Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Explanation

Let `m_(1)` = 12 kg and `m_(2)` = 8 kg
Then, a = downward acceleration of `m_(1)`
= upward acceleration of `m_(2)`
So, the force equations are,
for mass `m_(1) : _(1)g - T = m_(1) a ` ....(1)
and for mass `m_(1) : T - m_(2) g = m_(2) a`
Now adding equations (1) and (2), we get
`g(m_(1) - m_(2)) = a(m_(1) + m_(2))`
or, a = `(m_(1) - m_(2))/(m_(1) + m_(2)) g = (12 - 8)/(12 + 8)xx 9.8`
Then from (2),
`T = m_(2) g + m_(2) (m_(1) - m_(2))/(m_(1) + m_(2))g`
= `m_(2) g. (2m_(1))/(m_(1) + m_(2)) = (2m_(1) m_(2))/(m_(1) + m_(2)) g`
= `(2 xx 12 xx 8)/(12 + 8) xx 9.8 = 94.08 N`