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 7.85 m/s. The containers are initially at rest on the dock. The container has a mass of 150 kg and the total distance the container is lifted is 11.8 m. What is the tension force in the cable used to lift the container during the first half (accelerating upwards) part of the lift? Give you answer in Newtons to at least four digits. Do not include units in your answer.

Answer 1

The concepts involved in solving are

Newton's second law of motion and Kinematic relations.

Please find answers below, if any doubt plz feel free to ask.

The concepts involved in solving are Newton's second law of motion and Kinematic relations. Please find answers below, if any doubt plz feel free to ask m = 150 kg s = 11.8 m v_m = 7.85 m/s During first half of motion T - mg = ma - (1) Now, we have v^2 - u^2 = 2as v = v_max: s = 11.8/2: u = 0 (initially rest) a = + a (as body acceleration in first half) rightarrow 7.85^2 - 0 = 2(a) (11.8/2) rightarrow a = 5.22 m/s^2 put a = 5.22 in ef^n (1) rightarrow T = mg + ma = m(g + a) = 150(9.81 + 5.22) rightarrow T = 2254.8 N Therefore Tension T = mg + ma + m(g + a) = 150(9.81 + 5.22) rightarrow T = 2254.8 N rightarrow Tension in cable = 225 + 4.8 N