Dam Gate!

Find the hydrostatic pressure to get the force on the rectangular stripe. Then find the torque done at the stripe.

The torque is given by:

\begin{equation*}
\tau = F d.
\end{equation*}

The hydrostatic pressure at a depth \(y\) below the water is:

\begin{equation*}
P = P_{\text{atm}} + \rho_w g y.
\end{equation*}

The force is the product of the pressure with the area. The differential force on the rectangular stripe is:

\begin{equation*}
dF = ( P_{\text{atm}} + \rho_w g y ) L dy.
\end{equation*}

The differential force produced by the atmosphere is just proportional with \(P_{\text{atm}}\). Then, the total force is:

\begin{equation*}
dF = \rho_w g y L dy.
\end{equation*}

The torque is the last force multiplied by \(H-y\). As an integral is:

\begin{equation*}
\tau_T = \int_0^H L \rho_w g ( H y dy \, – y^2 dy).
\end{equation*}

By performing the integral we get:

\begin{equation*}
\tau_T = \frac{1}{6} L \rho_w g H^3.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

The problem is asking us to find the magnitude of the torque that the water exerts on the dam gate. Let’s recall that the magnitude of the torque produced by a force perpendicular to an object is defined as the force multiplied by the distance from the axis of rotation (in this case the hinge at the bottom of the gate). There are forces acting over the entire surface of the gate that arise from the hydrostatic pressure and the atmospheric pressure. Since the hydrostatic pressure depends on the depth, these forces will continuously change as a function of the depth. For this reason, we must first write the contributions to the total torque for a given depth as differentials that we can then integrate across the entire height of the gate to obtain the total torque.

The forces acting on the gate are produced by the hydrostatic pressure and the atmospheric pressure on the side of the gate that is in contact with the water, and by the atmospheric pressure only on the opposite side of the gate. These forces are perpendicular to the surface of the gate and are directed towards it. The magnitude of the torque \(\tau\) produced by a force \(F\) perpendicular to the axis of rotation of an object is given by

\begin{equation}
\label{EQ:torque}
\tau = F d,
\end{equation}

where \(d\) is the distance from the axis of rotation to the point in the gate at which the force is being exerted, which in this case is the hinge at the bottom of the gate. We should first write expressions for the forces on each side of the gate and then use eq. \eqref{EQ:torque} to find the torque produced by these forces.

Let’s first consider the side of the gate that is in contact with the water. The hydrostatic pressure at a point located at a depth \(y\) below the water surface can be written as

\begin{equation}
\label{EQ:hydro}
P = P_{\text{atm}} + \rho_w g y.
\end{equation}

On the other hand, by definition, pressure is defined as force per unit area i.e.

\begin{equation}
P = \frac{F}{A},
\end{equation}

and if we multiply by \(A\), we find that the force as a function of pressure is given by

\begin{equation}
\label{EQ:f_p}
F = PA.
\end{equation}

Hence, the force that produces the torque at a certain point on the gate will depend on the pressure at that point. Also, according to eq. \eqref{EQ:hydro}, this pressure depends on the distance from the water surface. Therefore, the contributions to the total torque will change according to the distance from the water surface. We should write these contributions as torque differentials as a function of the distance from the surface and then integrate over all these differentials to obtain the total torque.

Since the force associated with the hydrostatic pressure varies with depth, it should be constant over a small area of the gate at a certain depth. With this in mind, let’s consider a portion of the gate of height \(dy\), centered at a distance \(y\) from the water surface, and extending over the entire width \(L\) of the gate, as shown in figure 1.

Figure 1: We place the coordinate system such that the Y-axis points downwards in the same direction as gravity and its origin is on the free surface of water. In red, we see the area differential of length \(L\) and height \(dy\) placed at a depth \(y\). The differential force vectors \(d\vec{F}\) and \(d\vec{F}’\) are also shown.

The area \(dA\) of the rectangular stripe is given by

\begin{equation}
dA = L dy,
\end{equation}

and thus according to eq. \eqref{EQ:f_p}, the force \(dF\) exerted on the rectangular stripe is equal to the pressure multiplied by the area of the stripe i.e.

\begin{equation}
dF = P dA = P L dy,
\end{equation}

and substituting eq. \eqref{EQ:hydro} gives

\begin{equation}
\label{EQ:r_df}
dF = ( P_{\text{atm}} + \rho_w g y ) L dy.
\end{equation}

Now, let’s consider the force from the opposite side of the gate. Only the atmospheric pressure acts from this side, and so we can write that the pressure \(P’\) exerted on the gate from that side is

\begin{equation}
P’ = P_{\text{atm}}.
\end{equation}

Hence, according to eq. \eqref{EQ:f_p}, the magnitude of the force \(dF’\) produced by the atmosphere on this side of the gate is,

\begin{equation}
\label{EQ:r_df’}
dF’ = P_{\text{atm}} dA = P_{\text{atm}} L dy.
\end{equation}

