Question

Determine the force of water against the plate using Simpson’s rule. The density of water is\(\delta = 64{\rm{lb}}/{\rm{f}}{{\rm{t}}^3}\).

Short answer

The force of water against the plate is\(4148{\rm{lb}}\).

Step-by-step solution

The Simpson’s rule.

Simpson's Ruleis a numerical method that approximates the value of a definite integral by using quadratic functions.

Use the Simpson’s rule for calculation.

The density of water is\(\delta = 64{\rm{lb}}/{\rm{f}}{{\rm{t}}^3}\).

The depth and width of the plate as shown in table 1.

The number of subintervals as\({\rm{n = 7}}\).

The width of the plate is measured at equal interval of depth. Hence, the subinterval width\((\Delta x)is0.4.\)Calculate the force of the water against the plate using Simpson's Rule:

\(F = \frac{{\Delta x\delta }}{3}\left( {\begin{aligned}{}{{x_1}{w_1}\left( {{x_1}} \right) + 4{x_2}{w_2}\left( {{x_2}} \right) + 2{x_3}{w_3}\left( {{x_3}} \right) + }\\{4{x_4}{w_4}\left( {{x_4}} \right) + 2{x_5}{w_5}\left( {{x_5}} \right) + 4{x_6}{w_6}\left( {{x_6}} \right) + }\\{{x_7}{w_7}\left( {{x_7}} \right)}\end{aligned}} \right)\) …… (1)

Substitute\(0.4{\rm{ for }}\Delta x,7.0{\rm{ for }}{x_1},7.4{\rm{ for }}{x_2},7.8{\rm{ for }}{x_3},8.2{\rm{ for }}{x_4},8.6{\rm{ for }}{x_5},9.0{\rm{ for }}{x_6}{\rm{,}}\)and\(9.4{\rm{ for }}{x_7}\)in equation (1).

\(\begin{aligned}{}F = \frac{{0.4\delta }}{3}\left( {\begin{aligned}{}{7.0{w_1}(7.0) + 4 \times 7.4{w_2}(7.4) + 2 \times 7.8{w_3}(7.8) + }\\{4 \times 8.2{w_4}(8.2) + 2 \times 8.6{w_5}(8.6) + 4 \times 9.0{w_6}(9.0) + }\\{9.4{w_7}(9.4)}\end{aligned}} \right)\\ = \frac{{0.4\delta }}{3}\left( {\begin{aligned}{}{7.0{w_1}(7.0) + 29.6{w_2}(7.4) + 15.6{w_3}(7.8) + }\\{32.8{w_4}(8.2) + 17.2{w_5}(8.6) + 36{w_6}(9.0) + }\\{9.4{w_7}(9.4)}\end{aligned}} \right)\end{aligned}\) …… (2)

Substitute,\(64{\rm{lb}}/{\rm{f}}{{\rm{t}}^3}for\delta ,1.2for{w_1}(7.0),1.8for{w_2}(7.4),2.9for{w_3}(7.8),3.8for{w_4}(8.2),3.6for{w_5}(8.6),4.2for{w_6}(9.0),\)and\(4.4for{w_7}(9.4)\)in equation (1).

\(\begin{aligned}{}F = \frac{{0.4 \times 64}}{3}\left( {\begin{aligned}{}{7.0(1.2) + 29.6(1.8) + 15.6(2.9) + }\\{32.8(3.8) + 17.2(3.6) + 36(4.2) + 9.4(4.4)}\end{aligned}} \right)\\ = \frac{{25.6}}{3}(8.4 + 53.28 + 45.24 + 124.64 + 61.92 + 151.2 + 41.36)\\ = \frac{{25.6}}{3}(486.04)\\ = 4147.54{\rm{lb}}\\ \approx 4148{\rm{lb}}\end{aligned}\)

Hence, the force of water against the plate is\(4148{\rm{lb}}\).