Question

Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment, 2.0 s after the system is set into motion. Find the time that elapses before the string is tight again.

Answer 1

Given :

a=3.26 m/s²

T= 3.9 N

After 2 sec mass m₁ has velocity

v=u+at

=0+3.26 x 2

= 6.52 m/s upward

at this time m₂ is moving 6.52 m/s downward

At time 2 s, m₂ stops for a moment. But m₁ is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.

Here, v = 0, u = 6.52 m/s

a = −g = − 9.8 m/s²

v = u + at = 6.52 + (−9.8)t

⇒t=6.52/9.8≈2/3 sec

After this time, the mass m1 also starts moving downward.

So, the string becomes tight again after 2/3 s.

Answer 2

Answer:

Let the upward acceleration of both platform and man be a and the normal reaction pair acting on the system have magnitude N.

Considering the forces acting on man,

2T+N−mg=ma

Considering the forces acting on platform,

2T−N−mg=ma

⟹N=0

For T=

2

mg

a=0