Math Problem Statement

A vertical dam retains 10 m of water. It is pierced at the base by a rectangular door 4 m wide and 3 m high, as indicated in the figure below. Maple plot a) Let y denote the height in meters measured from the base of the dam. The hydrostatic force exerted by the water on the portion of the door comprised between y m and y+Δy m is approximately f(y)Δy N . What is f(y) ? Note that the density of water is ρ=1000 Kg/m3 and the acceleration due to gravity on the Earth’s surface is g=9.8 m/s2 . Express your answer as a formula. Answer: f(y)= Preview Change entry mode b) In Newtons, what is the total hydrostatic force exerted on the door? Give the exact value. Answer:

Solution

To solve these parts, let's break down the question into the components provided:

Part (a)

We need to find the hydrostatic force f(y)f(y) exerted by the water on the rectangular door between the height yy and y+Δyy + \Delta y.

Hydrostatic Force Formula

The hydrostatic force on a small area at a depth is given by the formula: f(y)=PAf(y) = P \cdot A where:

  • PP is the pressure at a depth yy,
  • AA is the area of the door segment at depth yy.

Pressure at Depth yy

The pressure at depth yy below the surface of the water is given by the hydrostatic pressure formula: P=ρgyP = \rho g y where:

  • ρ=1000kg/m3\rho = 1000 \, \text{kg/m}^3 is the density of water,
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2 is the acceleration due to gravity,
  • yy is the depth from the surface of the water.

Area of the Door Segment

The area AA of the door segment between yy and y+Δyy + \Delta y is the width of the door (which is 4 meters) multiplied by the small height Δy\Delta y: A=4m×ΔyA = 4 \, \text{m} \times \Delta y

Force f(y)f(y)

Now, the force f(y)f(y) exerted on the small segment is: f(y)=PA=(ρgy)(4Δy)f(y) = P \cdot A = (\rho g y) \cdot (4 \Delta y) Substitute the known values for ρ\rho and gg: f(y)=(1000×9.8×y)(4Δy)=39200yΔyf(y) = (1000 \times 9.8 \times y) \cdot (4 \Delta y) = 39200 y \Delta y Thus, the expression for the hydrostatic force is: f(y)=39200yΔyNf(y) = 39200 y \Delta y \, \text{N}

Part (b)

Now, we need to calculate the total hydrostatic force exerted on the entire door. This is obtained by integrating the force over the height of the door, from y=0y = 0 (at the base) to y=3y = 3 meters (the top of the door).

Total Force

The total hydrostatic force FF is the integral of f(y)f(y) from y=0y = 0 to y=3y = 3: F=0339200ydyF = \int_0^3 39200 y \, dy To calculate this: F=3920003ydy=39200[y22]03F = 39200 \int_0^3 y \, dy = 39200 \left[ \frac{y^2}{2} \right]_0^3 F=39200×92=39200×4.5=176400NF = 39200 \times \frac{9}{2} = 39200 \times 4.5 = 176400 \, \text{N}

Final Answers:

  • (a) The hydrostatic force exerted on the door between yy and y+Δyy + \Delta y is: f(y)=39200yΔyNf(y) = 39200 y \Delta y \, \text{N}
  • (b) The total hydrostatic force exerted on the door is: F=176400NF = 176400 \, \text{N}

Would you like more details on the steps or have any questions about the calculations? Here are some related questions:

  1. What is the effect of increasing the width of the door on the total hydrostatic force?
  2. How would the force change if the water was deeper, say 20 m instead of 10 m?
  3. What happens to the force if the door were positioned at different heights from the base of the dam?
  4. How would the hydrostatic force be affected if the density of water were to change (e.g., salty water)?
  5. How is the hydrostatic pressure formula derived in terms of fluid statics?

Tip: Remember that the force increases linearly with depth, so the bottom of the door experiences the highest pressure!

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Math Problem Analysis

Mathematical Concepts

Hydrostatics
Integral Calculus
Fluid Mechanics

Formulas

Hydrostatic Pressure: P = ρgy
Force on a Small Area: f(y) = P * A
Total Force: F = ∫ f(y) dy

Theorems

Fundamental Theorem of Calculus
Hydrostatic Pressure Law

Suitable Grade Level

University (Engineering, Physics, or Advanced Calculus)