Cargo Elevator

Classical Mechanics Level 2

A heavy cargo elevator is moving downward at 5 m/s at t=0 when the motor is engaged to bring the elevator to rest. At t=0, F(t)=50,000 N, and at t=5 s, F(t)=100,000 N. The cargo and elevator have a combined mass of 50,000 Kg. How much cable must be unfurled during the time it takes to bring the elevator to v=0 m/s? Side rail friction is negligible and g=9.81 m/s^2.

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2 solutions

Steven Chase
Mar 26, 2020 
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import math

m = 50000.0
g = 9.81

dt = 10.0**(-5.0)

######################################################

t = 0.0

x = 0.0       # elevator pos,vel,acc
xd = 5.0      # down is the positive direction
xdd = 0.0

while xd > 0.0:

    x = x + xd*dt
    xd = xd + xdd*dt

    # Assume linear force progression
    # Pulleys amplify force by factor of 10

    F = m*g - 10.0*(50000.0 + 10000.0*t)

    xdd = F/m

    t = t + dt

######################################################

print "time step"
print dt
print ""

print "elapsed time"
print t
print ""

print "cable unfurled"   # must account for pulley here
print (10.0*x)

######################################################

#time step
#1e-05

#elapsed time
#2.14310000001

#cable unfurled
#69.9827592297

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