Question

Explain the ways in which the hydrostatic force is approximated against one side of the plate by a Riemann sum, also express the force and evaluate it.

Short answer

The integral for the hydrostatic force is \({\rm{ }}\int_0^5 2 5{y^2}dy\){and is evaluated\(1041\frac{1}{3}{\rm{lb}}{\rm{. }}\)

Step-by-step solution

Formula of pressure and force.

The pressure on the object placed in a medium of density\(\rho \)at a depth \(x\)is obtained as\(\rho x\), that is\(P = \rho x\).

The force exerted by the medium on the object is obtained as\(F = PA\)where \(A\)is the area of the object on which the pressure is exerted.

Use the formula of pressure and force for calculation.

It is given that the vertical triangular plate has a height of \(5m\) and the base of\(4m\).

The whole plate is under the water.

Assume the horizontal axis at the tip of the plate.

Let the vertical axis passes through the middle of the plate.

Consider the \({\rm{ith}}\) strip of width \(\Delta y\) be at the distance \(y_i^*\) below the horizontal axis such that the area is \(2 \times \frac{1}{2} \times \left( {\frac{2}{5}y} \right) \times \Delta y = \frac{2}{5}y\Delta y\).

The pressure of the \(ith\) strip is\(2 \times \frac{1}{2} \times \left( {\frac{2}{5}y} \right) \times \Delta y = \frac{2}{5}y\Delta y\).

The force is computed as

\({F_i} = {P_i}{A_i}\)

\( = \rho {d_i}{A_i}\)

\( = 62.5y_i^*\left( {\frac{2}{5}y_i^*\Delta y} \right)\)

\( = 25{\left( {y_i^*} \right)^2}\Delta y\)

If there are \(n\) strips the sum of forces becomes\(\sum\limits_{i = 1}^n {{F_i}} = \sum\limits_{i = 1}^n 2 5{\left( {y_i^*} \right)^2}\Delta y\).

Apply limit\(n\) tends to\(\infty \)in order to obtain the integral as follows.

\(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{F_i}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n 2 5{\left( {y_i^*} \right)^2}\Delta y = \int_0^5 2 5{y^2}dy\)

Compute the integral:

\(\int_0^5 2 5{y^2}dy = 25\left[ {\frac{{{y^3}}}{3}} \right]_0^5\)

\( = 25\left[ {\frac{{{{(5)}^3}}}{3} - 0} \right]\)

\( = \frac{{3125}}{3}\)

\( = 1041\frac{1}{3}{\rm{lb}}\)

Therefore, the integral for the hydrostatic force is \({\rm{ }}\int_0^5 2 5{y^2}dy\){and is evaluated\(1041\frac{1}{3}{\rm{lb}}{\rm{. }}\)