Answer :
Answer:
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