An Atwood's machine consists of blocks of masses
m1 = 11.0 kg
and
m2 = 19.0 kg
attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass
M = 7.90 kg
and radius
r = 0.200 m.
The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.

1) Provided that the mass in [tex]m_2[/tex] is greater than the mass we have in [tex]m_1[/tex], then likewise the tension we have in [tex]T_2[/tex] will be greater than the tension in [tex]T_1[/tex]

Using Newton's second law to mass [tex]m_1[/tex], we have:

[tex]m_2g - T_2 = m_2 a \\ \\ T_1 = m_1 g +m_1 a \\ \\ T_1= m_1 (g+a) --- (1)[/tex]

By using the second law, we have:

[tex]m_2g - T_2 = m_2a \\ \\ T_2 = m_2 (g-a)---(2)[/tex]

For the pulley, let's use the torque equation, so we have:

[tex]T_2 r -T_1 r = I \alpha \\ \\ T_2r -T_1r = \Big ( \dfrac{Mr^2}{2}\Big) \dfrac{a}{r} \\ \\ T_2 -T_1 = \dfrac{Ma}{2} ---- (3)[/tex]

Altogether, from equation (1)(2) and (3), we have:

[tex]m_2(g-a) -m_1 (g+a) = \dfrac{Ma}{2} \\ \\ m_2g -m_2a -m_1g-m_1a = \dfrac{Ma}{2} \\ \\ a = \dfrac{(m_2 -m_1) g}{(m_1 + m_2 + \dfrac{M}{2})} \\ \\ a = \dfrac{(19.0 \ kg - 11.0 \ kg ) ( 9.8 \ m/s^2)}{(19.0 \ kg + 11.0 \ kg + \dfrac{7.90 \ kg }{2} )}[/tex]

[tex]a = 2.31 \ m/s^2[/tex]

Also; from equation (1), the tension in the string is:

[tex]T_1[/tex] = (11.0 kg ) ( 9.8 + 2.31) m/s²

[tex]T_1[/tex] = 133.21 N

[tex]T_1[/tex] ≅ 133 N

From equation (2):

[tex]T_1[/tex] = m_2(g-a)

[tex]T_1[/tex] = (19.0 kg) ( 9.8 - 2.31) m/s²

[tex]T_1[/tex] = 142.31 N

[tex]T_1[/tex] = 142 N