Now, the force \(d \vec{F}\) and the force \(d \vec{F}’\) act in opposite directions on the gate. In the reference frame shown in the previous figure, \(d \vec{F}\) points towards the \(+\hat{i}\) direction, and the force \(d \vec{F}’\) points towards the -hat{i} direction. Hence, the total force \(d \vec{F}_T\) on the stripe of area \(dA\) is given by

\begin{equation}
d \vec{F}_T = d \vec{F} + d \vec{F}’
= d F \hat{i} – d F’ \hat{i},
\end{equation}

and substituting eqs. \eqref{EQ:r_df} and \eqref{EQ:r_df’} gives

\begin{equation}
d \vec{F}_T = ( P_{\text{atm}} + \rho_w g y ) L dy \hat{i} – P_{\text{atm}} L dy \hat{i}.
= P_{\text{atm}} L dy \hat{i} + \rho_w g y L dy \hat{i} – P_{\text{atm}} L dy \hat{i}.
\end{equation}

The first and third terms cancel out yielding

\begin{equation}
d \vec{F}_T = \rho_w g y L dy \hat{i},
\end{equation}

and thus the magnitude \(d F_T\) is

\begin{equation}
dF_T = \rho_w g y L dy.
\end{equation}

Intuitively, the fact that the contribution to the force produced by the atmospheric pressure cancels out is expected because it acts from both sides of the gate, producing forces that are equal in magnitude but have opposite directions.

Now, according to the second figure, the distance between the stripe and the hinge at the bottom of the gate is \(H – y\). Therefore, from eq. \eqref{EQ:torque} the magnitude of the total torque \(d\tau_T\) produced by the force \(dF_T\) is given by

\begin{equation}
d\tau_T = dF_T (H – y) = \rho_w g y L dy (H – y),
\end{equation}

which we can rewrite as

\begin{equation}
d \tau_T = L \rho_w g (H – y) y dy
= L \rho_w g ( H y dy \, – y^2 dy).
\end{equation}

Finally, in order to obtain the total magnitude of the torque, we should sum the contributions from all the stripes at all depths \(y\) from \(0\) to \(H\). This corresponds to integrating \(d\tau_T\) with respect to \(y\) from \(y = 0\) to \(y = H\), and it can be written as

\begin{equation}
\tau_T = \int_0^H d\tau_T = \int_0^H L \rho_w g ( H y dy \, – y^2 dy).
\end{equation}

Now, the constant terms can be taken out of the integral. Hence, we can rewrite it as

\begin{equation}
\tau_T = L \rho_w g \int_0^H ( H y dy \, – y^2 dy).
\end{equation}

Also, if we distribute the integral over both terms, we obtain

\begin{equation}
\label{EQ:ints}
\tau_T = L \rho_w g \left( \int_0^H H y dy \, – \int_0^H y^2 dy\right)
= L \rho_w g \left( H \int_0^H y dy \, – \int_0^H y^2 dy\right).
\end{equation}

We can use the power rule for integration in order to solve both integrals. The power rule states that for all \(n \neq -1\), we have

\begin{equation}
\int_a^b y^{n} dy = \frac{y^{n+1}}{n+1}\Big|_a^b = \frac{b^{n+1}}{n+1} – \frac{a^{n+1}}{n+1}.
\end{equation}

We can apply this rule for \(n = 1\) to solve the first integral in eq. \eqref{EQ:ints}:

\begin{equation}
\label{EQ:y1}
\int_0^H y dy = \int_0^H y^{1} dy = \frac{y^{1+1}}{1+1}\Big|_0^H = \frac{y^{1}}{1}\Big|_0^H = \frac{H^2}{2} – \frac{0^2}{2} = \frac{H^2}{2}.
\end{equation}

In addition, apply the power rule for \(n = 2\) to solve the second integral in eq. \eqref{EQ:ints} as follows:

\begin{equation}
\label{EQ:y2}
\int_0^H y^2 dy = \frac{y^{2+1}}{2+1}\Big|_0^H = \frac{y^{3}}{3}\Big|_{0}^H = \frac{H^3}{3} – \frac{0^3}{3} = \frac{H^3}{3}.
\end{equation}

Substituting eqs. \eqref{EQ:y1} and \eqref{EQ:y2} in eq. \eqref{EQ:ints} gives

\begin{equation}
\tau_T = L \rho_w g \left( H \cdot \frac{H^2}{2} – \frac{H^3}{3}\right) = L \rho_w g \left( \frac{H^3}{2} – \frac{H^3}{3}\right) = L \rho_w g H^3 \left( \frac{1}{2} – \frac{1}{3}\right),
\end{equation}

which is equivalent to

\begin{equation}
\tau_T = \frac{1}{6} L \rho_w g H^3.
\end{equation}

You need to be registered and logged in to take this quiz. Log in or Register