Question

How to do 187th question?

Answer 1

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

`2y_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`

`{(T-m_1(a_0+g)=m_1a_1),(T-m_2(a_0+g)=m_2a_2),(2a_0=a_1+a_2):}`

Solving those equations we obtain

`{(T = 2(2a_0+g)(m_1m_2)/(m_1+m_2)),(a_1 = 3 a_0 + g - (2 (2 a_0 + g) m_1)/(m_1 + m_2)),(a_2=((3 a_0 + g) m_1 - (a_0 + g) m_2)/(m_1 + m_2)):}`

we leave the conclusions to the reader.

Answer 2

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_2g+m_2a_0`
`T= m_1g-m_1a_0`

Substitute for `a_0`:
`T= m_2g+m_2a_2`
Number 2 is correct

Substitute for `a_0`:
`T= m_1g-m_1(-a_1)`
`T= m_1g+m_1a_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_1g-m_1(a_r)`
`T= m_1g-m_1a_r`
`T= m_1(g-a_r)`
Number 3 is also correct

Number 4 is incorrect because:
going back to:
`T= m_2g+m_2a_0`
`T= m_1g-m_1a_0`

Solve for `a_0` in both and set them equal:
`a_0= (T-m_2g)/m_2`
`a_0= (T-m_1g)/-m_1`

So:
`(T-m_2g)/m_2= (T-m_1g)/-m_1`

`(-m_1T+m_2m_1g)/m_2= (T-m_1g)`

`-m_1T+m_2m_1g= m_2T-m_1m_2g`

`m_2m_1g+m_1m_2g= m_2T+m_1T`

`m_2m_1g+m_1m_2g= T(m_2+m_1)`

`(2m_1m_2g)/(m_2+m_1)= T`