IndietroMAC 2312 Exam 1 Review – Calculus Study Guidance
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Q4. The flat surface shown is submerged vertically in a liquid of weight density 31.2 lb/ft3. Find the fluid force against the surface.
Background
Topic: Applications of Integration – Fluid Force
This question tests your ability to use calculus to compute the force exerted by a fluid on a submerged surface. The force depends on the depth, area, and weight density of the fluid.
Key Terms and Formulas
Fluid Force ($F$): The total force exerted by a fluid on a submerged surface.
Weight Density ($w$): The weight per unit volume of the fluid, given as $31.2$ lb/ft3.
Depth ($y$): The vertical distance below the surface of the liquid.
Formula for fluid force on a vertical surface: $F = \int_{a}^{b} w \cdot y \cdot L(y) \, dy$ where $L(y)$ is the width of the surface at depth $y$.
Step-by-Step Guidance
Examine the diagram to determine the shape and dimensions of the submerged surface. For part (a), the surface is a right triangle with legs of $6$ ft and $8$ ft, and a hypotenuse of $10$ ft. The top of the triangle is $2$ ft below the surface of the liquid.
Set up a coordinate system with $y = 0$ at the surface of the liquid, increasing downward. The triangle extends from $y = 2$ ft to $y = 8$ ft below the surface.
Express the width $L(y)$ of the triangle at depth $y$ in terms of $y$. Use similar triangles or linear relationships to find $L(y)$ as a function of $y$.
Write the integral for fluid force: $F = \int_{2}^{8} 31.2 \cdot y \cdot L(y) \, dy$ Plug in the expression for $L(y)$ from the previous step.
Set up the integral, but do not evaluate it yet. Make sure all units are consistent (feet).
Try solving on your own before revealing the answer!
Final Answer:
For part (a), the fluid force is $F = 31.2 \int_{2}^{8} y \cdot \frac{6}{6} (y-2) \, dy = 31.2 \int_{2}^{8} y \cdot \frac{6}{6} (y-2) \, dy$ (after expressing $L(y)$ as $\frac{6}{6}(y-2)$). Evaluating the integral gives $F \approx 2,246.4$ lb.
The calculation uses the weight density, depth, and width function to find the total force.
Q4b. The flat surface shown is submerged vertically in a liquid of weight density 31.2 lb/ft3. Find the fluid force against the surface (semicircle).
Background
Topic: Applications of Integration – Fluid Force
This question tests your ability to compute the fluid force on a submerged semicircular surface using calculus.
Key Terms and Formulas
Fluid Force ($F$): The total force exerted by a fluid on a submerged surface.
Weight Density ($w$): The weight per unit volume of the fluid, given as $31.2$ lb/ft3.
Depth ($y$): The vertical distance below the surface of the liquid.
Formula for fluid force on a vertical surface: $F = \int_{a}^{b} w \cdot y \cdot L(y) \, dy$ where $L(y)$ is the width of the surface at depth $y$.
Step-by-Step Guidance
Examine the diagram to determine the shape and dimensions of the submerged surface. For part (b), the surface is a semicircle with a radius of $3$ in. Convert units to feet if needed.
Set up a coordinate system with $y = 0$ at the surface of the liquid, increasing downward. The semicircle extends from $y = 0$ to $y = 3$ in below the surface.
Express the width $L(y)$ of the semicircle at depth $y$ in terms of $y$. Use the equation of a circle to find $L(y)$ as a function of $y$.
Write the integral for fluid force: $F = \int_{0}^{3} 31.2 \cdot y \cdot L(y) \, dy$ Plug in the expression for $L(y)$ from the previous step.
Set up the integral, but do not evaluate it yet. Make sure all units are consistent (convert inches to feet if necessary).
Try solving on your own before revealing the answer!
Final Answer:
For part (b), the fluid force is $F = 31.2 \int_{0}^{0.25} y \cdot 2 \sqrt{(0.25)^2 - (y-0.25)^2} \, dy$ (after converting $3$ in to $0.25$ ft and expressing $L(y)$ for the semicircle). Evaluating the integral gives $F \approx 0.51$ lb.
The calculation uses the weight density, depth, and width function to find the total force.
