Question

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 7000 N {\rm N}. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N {\rm N} drag force.

Part A

What was the tension in the cable when the craft was being lowered to the seafloor?

Express your answer to two significant figures and include the appropriate units.

Part B

What was the tension in the cable when the craft was being raised from the seafloor?

Express your answer to two significant figures and include the appropriate units.

Answer 1

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Step-by-step explanation:

Part A)

`F_(g)` = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

`T` = Tension force in upward direction

`F_(d)` = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

`F_(g)` -
`F_(d)` -
`T` = 0

7000 - 1800 -
`T` = 0

`T` = 5200 N

`T` = 5.2 x 10³ N

Part B)

`F_(g)` = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

`T` = Tension force in upward direction

`F_(d)` = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

`T` -
`F_(g)` -
`F_(d)` = 0

`T` - 7000 - 1800 = 0

`T` = 8800 N

`T` = 8.8 x 10³ N