A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. How do you express the hydrostatic force against one side of the plate as an integral and evaluate it?

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A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. How do you express the hydrostatic force against one side of the plate as an integral and evaluate it?

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ $f t^{3} .$ $ft^3.$ )

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Answer 1 · 2017-10-03T20:00:13

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Explanation:

It's $\int_{3}^{9}$ $int_3^9$ $\frac{4}{3}$ $4/3$ $δ$ $delta$ (9 - x)x dx. Solving it gives us:

= $\frac{4}{3}$ $4/3$ $\int_{5}^{9}$ $int_5^9$ $δ$ $delta$ (9 - x)x dx

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ $\int_{3}^{9}$ $int_3^9$ 9x - $x^{2}$ $x^2$ dx

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [ $\frac{9}{2}$ $9/2$ $x^{2}$ $x^2$ - $\frac{1}{3}$ $1/3$ $x^{3}$ $x^3$ $]_{3}^{9}$ $]_3^9$ dx

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [ $\frac{9}{2}$ $9/2$ $9^{2}$ $9^2$ - $\frac{1}{3}$ $1/3$ $9^{3}$ $9^3$ - ( $\frac{9}{2}$ $9/2$ $3^{2}$ $3^2$ - $\frac{1}{3}$ $1/3$ $3^{3}$ $3^3$ )]

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [ $\frac{9}{2}$ $9/2$ $81$ $81$ - $\frac{1}{3}$ $1/3$ $729$ $729$ - ( $\frac{9}{2}$ $9/2$ $9$ $9$ - $\frac{1}{3}$ $1/3$ $27$ $27$ )]

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [ $\frac{729}{2}$ $729/2$ - $\frac{729}{3}$ $729/3$ - ( $\frac{81}{2}$ $81/2$ - $\frac{27}{3}$ $27/3$ )]

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [ $\frac{729}{2}$ $729/2$ - $\frac{729}{3}$ $729/3$ - $\frac{81}{2}$ $81/2$ + $\frac{27}{3}$ $27/3$ ]

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [ $\frac{729}{2}$ $729/2$ - $\frac{81}{2}$ $81/2$ - $\frac{729}{3}$ $729/3$ + $\frac{27}{3}$ $27/3$

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [ $\frac{648}{2}$ $648/2$ - $\frac{702}{3}$ $702/3$ ]

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [324 - 234]

= $\frac{4}{3}$ $4/3$ $δ$ $delta$ [90]

= $\frac{360}{3}$ $360/3$ $δ$ $delta$

= 120 $δ$ $delta$

Note that $δ$ $delta$ is the weight of density of water. Now we must use $δ$ $delta$ = 62.5

= 120(62.5)

= 7500

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A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. How do you express the hydrostatic force against one side of the plate as an integral and evaluate it?

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ $f t^{3} .$ $ft^3.$ )

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. How do you express the hydrostatic force against one side of the plate as an integral and evaluate it?

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft^3.ft3.ft^3.)

1 Answer

Explanation:

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Impact of this question

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ $f t^{3} .$ $ft^3.$ )