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Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge wall

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Presentation on theme: "Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge wall"— Presentation transcript:

1 Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge 156 377 wall
752 1,234 Find the reaction force on the door, F R, in kiloNewtons. [pause]. In this problem, --- wall liquid concrete SG = 2.4

2 Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge 156 377 wall
752 1,234 liquid concrete with a known specific gravity is held back by --- wall liquid concrete SG = 2.4

3 Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge 156 377 wall
752 1,234 a vertical door, which is positioned above a hinge, --- wall liquid concrete SG = 2.4

4 Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge 156 377 wall
752 1,234 which is on top of, a vertical wall segment. wall liquid concrete SG = 2.4

5 Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge 156 377 wall
752 1,234 [pause] The door is triangular in shape, --- wall liquid concrete SG = 2.4

6 Find: FR [kN] door 0.4 [m] 0.4 [m] door FR 0.3 [m] hinge 156 377 wall
752 1,234 measuring 0.4 meters tall and 0.3 meters wide. We’ll assume that the wall --- wall liquid concrete SG = 2.4

7 Find: FR [kN] wall wall door 0.4 [m] door FR wall hinge 156 377 752
1,234 is located around the sides of the door as well, to hold back the liquid concrete. [pause] The reaction force, F R, --- wall liquid concrete SG = 2.4

8 Find: FR [kN] FR= P * A 0.4 [m] door reaction 0.3 [m] force 0.4 [m]
equals P times A, where P represents, --- FR hinge liquid SG = 2.4 concrete wall

9 Find: FR [kN] FR= P * A 0.4 [m] door static pressure reaction 0.3 [m]
force 0.4 [m] door the static pressure exerted against the door by the liquid concrete, and A represents, --- FR hinge liquid SG = 2.4 concrete wall

10 Find: FR [kN] area FR= P * A 0.4 [m] door static pressure reaction
force 0.4 [m] door the area of the door. [pause] Also, we must recognize that, the static pressure, P, --- FR hinge liquid SG = 2.4 concrete wall

11 Find: FR [kN] area FR= P * A 0.4 [m] door static pressure reaction
force 0.4 [m] door is not a contant, and varies with depth, in which case, the reaction force, F R, equals, --- FR hinge liquid SG = 2.4 concrete wall

12 Find: FR [kN] FR= P * A 0.4 [m] door FR= ∫P * A 0.3 [m] reaction force
the INTEGRAL of P times A, over for the entire area of the door, A. In this equation, P equals, --- FR hinge liquid SG = 2.4 concrete wall

13 Find: FR [kN] FR= P * A 0.4 [m] door FR= ∫P * A 0.3 [m] P = ρ * g * y
rho, times g, times y. Where rho, represents, --- FR hinge liquid SG = 2.4 concrete wall

14 Find: FR [kN] FR= P * A 0.4 [m] door FR= ∫P * A 0.3 [m] P = ρ * g * y
fluid 0.4 [m] door the density of the liquid, g, represents, --- density FR hinge liquid SG = 2.4 concrete wall

15 Find: FR [kN] FR= P * A 0.4 [m] door FR= ∫P * A acceleration constant
P = ρ * g * y fluid 0.4 [m] door 9.81 [m/s2] the gravitational accelration constant, 9.81 meters per second squared, and y, represents, ---- density FR hinge liquid SG = 2.4 concrete wall

16 Find: FR [kN] FR= P * A 0.4 [m] door FR= ∫P * A acceleration constant
depth P = ρ * g * y fluid 0.4 [m] door 9.81 [m/s2] the depth, as defined as the distance from the surface of the liquid concrete, --- density FR hinge liquid SG = 2.4 concrete wall

17 Find: FR [kN] y FR= P * A 0.4 [m] door FR= ∫P * A acceleration
constant 0.3 [m] depth P = ρ * g * y fluid 0.4 [m] door 9.81 [m/s2] measured downward to the point of interest. [pause] The fluid density, rho, --- density FR hinge liquid SG = 2.4 concrete wall

