Question

a 300 kg has passenger with an unknown mass. From rest, the elevator accelerates upward at 1.2m/s/s. The cable supporting the elevator exerts and upward force of 8000 N. What is the maximum mass of the passenger that the elevator can carry?

Answer 1

The tension in the cable

$$T=m(g+a)$$

Hence, the net mass of system (elevator and passengers) is equal

$$m=T/(g+a)\\ =8000/(9.8+1.2)=727\:\rm kg$$

The limit mass of passengers

$$m_p=727-300=427\:\rm kg$$