Hydrostatic Force

Question 1

What is the general force for hydrostatic force

Answer 1

F = pw(x)s(x)dx, where w(x) is the width at a certain point, and s(x) is the depth at the same corresponding point, and p is the weight density of water.

Question 2

What are the problem solving steps for hydrostatic force?

Answer 2

1. Graph it/ Pick a system of coordinates
2. Identify width and width function
3. Determine depth function, and make sure that it properly matches up with the width function
4. Determine the integration bounds
5. Integrate using the proper integration techniques, make sure to be creative.

Question 3

A vertical rectangular gate in a dam is 5m wide and 10m high. The top of the gate is 2m below the water surface. Find the hydrostatic force (w = 9800 N/m^3).

Answer 3

F = ∫₂¹² 9800 * y * 5 dy = 3,430,000 N

Question 4

A vertical triangular plate with base 4ft and height 6ft is submerged in water with its vertex at the surface and base at the bottom. Find the force (w = 62.4 lb/ft^3).

Answer 4

F = ∫₀⁶ 62.4 * y * (4/6 * y) dy = 2,995.2 lb

Question 5

A vertical circular plate of radius 3m is submerged so that its center is 10m below the surface. Find the force.

Answer 5

F = ∫₋₃³ 9800 * (10 - y) * 2√(9 - y²) dy = 2,770,884 N

Question 6

A vertical plate shaped by y = x² (0 ≤ x ≤ 2m) is submerged with the top (y=4) at the water surface. Find the force.

Answer 6

F = ∫₀⁴ 9800 * (4 - y) * 2√y dy = 167,253 N

Question 7

A semicircular plate of radius 2ft is submerged vertically with its diameter at the surface. Find the force.

Answer 7

F = ∫₀² 62.4 * y * 2√(4 - y²) dy = 332.8 lb

Question 8

An inverted isosceles triangular gate has height 4m and top width 6m. The top is at the surface. Find the force.

Answer 8

F = ∫₀⁴ 9800 * y * (6 - 1.5y) dy = 156,800 N

Question 9

A vertical elliptical plate x²/4 + y²/9 = 1 is submerged with its center at 5m depth. Find the force.

Answer 9

F = ∫₋₃³ 9800 * (5 - y) * 2√(4(1 - y²/9)) dy = 588,000π N

Question 10

A trapezoidal plate (top 10ft, bottom 6ft, height 4ft) is submerged with the top 2ft below the surface. Find the force.

Answer 10

F = ∫₂⁶ 62.4 * y * (10 - (y-2)) dy = 8,320 lb

Question 11

A square plate (side 2m) is submerged at a 45° angle. The top edge is 1m below the surface. Find the force.

Answer 11

F = ∫₀^(2sin45) 9800 * (1 + y) * 2 dy = 37,716 N

Question 12

A vertical plate is bounded by y = sin(x) and y = 0 for 0 ≤ x ≤ π. The surface is at y = 2. Find the force.

Answer 12

F = ∫₀¹ 9800 * (2 - y) * (π - 2arcsin(y)) dy (Requires IBP)

Question 13

A tank end is bounded by y = x⁴ and y = 16m. Oil density is 900 kg/m³. Find the force when full.

Answer 13

F = ∫₀¹⁶ (900*9.8) * (16 - y) * 2y^(1/4) dy = 1,185,408 N

Question 14

An equilateral triangle (side 2m) is submerged vertically with one side at the surface. Find the force.

Answer 14

F = ∫₀^√3 9800 * y * (2 - (2/√3)y) dy = 9,800 N

Question 15

A vertical plate is bounded by y = eˣ, y = 1, and x = 1. The surface is at y = e. Find the force.

Answer 15

F = ∫₁^e 9800 * (e - y) * (1 - ln y) dy = 9800(e/2 - 1) N

Question 16

A circular window (radius 1ft) in a tank is centered 8ft below the water surface. Find the force.

Answer 16

F = ∫₋₁¹ 62.4 * (8 - y) * 2√(1 - y²) dy = 499.2π lb

Question 17

A vertical gate is bounded by y = 4 - x² and y = 0. The surface is at y = 5. Find the force.

