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Assume a pulley and a massless rope. On one end there is a $10.0$ kg mass. On the other there is a $34.5$ kg mass. What is the acceleration of the $34.5$ kg mass?

I created a free-body diagram with $m_1g_1$ pointing up and $m_2g_2$ pointing down, so the total tension is $m_1g_1 - m_2g_2$. This results in $240$ N downwards, so I divide by $34.5$ kg to get $6.96$ $m/s^2$. However that is incorrect. What is my mistake?

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    $\begingroup$ This is called the Atwood apparatus, a full explanation is on Wikipedia (as always). en.wikipedia.org/wiki/Atwood_machine $\endgroup$
    – user38491
    Jan 29 '14 at 1:31
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What I am assuming from your question is that these two blocks are hanging vertically from the pulley. Therefore I don't know what you mean by $m_1g_1$ pointing up. The force of gravity is acting downward on both of the blocks. The easiest way to solve these is to create two free body diagrams, one for each block. Note that there is a tension force pulling upward on each block, and because the string and pulley are massless, these tensions are equal.

Once you have created your two free body diagrams, make a guess as to which direction you think the acceleration is going to go and define that direction as positive. Once you have this definition of the positive direction, then use $F_{net} = ma$ for each block (paying attention to what forces are in the positive and which forces are in the negative direction). This will result in two equations with $T$ and $a$ as unknowns. As an example, say that I believe the blocks will accelerate clockwise around the pulley. I will define this direction as positive and my free body diagrams will look like so (I apologize, my force vectors are not to scale):

enter image description here

Therefore, for $m_1$ I have $F_{net} = T - m_1g$, whereas for $m_2$ I have $F_{net} = m_2g - T$. The signs here are important. I would then plug both of these $F_{net}$ expressions into their own Newton's law equation, $F_{net} = ma$. You can solve the resulting system to find $a$ (note that the acceleration for both blocks must be the same number). If your answer comes out positive, then this means your initial guess for the direction of acceleration was correct. If it comes out negative, then your guess was incorrect and the acceleration moves in the opposite direction.

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Write 2 equations for $f=ma$. Tension and acceleration are the 2 unknowns. Equation 1 : $$T-m_1g=m_1a$$ Equation 2:
$$m_2g-T = m_2a$$ where $T$ is tension and $a$ is acceleration of system. Solve simultaneously.

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  • $\begingroup$ Draw a FBD for each mass. First FBD has (m1)g acting down, tension T pulling up, and body accelerating upwards with value a. 2nd FBD has (m2)g acting down, tension T pulling up, and body accelerating with value a in the down direction. a and T values are the same for both diagrams. 2 equations, 2 unknowns. Just do the math. $\endgroup$
    – user38488
    Jan 29 '14 at 1:11
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In FBD of $m_2$ you have taken two forces $m_2g$ acting downward and $m_1g$ acting upward but here the upward force is tension $T$ not $m_1g$ , also note that by your method you will get different accelerations for $m_1$ and $m_2$, right equations for $m_1$ and $m_2$ are $m_1g-T=m_1a'$ -<1> and $m_2g-T=m_2a$, here $a=-a'$ so equation for $m_2$ will be $T-m_2g=m_2a'$ <2> by <1> and <2> $a'$ and $a$ can be found

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We have two different masses attached to the same rope, meaning that their acceleration will be the same. Also, tension is equal throughout the rope, meaning that is the same as well. When we consider two free body diagrams of both masses, we find a system of simultaneous equations.

Let $M>m$, then: $$T-mg=ma$$ and, $$Mg-T=Ma$$ Solving this system with the given values gives us: $a=5.51 \text{ m}/\text{s}^2$ and $T=155.1\text{ N}$.

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