A light, inextensible cord passes over a light, frictionless pulley
with a radius of 7.2 cm
It has a(n) 16 kg mass on the left and a(n)
6.7 kg mass on the right, both hanging freely.
Initially their center of masses are a vertical
distance 4.2 m apart.
The acceleration of gravity is 9.8 m/s^2

The question tied to it is…

What is the tension
in the cord when they
pass each other?
Answer in units of N.

         T = 93 N

Explanation:

Assuming the pulley is ideal meaning frictionless as mentioned and also negligible mass.

         ΣF = Σma

Mg - mg = Ma + ma

           a = g(M - m) / (M + m)

Now looking only at the larger mass as it falls

Mg - T = Ma

       T = Mg - Ma

       T = Mg - Mg(M - m) / (M + m)

       T = Mg(1  -(M - m) / (M + m))

       T = 16(9.8)(1 - (16 - 6.7) / (16 + 6.7))

       T = 93 N

or looking only at the smaller mass

T - mg = ma

T = m(g + a)

T = m(g +  g(M - m) / (M + m))

T = mg(1 +  (M - m) / (M + m))

T = 6.7(9.8)(1 +  (16 - 6.7) / (16 + 6.7))

T = 93 N

increase masses, decrease separation