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

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

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