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An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary, the tension in the cable was 5500 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force.What was the tension in the cable when the craft was being lowered to the seafloor?

User Jagan
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Answer:


T = 12910.5\,N for a craft with a mass of 1500 kg.

Step-by-step explanation:

Let consider that craft has a mass of 1500 kg. The submersible craft is modelled after the Newton's Laws, whose equation of equilibrium is:


\Sigma F = T - W +F_(D) = 0

The tension experimented by the cable while the craft is lowering to the seafloor is:


T = W - F_(D)


T = (1500\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right)-1800\,N


T = 12910.5\,N

User Josh Close
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