# Dynamics: Tension (Elevator)

1. The problem statement, all variables and given/known data
An elevator(4850kg) is to be designed so that the maximum acceleration is 0.068m/s2. What are the minumum and maximum forces the motor must exert on the supporting cable?

2. Homework Equations

Force = mass(acceleration)
Forcevertical = Ftens + Fgrav

## The Attempt at a Solution

Forcevertical = Ftens + Fgrav
Force = 4850kg(0.068m/s2)
= 329.8N

Forcevertical = 329.8N - 47530N
= -47200N

I'm not so sure about both answers but i think the first solution is kinda right?

Not quite. Think about what is happening. It also has to account for the force of gravity. So the minimum amount of force needed would be to support the force of gravity, and the maximum would be to not only support the force of gravity, but to accelerate the elevator upward at 0.068m/s2. So How would you express those forces with mathematics?

haruspex
Homework Helper
Gold Member
Forcevertical = Ftens + Fgrav
Force = 4850kg(0.068m/s2)
Which of the forces in the first equation corresponds to "Force" in this equation?

Forcevertical means forces on the y-axis

Which of the forces in the first equation corresponds to "Force" in this equation?
Ftens is the one that corresponds to "Force"

haruspex