find the tension in an elevator cable which must support a 4000 N elevator with a maximum upward acceleration of 2 g's. What would be the tension if the elevator were allowed to accelerate downward at .25g's?

Upwards:

-ma = mg –T1,
T1 = m(g+a).
Downwards:
ma = mg –T2,
T2 =m(g-a).

To find the tension in the elevator cable, we first need to calculate the net force acting on the elevator.

1. When the elevator is accelerating upward at 2 g's:

Given:
- Elevator mass, m = 4000 N / 9.8 m/s^2 = 408.16 kg
- Upward acceleration, a = 2 g's = 2 * 9.8 m/s^2 = 19.6 m/s^2

The net force acting on the elevator can be calculated using Newton's second law of motion:

Net force (F_net) = mass (m) * acceleration (a)

F_net = 408.16 kg * 19.6 m/s^2
F_net = 7996.74 N

Therefore, the net force acting on the elevator when it accelerates upward at 2 g's is 7996.74 N.

Since the cable tension needs to balance this force, the tension in the cable will be 7996.74 N.

2. When the elevator is accelerating downward at 0.25 g's:

Given:
- Downward acceleration, a = 0.25 g's = 0.25 * 9.8 m/s^2 = 2.45 m/s^2

The net force acting on the elevator can be calculated using the same formula:

F_net = mass (m) * acceleration (a)

F_net = 408.16 kg * 2.45 m/s^2
F_net = 998.41 N

Therefore, the net force acting on the elevator when it accelerates downward at 0.25 g's is 998.41 N.

Since the cable tension needs to balance this force, the tension in the cable will be 998.41 N.