2 is concerned it only feels the string, its weight, and the surface it's resting on $\endgroup$ – danimal Mar 20 '15 at 9:25
  • $\begingroup$ So, tension on acting m3 shuld be (m1+m2).g ? $\endgroup$ – alim Mar 20 '15 at 9:46
  • $\begingroup$ the y-axis forces on m2 are its weight (down), the reaction from the surface (up), and the reaction from the block on top of it (down), which will not necessarily be equal to the weight of the m1 $\endgroup$ – danimal Mar 20 '15 at 10:11
  • $\begingroup$ @alim Note, you have a sign error in your $m_3$ equation. $\endgroup$ – Steeven Mar 20 '15 at 11:56
  • $\begingroup$ I saw it now, okay you are right. $$T = m_2a$$ Seems correct. $\endgroup$ – alim Mar 20 '15 at 11:59
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If there is no friction in the system, as the question states, then there is no horizontal force between the boxes $m1$ and $m2$ and also no horizontal force between the floor and $m2$. The only horizontal force remaining is the string tension $T$ that pulls to the right. That's it.

If you wish to find the tension $T$, you can simply set up the equations that involve it. You have already set it up for $m_3$, and for $m_2$ along the x-axis it is simply:

$$T=m_2a$$

Now just solve these two equations for $T$.

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  • $\begingroup$ Thanks for answer I get it the force applied but still I can't figure out the tension equation in my head. Tension should be (m1 + m2).g or what it should be ? $\endgroup$ – alim Mar 20 '15 at 10:59
  • $\begingroup$ I added an edit. If your real question is how to find the tension, you should write that in the question above. $\endgroup$ – Steeven Mar 20 '15 at 11:11
  • $\begingroup$ Ah I'm sorry about it, my question first forces on the mass but after that I can't figure out the T as well. Sorry about it. $\endgroup$ – alim Mar 20 '15 at 11:25
  • $\begingroup$ In the end, when I try to solve equation for question one m2a - m3g = m3a, result a = 20m/s^2 so answer can't be right. Do you have any clue ? $\endgroup$ – alim Mar 20 '15 at 11:30
  • $\begingroup$ Can you please show your calculation that gave you $20 \mathrm{m/s^2}$? $\endgroup$ – Steeven Mar 20 '15 at 11:43
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I solved for $ a $. You can solve for $ T $ on your own.

Observe $ m_{3} $, $$ \sum F = m_{3}a $$ $$ m_{3}g - T = m_{3}a $$



Observe $ m_{2} $, $$ \sum F = m_{2}a $$ $$ T = m_{2}a $$



Substitute $T$ from the second equation to the first equation, $$ m_{3}g - m_{2}a = m_{3}a $$ $$ m_{3}g = m_{2}a + m_{3}a $$ $$ m_{3}g = (m_{2} + m_{3})a $$ $$ a = \dfrac{m_{3}g}{m_{2} + m_{3}} $$



Plugging in the numbers, $$ a = \dfrac{4\times 10}{4+6} $$ $$ a = 4\;\frac{m}{s^{2}}$$

enter image description here

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