Hydrostatic Pressure and Force — Question 7

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Question 7

A rectangular dam face is bb meters wide and extends from the surface to depth hh. Derive F(h)F(h). If hh doubles, by what factor does FF change?

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Original worksheet page 1: question and worked solution for 2-4-007
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Question 7 – Solution

See the diagram in the original worksheet below.

Step 1: Choose the depth variable. Let yy measure depth below the surface. The dam face extends from y=0y=0 to y=hy=h.

Step 2: Write the force on a horizontal strip. The pressure is p(y)=γyp(y)=\gamma y, and a strip of width bb and thickness dydy has area dA=bdydA=b\,dy. Therefore, dF=p(y)dA=γbydy.dF=p(y)\,dA=\gamma by\,dy.

Step 3: Integrate from the surface to depth hh. F(h)=γb∫0hydy=γb[y22]0h=12γbh2.\begin{align*} F(h)&=\gamma b\int_0^h y\,dy\\ &=\gamma b\left[\frac{y^2}{2}\right]_0^h =\frac12\gamma bh^2. \end{align*} F(h)=12γbh2\boxed{F(h)=\frac12\gamma bh^2}

Step 4: Determine the effect of doubling the depth. F(2h)=12γb(2h)2=4(12γbh2)=4F(h).F(2h)=\frac12\gamma b(2h)^2 =4\left(\frac12\gamma bh^2\right)=4F(h). Doubling h multiplies the force by 4.\boxed{\text{Doubling }h\text{ multiplies the force by }4.}

Original worksheet page 2: question and worked solution for 2-4-007

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