Tension in an elevator cable
An elevator has a mass of 1400kg. What is the tension in the supporting cable when the elevator traveling down at 10 m/s is brought to rest in a distance of 40 m. Assume a constant acceleration.
m =1400 kg mass of elevator,
v = 10m/s initial speed of the elevator,
D = 40 m distance required to stop the elevator.
g = 9.81 m/s2 gravitational acceleration, as usual is assumed to be known.
T = ? magnitude of tension in the cable while bringing the elevator to rest.
To find T we must calculate:
a = ? acceleration while stopping the elevator,
t = ? time required to stop elevator.
It is convenient to draw a free-body diagram, as in Figure below.
is the tension in the cable of the elevator, is the gravity force. The resultant force is the force producing acceleration (deceleration in this case) of our elevator.
This can be written in the form of the equation
if we chose the upward direction as positive. Solving for tension gives
For further calculations we can drop the vector notation as all the forces are acting along one line. To calculate the magnitude of the tension T, we must find the magnitude a of the acceleration. It can be found from kinematics equations
a = v/t (2)
D = vt (1/2)a t2 (3)
Equation (2) is based on the fact that elevator final speed is zero. Equation (3) is a standard formula for distance traveled in motion with constant acceleration (negative in this case as directed opposite to the initial speed).
Solving the equations (2) and (3) with respect to acceleration a, we find
Magnitude of tension T can be found from formula (1) taken without the vector notation (magnitude only!!)
Substituting numbers given in the problem we get
T = 15484 N.