Question

Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward weight, there are buoyant forces and forces from the flowing water that allow them to travel in a horizontal path. One such submersible has a cross section area of 1.3m2 , a drag coeffecient of 1.2, and when towed at 4.3 m/s, the tow cable makes an angle of 30 degrees with the horizontal. What is the tension in the cable? Take the water density to be 1000 kg / m3

Answer 1

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density $\rho \ of\ water= 1000 kg/m3$

 Drag force FD is given as

$F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A$

        $= 0.5\times 1000\times 4.32\times  1.2\times 1.3$

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

$T = \frac{F_D}{cos30}$

$T =\frac{ 14422.2}{cos 30}$

T = 16653.32 N

Answer 2

This question involves the concepts of the drag force and the tension force.

The tension in the cable is "3.87 KN".

First, we will calculate the drag force on the submersible:

$F_D=\frac{1}{2}\rho A v^2C$

where,

$F_D$ = Drag Force = ?

$\rho$ = density = 1000 kg/m³

A = Area of cross-section = 1.3 m²

v = speed = 4.3 m/s

C = drag coefficient = 1.2

Therefore,

$F_D = \frac{1}{2}(1000\ kg/m^3)(1.3\ m^2)(4.3\ m/s)^2(1.2)\\\\F_D=3354\ N$

Now, this drag force must be equal to the horizontal component of the tension force:

$F_D=TCos30^o\\\\T=\frac{F_D}{Cos30^o}\\\\T=\frac{3354\ N}{Cos30^o}$

T = 3872.86 N = 3.87 KN

Learn more about drag force here:

brainly.com/question/5815479?referrer=searchResults

The attached picture shows drag force.