Great White Shark Attack In the classic thriller movie Jaws, Captain Quint manages to shoot harpoons into the great white shark that is attacking his boat. Each harpoon is attached to a cable, which in turn is tied to an empty watertight barrel. Quint’s intention is to the shark by forcing it

Question

Great White Shark Attack

In the classic thriller movie Jaws, Captain Quint manages to shoot harpoons into the great white shark that is attacking his boat. Each harpoon is attached to a cable, which in turn is tied to an empty watertight barrel. Quint’s intention is to fatigue the shark by forcing it to drag the barrels through the water. For a sealed 55-gal barrel weighing 35 lb, what force must the shark overcome when it dives beneath the boat and fully submerges the barrel? (See Figure 6.12.)

Approach
To find the force that the shark must overcome, we must consider the three forces acting on the barrel: its weight w, the tension T in the cable, and the buoyancy force [latex]F_B [/latex](Figure 6.13). The shark must overcome the cable’s tension, which depends on the other two forces. We begin by drawing a free body diagram of the barrel and indicate on it that we have chosen upward as the positive direction. The buoyancy force is proportional to the density of seawater, which is listed as 1.99 [latex]slugs/ft^3[/latex] in Table 6.1.

Table 6.1 Density and Viscosity Values for Several Gases and Liquids at Room Temperature and Pressure

	Density,ρ		Viscosity, μ
Fluid	[latex]kg/m^3[/latex]	[latex]slug/ft^3[/latex]	kg/(m . s)	slug/(ft . s)
Air	1.2	[latex]2.33×10^{-3}[/latex]	[latex]1.8×10^{-5}[/latex]	[latex]3.8×10^{-7}[/latex]
Helium	0.182	[latex]5.53×10^{-4}[/latex]	[latex]1.9×10^{-5}[/latex]	[latex]4.1×10^{-7}[/latex]
Freshwater	1000	1.94	[latex]1.0×10^{-3}[/latex]	[latex]2.1×10^{-5}[/latex]
Seawater	1026	1.99	[latex]1.2×10^{-3}[/latex]	[latex]2.5×10^{-5}[/latex]
Gasoline	680	1.32	[latex]2.9×10^{-4}[/latex]	[latex]6.1×10^{-6}[/latex]
SAE 30 oil	917	1.78	0.26	[latex]5.4×10^{-3}[/latex]

Accepted Answer

Since the barrel’s weight is given, we will first calculate the magnitude of the buoyancy force by applying Equation (6.5).

[latex]F_{B} = ho _{fluid} g V_{object}[/latex] (6.5)

The volume of the barrel is converted to the dimensionally consistent units of [latex]ft^3[/latex] by using the conversion factor in Table 3.5:

Table 3.5 Certain Derived Units in the USCS

Quantity	Derived Unit	Abbreviation	Definition
Length	mil	mil	1 mil = 0.001 in.
	inch	in.	1 in. = 0.0833 ft
	mile	mi	1 mi = 5280 ft
Volume	gallon	gal	[latex] 1 gal = 0.1337 ft^3 [/latex]
Mass	slug	slug	[latex]1 slug = 1 (lb . S^2)/ft[/latex]
Mass	pound-mass	lbm	[latex] 1 lbm = 3.1081 imes 10^{-2}(lb s^{2})/ft [/latex]
Force	ounce	oz	1 oz = 0.0625 lb
Force	ton	ton	1 ton = 2000 lb
Torque, or moment of a force	foot-pound	ft · lb	—
Pressure or stress	[latex]pound/inch^2[/latex]	psi	[latex]1 psi = 1 lb/in^2[/latex]
Energy, work, or heat	foot-pound	ft · lb	—
Energy, work, or heat	British thermal unit	Btu	1 Btu = 778.2 ft . lb
Power	horsepower	hp	1 hp = 550 (ft . lb)/s
Temperature	degree Fahrenheit	[latex]°F[/latex]	[latex]°F = °R-459.67[/latex]
Although a change in temperature of [latex] 1°[/latex] Rankine also equals a change of 1 [latex]°F[/latex], numerical , numerical values are converted using the formula.

[latex] V=(55 gal)\left(0.1337 \frac{ft^3}{gal} ight) [/latex]

[latex]=7.354(\cancel{gal})(\frac{ft^3}{\cancel {gal}})[/latex]

[latex]=7.354 ft^3[/latex]

The buoyancy force becomes

[latex]F_{B}=\left(1.99\frac{slugs}{ft^3} ight)\left(32.2\frac{ft}{s^2} ight)(7.354 ft^3) \longleftarrow F_{B} = ho _{fluid} g V_{object}[/latex]
[latex]=471.2\left(\frac{slugs}{\cancel{ft^3}} ight)(\frac{ft}{s^2} )(\cancel{ft^3})[/latex]

[latex]=471.2\frac{slug.ft}{s^2} [/latex]

[latex]= 471.2 lb[/latex]

where we have used the definition of the pound from Equation (3.3).

[latex]1 lb = (1 slug)\left(1\frac{ft}{s^2} ight) =1\frac{slug.ft}{s^2} [/latex] (3.3)

By referring to the free body diagram and the positive sign convention shown on it, the equilibrium-force balance for the barrel becomes

[latex]F_B-T-\omega =0 \longleftarrow \left[\sum\limits_{i=1}^{N}{F_y,i} =0 ight] [/latex]

which we solve for the cable’s tension:

[latex]T = 471.2 lb - 35 lb[/latex]

=436.2 lb

Discussion
As far as the shark is concerned, it feels the cable’s tension, which is the difference between the buoyancy force and the barrel’s weight. If the barrel instead weighed the same amount as FB, then the cable’s tension would be zero. In that event, the barrel would be neutrally buoyant and the shark would have to work much less to drag the barrel.

[latex]T = 436.2 lb[/latex]

Question 6.4: Great White Shark Attack In the classic thriller movie Jaws,...

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