User:Phys1csf4n

Welcome to my physics page! I will put any interesting proofs that I run across in my experiences. If you have questions feel free to comment. Thx!

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Double Atwood machine accelerations!

A Double Atwood machine with three masses

We are able to derive an equations for the acceleration by using Newton's Laws. If we consider a massless, inelastic string and an ideal massless pulley the only forces we have to consider are: tension forces ($T_{1}$ = $T_{2}$, $T_{3}$ = $T_{4}$), and the weight of the three masses ($m_{A}$, $m_{B}$, $m_{C}$). To find the accelerations we need to consider the forces affecting each individual mass and how the accelerations are related.

Using Newton's laws we can create this set of equations:

 $\;{T_{1}-m_{A}*g=m_{A}*a_{A}}$
 $\;T_{2}-m_{B}*g=m_{B}*a_{B}$

And since we can assume the pulley and sting are massless and frictionless:

 $\;T_{1}=T_{2}$

The above equation helps you to find $a_{B}$ in terms of $a_{A}$ to substitute in later on.

Since we can assume that $T_{1}=T_{2}$ we can then also assume that since $\;T_{3}=T_{1}+T_{2}$ that the equation:

 $\;T_{3}=2*T_{1}$

easily follows.

As we assumed in part 1 that $T_{1}=T_{2}$, we can also assume that $\;T_{3}=T_{4}$ and can thus come up with an equation in terms of $m_{C}$ and $T_{3}$:

 $\;T_{3}-m_{C}*g=m_{C}*a_{C}$

The above equation helps you to find $a_{C}$ in terms of $a_{A}$ to substitute later on.

The Last equation is by far the most difficult equation to understand and it is needed to relate the accelerations to each other. Essentially we know by logic that:

if $a_{A}=1{m \over s^{2}}$ down and $a_{B}=1{m \over s^{2}}$ up then: $a_{C}=0{m \over s^{2}}$. 

if $a_{A}=1{m \over s^{2}}$ down and $a_{B}=1{m \over s^{2}}$ down then: $a_{C}=1{m \over s^{2}}$ up. 

 it easily follows that $\;{a_{A}+a_{B}+2*a_{C}=0}$

More simply this can be summarized by the fact that the average of $a_{A}$ and $a_{B}$ is moving in the opposite direction of $a_{C}$.

Using Equations to get results!

First solve for $a_{B}$ in terms of $a_{A}$ to plug into $a_{B}+a_{A}=-2*a_{C}$:

Since you know that $T_{1}=T_{2}$ you can obtain $a_{B}$ from $a_{A}$ with this equation:

 $\;m_{A}*(a_{A}+g)=m_{B}*(a_{B}+g)$ 

Solving for $a_{B}$ you obtain:

 ${m_{A} \over m_{B}}*(a_{A}+g)-g=a_{B}$

Also you know that $2*T_{1}=T_{3}$ you can relate $a_{A}$ and $a_{C}$ to plug into $a_{B}+a_{A}=-2*a_{C}$:

 $\;2*m_{A}*(a_{A}+g)=m_{C}*(a_{C}+g)$

Which can be simplified to:

 $2*{m_{A} \over m_{C}}*(a_{A}+g)-g=a_{C}$

Then you plug in what you have solved to find $a_{A}$

 $a_{A}+{m_{A} \over m_{B}}*(a_{A}+g)-g=-4*{m_{A} \over m_{C}}*(a_{A}+g)+2*g$

Many Algebraic Errors Later after substitution you can solve for $a_{A}$ and obtain:

 $a_{A}=g*{-4*m_{A}*m_{B}-m_{A}*m_{C}+3*m_{B}*m_{C} \over m_{B}*m_{C}+m_{A}*m_{C}+4*m_{A}*m_{B}}$

You then use the above equation relating $a_{A}$ to $a_{B}$ which you can then use to solve for $a_{B}$:

 $a_{B}=g*{-4*m_{A}*m_{B}-m_{B}*m_{C}+3*m_{A}*m_{C} \over m_{B}*m_{C}+m_{A}*m_{C}+4*m_{A}*m_{B}}$

This equation makes perfect sense because all that is different is that the $m_{A}$'s and $m_{B}$'s are swapped.

The last acceleration to solve for is $a_{C}$. You can solve directly since you've found $a_{C}$ in terms of $a_{A}$ already.

GOOD LUCK!