How to do 187th question?

enter image source here enter image source here

2 Answers
Mar 30, 2018

Answer:

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.

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_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#