A crane is being used to lift a cargo container onto a ship. The crane needs to

Question

A crane is being used to lift a cargo container onto a ship. The crane needs to be programed so that it lifts the container with a constant positive acceleration, a, for the first half of the lift, and with a negative acceration of the same magnitude, -a, during the second half of the lift. You want to do this so that the max speed of the container during the lift is 3 m/s. The container starts at rest. The container has a mass of 1250 kg and it needs to be moved 6.5 m. What is the tension force in the cable to lift the container during the first half part of the lift? Answer in Newtons.

Explanation / Answer

If the maximum speed is 3.0 m/s and we begin and end "at rest," then the average velocity is Vavg = 1.5 m/s.
Then the total time lifting is t = h / Vavg = 6.5m /1.5 m/s = 4.33 s
and the time for accelerating and decelerating is 2.17 s each.

Then a = v / t = 3m/s /2.17s = 1.38 m/s²
and
tension T = m(g + a) = 1250kg * (9.8 + 1.38)m/s² = 13975 N

You could also reason as follows:
v² = u² + 2as for the first half of the ride, accelerating from rest (u = 0)
(3m/s)² = 0 + 2 * a * (6.5m / 2)
a = 1.38 m/s²
as before.

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