Question

A. A cord passing over an easily turned pulley (one that is both massless and frictionless) has 7kg mass hanging from one end and a 9kg mass hanging from the other end as shown below. G 7.0 kg 9.0 ks This arrangement is known as Atwood's machine. Calculate the (i). acceleration of the masses (ii). tension in the cord. [2.5 ​

Answer 1

Answer:

M g = M a = a (9 - 7) kg      with the two masses going in different directions

(7 + 9) kg * g = a * 2 kg       2 kg effective in prducing acceleration

a = 16 / 2 * g = 9.8 m/s^2 / 8 = 1.23 m/s^2

M1 a = T - M1 * g        T producing upward acceleration of 1.23 m/s^2

T = 7 (a + g) = 7 * 11 = 77 N

Check:

M2 a =  M2 * g - T          M2 producing downward acceleration

T = M2 (g - a) = 9 * (9.8 - 1.23) = 77 N