Question

The figure below shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0, container 1 has mass 1.30 kg and container 2 has mass 2.40 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.170 kg/s.
(a) At what rate is the acceleration magnitude of the containers changing at t = 0?

Answer 1

The force F is the difference between the two masses times g

F = (m2-m1) g

The acceleration is a=F/(m2+m1)

get dF/dt and da/dt and use dm1/dt = 0.170 kg/s

Answer 2

because m2 and g are constants, do you pull those out and only derive .17t ?

Answer 3

Almost, dF/dt = -(dm1/dt) g

a = (m2-m1) g / (m2+m1)

da/dt = (da/dm1)(dm1/dt) by chain rule

Answer 4

(.170[(1.3-.170t)+2.4]-[2.4-(1.3-.170t)g])/((1.3-.170t)+2.4)^2
??

Answer 5

No, acceleration is changing at t=0 because m1 is changing at t=0

Answer 6

Ok now I think I got it :
Call the (constant) mass of the heaviest container M (2.4 kg).
Newton's second law reads: F = dp/dt = d(mv)/dt = m dv/dt + v dm/dt.
, so m dv/dt = F - v dm/dt. Also, m(t) = m0 - 0.220 t, since m0 = 1.30kg and dm/dt = -0.170 kg/s

Since F = (M-m) g we have

m dv/dt = (M-m)g + 0.170 v
Differentiating once more with respect to time:

0.170 a+ m da/dt = -g dm/dt + .170 a
or
m da/dt = - g dm/dt
da/dt = 0.170 g /m

(a). At t=0 s: m= 1.30 kg, so da/dt = 0.170 kg/s * 9.81 m/s^2 /1.30 kg = 1.28 m/s^2 /s

Answer 7

Should of read "Also, m(t) = m0 - 0.170 t"