g A milk truck carries milk with density 64.6 lb/ft3 in a horizontal cylindrical tank with diameter 6 ft. (a) Find the force exerted by the milk on one end of the tank when the tank is full. (Round your answer to the nearest integer.) lb (b) What if the tank is half full

a) 12989 lb

b) 2756.26 lb

Explanation:

given the data in the question;

pressure on the strip is approximately;

Sdi = 64.6 ( 4 - yi )

a)

Total force is

F = Lim_n→∞  ∑^n_i=1  64.6 ( 4 - yi ) ( 2 ) ( √( 16 - (yi)²) Δy

F = 129.2 ₋₄∫⁴ (4 - y) ( √16 - y²) Δy

F = 129.2(4) ₋₄∫⁴ (4 - y) ( √16 - y²)  -  129.2 ₋₄∫⁴ y( √16 - y²) Δy

{area of semicircle of radius 4                 {add function and limits are -4 to 4

π/2(4²) = 8π}                                                    so this integral is 0}

F = (129.2) (4) ( 8π )

F = 12988.60 ≈ 12989 lb

b)

Tank is half full,

surface of the milk is y=0

so

Sdi = 64.6 ( 0 - yi )

total force will be

F = 129.2 ₋₄∫⁰ (0 - y) ( √16 - y²) Δy

now lwt 16-y²= t                                                              limits of t

-2yΔy = Δt                                                                        for y = -4

-yΔy = 1/2Δt                                                                          t = 0

                                                                                        for y=0, t = 16

so

F = 129.2 ₀∫¹⁶ (t)^1/2 ( 1/2 Δt)

F = 129.2/2 [ (t^3/2) / (3/2) ]₀¹⁶

F = 129.2/2 (2/3) [ 64 ]

F = 2756.26 lb

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thermal energy is the answer