A crane is being used to lift a container onto a cargo ship. You need to program

Question

A crane is being used to lift a container onto a cargo ship. You need to program the crane so that it lifts the container with a constant positive acceleration, a, during the first half of the lift and with a constant negative acceleration of the same magnitude, -a, during the second half of the lift. You want to do this so that the maximum speed of the container during the lift is 3.0 m/s. The containers are initially at rest on the dock.

The container has a mass of 1600 kg and the total distance the container is lifted is 5.5 m.

What is the tension force in the cable used to lift the container during the first half (accelerating upwards) part of the lift?

Explanation / Answer

As the lift is to be accelerated for the first part of the lift and then accelerated at the same rate, the distance travelled in both parts would be same.

That is, distance travelled while accelerating = 5.5 / 2 = 2.75 m

Also, the velocity achieved would be 3 m/s

Hence, we have v^2 = u^21 + 2aS

or 9 = 2a x 5.75

or, a = 1.6364 m/s^2

Now, we know that the lift accelerates at the rate of 1.6364, hence we will write the force equation for the first part of the motion to determine the required lift.

That is , T - Mg = Ma

or T = Mg + Ma

or T = 1600 (9.81 + 1.6364) = 18314.18182 N

Therefore the required tension is 18314.18182 Newtons

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