A pulley is suspended by a Cord (C). on one end of the pulley there is a 1.2 kg block and on the other is a 3.2 kg block.Determine the tension in this cord after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords

Answer 1

Solve these two simultaneous equations with unknowns a (acceleration) and T (tension).

M1*g - T = M1*a
T - M2*g = M2*a

This leads to
a = [(M1 -M2)/(M1 +M2)]*g
and
T = [2*M1*M2/(M1+M2)]*g

Answer 2

34N

Answer 3

17.1

Answer 4

I did the Ft=g[(2Mm)/(M+m)] equation and ended up with 17.1N

Which plugged in was
Ft=9.8[((2)(1.2)(3.2))/(1.2+3.2)]

Answer 5

if you use the formula as such as
ft=(2*m1*m2/m1+m2)2g
you will get the right answer which is 34.210