Question

In the dark knight, Batman (mass of 105 kg in armor) suspends The Joker (mass of 77 kg) from a rope off the side of a tall building, but he won't let go and kill him. Assume that the rope lies over a frictionless pulley. (a) what is the tension of the rope? (b) Joker let himself go and now has a mass of 120 kg. Batman cannot hold him up and is hoisted upward... What is Batmans acceleration? Show eqn. first!

Answer 1

(a) In the first case –

Suppose T is the tension in the rope and ‘a’ is the acceleration. Now write the Tension – mass equations-

105g – T = 105a

=> a = g – T/105----------------------------------(i)

Again –

T – 77g = 77a

Now put the values of ‘a’ from (i) –

T – 77g = 77*(g – T/105)

=> T = 154g – 0.73T

=> T = 154g/1.73 = 873 N

(b) In the second case, suppose ‘a’ is the acceleration of Batman in upward direction.

So, 120g – T = 120a

=> T = 120g – 120a

Again, T – 105g = 105a

Put the value of T –

120g – 120a – 105g = 105a

=> 15g = 225a

=> a = 15g/225 = 0.65 m/s^2