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