Answer 17

F = ∫₀⁴ 9800 * (5 - y) * 2√(4 - y) dy = 418,133 N

Question 18

A trough has a triangular end (base 3ft, altitude 2ft). Find the force on the end when full.

Answer 18

F = ∫₀² 62.4 * y * (1.5(2-y)) dy = 124.8 lb

Question 19

A vertical rhombus plate (diagonals 4m and 6m) has the top vertex at the surface. Find the force.

Answer 19

F = ∫₀³ 9800y(4/3y)dy + ∫₃⁶ 9800y(4 - 4/3(y-3))dy = 352,800 N

Question 20

A dam face is shaped by y = 100 - x²/25. Find the force if water is at y = 100.

Answer 20

F = ∫₀¹⁰⁰ 9800 * (100 - y) * 10√(100 - y) dy = 392,000,000 N

Question 21

A rectangular plate (2m x 4m) is submerged in fluid with variable density ρ = 1000 + 10y. Top is at y=0.

Answer 21

F = ∫₀⁴ (1000 + 10y) * 9.8 * y * 2 dy = 106,667 N

Question 22

A vertical plate is bounded by y = 1/x, x = 1, x = 2, and y = 0. Surface is at y = 2. Find the force.

Answer 22

F = ∫₁² 9800 * (2 - 1/x) * 1 dx = 9800(2 - ln 2) N

Question 23

A right triangle (legs 3m, 4m) has the 4m leg vertical and 3m leg at the surface. Find the force.

Answer 23

F = ∫₀⁴ 9800 * y * (3 - 0.75y) dy = 78,400 N

Question 24

A semi-ellipse x²/16 + y²/4 = 1 (y ≤ 0) is submerged with diameter at the surface. Find the force.

Answer 24

F = ∫₋₂⁰ 9800 * (-y) * 8√(1 - y²/4) dy = 209,067 N

Question 25

A vertical plate is bounded by y = √x, y = 0, and x = 4. The surface is at y = 3. Find the force.

Answer 25

F = ∫₀² 9800 * (3 - y) * (4 - y²) dy = 104,533 N

Question 26

Find the force on a vertical square of side 's' if the top edge is at depth 'd'.

Answer 26

F = ∫_d^(d+s) w * y * s dy = ws(sd + s²/2)

Question 27

A vertical gate is shaped like the area between y = x² and y = 8 - x². Surface at y = 10.

Answer 27

Sum of ∫₀⁴ 9800(10-y)2√y dy and ∫₄⁸ 9800(10-y)2√(8-y) dy

Question 28

A vertical plate is defined by 0 ≤ y ≤ ln x, 1 ≤ x ≤ e. The surface is at y = 2. Find the force.

Answer 28

F = ∫₀¹ 9800 * (2 - y) * (e - eʸ) dy = 9800(e - 2) N

Question 29

A circle sector (r=2m, 60°) is submerged with one radius at the surface. Find the force.

Answer 29

F = ∫₀^(2sin60) 9800 * y * (√(4 - y²) - y/tan60) dy

Question 30

A vertical plate is bounded by y = cos x, y = -1, x = 0, x = π. Surface at y = 1. Find the force.

Answer 30

F = ∫₋₁¹ 9800 * (1 - y) * π dy = 19,600π N

Question 31

Find the force on a vertical circle of radius R whose top is tangent to the surface.

Answer 31

F = ∫₀²ᴿ w * y * 2√(R² - (y-R)²) dy = wπR³

Question 32

A vertical arch y = 4 - x² has fluid surface at y = 4. Find the force.

Answer 32

F = ∫₀⁴ ρg * (4 - y) * 2√(4 - y) dy = 12.8ρg

Question 33

What is the formula for pressure/

Answer 33

F = pgs; where p is density, g is gravity, and s is depth

Question 34

What is the formula for force on an object with constant depth?

Answer 34

F = pg(inta-b of A(x)dx)s, where A is the total area of the object, and s is a constant.