# How to do 187th question?

##  Mar 30, 2018

See below.

#### Explanation:

Considering the rope length $L$ and calling

${y}_{1} = {m}_{1}$ coordinate
${y}_{2} = {m}_{2}$ coordinate
${y}_{c} =$ pulley center coordinate

We have

${y}_{c} - {y}_{1} + {y}_{c} - {y}_{2} = L$ or

$2 {y}_{c} = {y}_{1} + {y}_{2} + L$

now deriving twice

$2 {\ddot{y}}_{c} = {\ddot{y}}_{1} + {\ddot{y}}_{2}$

but ${\ddot{y}}_{c} = {a}_{0} , {\ddot{y}}_{1} = {a}_{1} , {\ddot{y}}_{2} = {a}_{2}$

$\left\{\begin{matrix}T - {m}_{1} \left({a}_{0} + g\right) = {m}_{1} {a}_{1} \\ T - {m}_{2} \left({a}_{0} + g\right) = {m}_{2} {a}_{2} \\ 2 {a}_{0} = {a}_{1} + {a}_{2}\end{matrix}\right.$

Solving those equations we obtain

$\left\{\begin{matrix}T = 2 \left(2 {a}_{0} + g\right) \frac{{m}_{1} {m}_{2}}{{m}_{1} + {m}_{2}} \\ {a}_{1} = 3 {a}_{0} + g - \frac{2 \left(2 {a}_{0} + g\right) {m}_{1}}{{m}_{1} + {m}_{2}} \\ {a}_{2} = \frac{\left(3 {a}_{0} + g\right) {m}_{1} - \left({a}_{0} + g\right) {m}_{2}}{{m}_{1} + {m}_{2}}\end{matrix}\right.$

we leave the conclusions to the reader.

Mar 31, 2018

#### Answer:

The question asks for each one is Incorrect...
(4)

#### Explanation:

As general statement:
$- {a}_{1} = {a}_{0} = {a}_{2}$

Since ${m}_{1}$ is heavier it till go down by some $- {a}_{1}$, now that value has to have the same magnitude as ${a}_{2}$ because that is the acceleration ${m}_{2}$ upwards and as shown by the picture that is equal to ${a}_{0}$

Here's the formula for tension of two masses:
$T = {m}_{2} g + {m}_{2} {a}_{0}$
$T = {m}_{1} g - {m}_{1} {a}_{0}$

Substitute for ${a}_{0}$:
$T = {m}_{2} g + {m}_{2} {a}_{2}$
Number 2 is correct

Substitute for ${a}_{0}$:
$T = {m}_{1} g - {m}_{1} \left(- {a}_{1}\right)$
$T = {m}_{1} g + {m}_{1} {a}_{1}$
Number 1 is also correct

Number 3 however refers to the MAGNITUDE of relative acceleration in comparison to the pulley, so we can add to our statement: $- {a}_{1} = {a}_{0} = {a}_{2} = {a}_{r}$
Substitute for ${a}_{0}$:
$T = {m}_{1} g - {m}_{1} \left({a}_{r}\right)$
$T = {m}_{1} g - {m}_{1} {a}_{r}$
$T = {m}_{1} \left(g - {a}_{r}\right)$
Number 3 is also correct

Number 4 is incorrect because:
going back to:
$T = {m}_{2} g + {m}_{2} {a}_{0}$
$T = {m}_{1} g - {m}_{1} {a}_{0}$

Solve for ${a}_{0}$ in both and set them equal:
${a}_{0} = \frac{T - {m}_{2} g}{m} _ 2$
${a}_{0} = \frac{T - {m}_{1} g}{-} {m}_{1}$

So:
$\frac{T - {m}_{2} g}{m} _ 2 = \frac{T - {m}_{1} g}{-} {m}_{1}$

$\frac{- {m}_{1} T + {m}_{2} {m}_{1} g}{m} _ 2 = \left(T - {m}_{1} g\right)$

$- {m}_{1} T + {m}_{2} {m}_{1} g = {m}_{2} T - {m}_{1} {m}_{2} g$

${m}_{2} {m}_{1} g + {m}_{1} {m}_{2} g = {m}_{2} T + {m}_{1} T$

${m}_{2} {m}_{1} g + {m}_{1} {m}_{2} g = T \left({m}_{2} + {m}_{1}\right)$

$\frac{2 {m}_{1} {m}_{2} g}{{m}_{2} + {m}_{1}} = T$