# Elevator Tension With Upward Acceleration

1. Oct 15, 2012

### SigFig

1. The problem statement, all variables and given/known data

An elevator weighing 10 000N is supported by a steel cable. Determine the tension in the cable when the elevator is accelerated upward at 3.0 m/s2

A) 7.0kN
B) 10.0 kN
C)11.6 kN
D) 13.1 kN
E) 40.0 kN

2. Relevant equations

F=ma W=mg (g=9.8m/s^2)

3. The attempt at a solution
W=mg
10 000=m(9.8)
m=1020

2. Oct 15, 2012

### tachyon_man

The net force (i.e. the total of all the forces) is what causes acceleration in the equation F=ma. So draw a diagram and label the forces acting on it. (Gravity and Tension) so now use the F=ma equation and plug everything in and off you go.

3. Oct 15, 2012

### SigFig

So it would be
W=mg
10 000=m(9.8)
m=1020

F=ma
F=1020(3)
F=3060

3060+10 000 = F
F= 13.1 kN?

4. Oct 15, 2012

### tachyon_man

Remember that the force of gravity is opposite the force of tension. One's up and one's down.

5. Oct 15, 2012

### SigFig

So the 3060 would be subtracted from the elevators weight because it's accelerating upwards (against gravity) making it 7000?

6. Oct 15, 2012

### tachyon_man

Oh my goodness, I'm sorry, you were right in the first place, I read it wrong. Good work! Sorry to confuse.