Figure 5-53 shows Atwood\'s machine, in which two containers are connected by a

Q: Figure 5-53 shows Atwood's machine, in which two containers are connected by a

m2g-T = m2*a T-m1g = m1*a a=(m2-m1)g/(m1+m2) given dm1/dt = -r da/dt = g{-(m1+m2)(-r)-(m2-m1)(-r)/(m1+m2)^2} = 2*gr*m2 / (m1+m2)2 so the accelaration will reach maximum at t = m1/r when container m1 becomes empty

ID: 2024001 • Letter: F

Question

Figure 5-53 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 m1 and container 2 has mass m2 with m1 < m2, but container 1 is losing mass (through a leak) at the constant rate r. (a) At what rate is the magnitude of acceleration of the containers changing at time t? (b) When will container 2 reach maximum acceleration? Express your answers in terms of the variables given and g.

Explanation / Answer

m2g-T = m2*a

T-m1g = m1*a

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

given dm1/dt = -r

da/dt = g{-(m1+m2)(-r)-(m2-m1)(-r)/(m1+m2)^2}

= 2*gr*m2 / (m1+m2)2

so the accelaration will reach maximum at t = m1/r when container m1 becomes empty

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Figure 5-5 shows a man sitting in a bosun\'s chair that dangles from a massless

Figure 5-53 shows a man sitting in a bosun\'s chair that dangles from a massless

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Figure 5-53 shows Atwood\'s machine, in which two containers are connected by a

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