Calculating Forces on a 2000kg Elevator Suspended by Single Cable

[Power24]

Elevator of 2000kg is suspended by a single cable calculate the forces on it while it is at rest.

F=ma
well it's at rest so there is 0 acceleration, does that mean the formula is not usable or is it simply 2000N?

partc asks what the value of all forces is at 4.0 m/s^2, would that just be F=2000(4) then F=8000N

 

[Werg22]

Actually gravity exerts a force of 2000g downward while the roof pulls it up with the same force, and the total net force=0.

 

[Power24]

ok well the question is, calculate all the forces acting on the elevator

c) calculate all the forces acting on the elevator when descending 4.0 m/s^2

 

[Werg22]

Gravity and normal force which are equal.

 

[Power24]

it's actually tension and gravity but I need to find the value in Newtons

 

[Werg22]

9.81*2000=Fg... Same for tension.

 

[Power24]

much appreciated but the value 9.8, I remember it from an example, it has to do with gravity, gravities acceleration is 9.8m/s^2 so assuming that's where its coming from a value of 4.0m/s^2 descending would give me an equation of F=2000(5.8)

 

[Werg22]

You mean the Elevetor is accelerating 4 m/s^2 upward? Gravity stays the same, 9.81(2000) and then torsion (I call it normal force, it is just a matter of what the force is going trought) will be equal to 2000(9.81+4)

 

[Power24]

yes it accelerating but it's going downward so does that make a difference?

 

[Werg22]

If it accelerates downward then its substracted from gravity so 5.8 upward is right.

 

[Power24]

Perfect, thanks for the help.