Question

A system of two objects suspended over a pulley by a flexible cable is sometimes referred to as an Atwood’s machine. Here, let the mass of the counterweight be 1000 kg. Assume the mass of the empty elevator is 850 kg, and its mass when carrying four passengers is 1150 kg. For the latter case calculate (a) the acceleration of the elevator and (b) the tension in the cable. Is the tension different when you calculate the tension in the cable of the elevator and the counterweight? If not, explain why.

Answer 1

From the given question,

Here m2>m1

From the free body diagram,

m2g-T=m2a..................(1)

T-m1g=m1a..................(2)

adding eq(1) and (2) we get,

m2g-m2g=(m2+m1)a

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

here m1=1000 kg, m2=1150 kg

a= (1150-1000)*9.8/(1150+1000)

=0.68 m/s^2

Acceleration of the elevator is 0.68 m/s^2

Replacing 'a' in eq (1) we get

1150*9.8-T=1150*0.68

T=10488 N

The tension in the cable is same for lighter and heavier mass.

We can treat the entire system as one object.

The acceleration for each object is equal in magnitude but opposite in direction.