Hydrostatic Pressure and Force — Question 10

PDF ↗

Question 10

A horizontal hatch of constant area A>0A>0 is slowly lowered from depth dd to d+hd+h, where d,h>0d,h>0. Assume an upward resisting hydrostatic force F(y)=γAyF(y)=\gamma Ay at depth yy, with constant weight density γ>0\gamma>0. Find the work done against this force, ignoring other forces.

See the diagram in the original worksheet below.

Original worksheet page 1: question and worked solution for 2-4-010
Show solutionHide solution

Question 10 – Solution

See the diagram in the original worksheet below.

Step 1: State the force model. Assume the hydrostatic force resisting the downward motion has magnitude F(y)=p(y)A=γyA,F(y)=p(y)A=\gamma yA, where yy is the hatch depth. The area AA is constant.

Step 2: Write a differential amount of work. Moving the hatch through a small downward distance dydy requires dW=F(y)dy=γAydy.dW=F(y)\,dy=\gamma Ay\,dy.

Step 3: Integrate over the change in depth. The hatch moves from y=dy=d to y=d+hy=d+h, so W=∫dd+hγAydy=γA[y22]dd+h=γA2((d+h)2−d2).\begin{align*} W&=\int_d^{d+h}\gamma Ay\,dy\\ &=\gamma A\left[\frac{y^2}{2}\right]_d^{d+h}\\ &=\frac{\gamma A}{2}\left((d+h)^2-d^2\right). \end{align*}

Step 4: Simplify. W=γA2(d2+2dh+h2−d2)=γA(dh+h22).\begin{align*} W&=\frac{\gamma A}{2}\left(d^2+2dh+h^2-d^2\right)\\ &=\gamma A\left(dh+\frac{h^2}{2}\right). \end{align*} W=γA(dh+h22)\boxed{W=\gamma A\left(dh+\frac{h^2}{2}\right)} This is the external work under the stated resisting-force assumption.

Original worksheet page 2: question and worked solution for 2-4-010

Original worksheet layout. Use Enlarge or open the PDF for a closer view.