18 Find: FR [kN] y FR= P * A 0.4 [m] door FR= ∫P * A acceleration
constant 0.3 [m] depth P = ρ * g * y fluid 0.4 [m] door 9.81 [m/s2] can be calcualted by multiplying the density of water, --- density FR hinge liquid SG = 2.4 concrete wall

19 Find: FR [kN] y FR= P * A 0.4 [m] door FR= ∫P * A acceleration
constant 0.3 [m] depth P = ρ * g * y fluid 0.4 [m] door 9.81 [m/s2] rho w, by the specific gravity of the fluid. [pause] Using 1000 kilograms --- density FR ρ = ρw * SG hinge liquid SG = 2.4 concrete wall

20 Find: FR [kN] y FR= P * A 0.4 [m] door FR= ∫P * A acceleration
constant 0.3 [m] depth P = ρ * g * y fluid 0.4 [m] door 9.81 [m/s2] per meter cubed for rho W, and 2.4 for the specific gravity, ---- density FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

21 Find: FR [kN] y FR= P * A 0.4 [m] door FR= ∫P * A acceleration
constant 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] the density equals, 2,400 kilograms per meters cubed. [pause] The area of the door, --- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

22 Find: FR [kN] y FR= P * A 0.4 [m] area door FR= ∫P * A acceleration
constant 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] A, equals, the width of the door, w, --- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

23 Find: FR [kN] dy y FR= P * A 0.4 [m] A=w * dy w FR= ∫P * A
acceleration constant 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] times the height of the door, dy. [pause] Furthermore, we can define the width, w, --- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

24 Find: FR [kN] w=f(y) dy y FR= P * A 0.4 [m] A=w * dy w FR= ∫P * A
acceleration constant 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] as a function of the depth, y. [pause] From observation, --- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

25 Find: FR [kN] 0[m] = f(0[m]) w=f(y) dy y FR= P * A 0.4 [m] A=w * dy w
acceleration constant 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] we notice at a depth 0 meters, the width equals 0 meters, and at a depth of 0.4 meters, ----- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

26 Find: FR [kN] 0[m] = f(0[m]) w=f(y) dy y FR= P * A 0.4 [m] A=w * dy w
depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] the width equals 0.3 meters. Since this function is linear we know the function f(y), equals, --- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

27 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] 0.75 times y. Which makes our revised equation for the area of the door equal to --- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

28 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] 0.75 times y, times dy. [pause] FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

29 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y 9.81 [m/s2] 0.4 [m] door ρ = 2,400 [kg/m3] This makes our equation for F R, equal to the integral of, --- FR ρ = ρw * SG hinge ρw = 1,000 [kg/m3] SG = 2.4 wall

30 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y FR= ∫ρ * g * y * 0.75 * y * dy 0.4 [m] door rho, g, y, 0.75, y, dy, and after simplifying, F R equals, --- FR hinge SG = 2.4 wall

31 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y FR= ∫ρ * g * y * 0.75 * y * dy 0.4 [m] door 0.75 times rho, g, times the integral of y squared dy. And we’ll integrate from a depth of --- FR FR= * ρ * g *∫ y2 * dy hinge SG = 2.4 wall

32 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y FR= ∫ρ * g * y * 0.75 * y * dy 0.4 [m] door 0 meters, to, 0.4 meters. [pause] After integrating, --- 0.4 [m] FR FR= * ρ * g *∫ y2 * dy hinge 0 [m] SG = 2.4 wall

33 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y FR= ∫ρ * g * y * 0.75 * y * dy we’ll substitute in the limits of integration, --- 0.4 [m] FR= * ρ * g *∫ y2 * dy 0 [m] y2=0.4 [m] FR= * g * ρ *[ 1/3 * y3 ] y1=0 [m]

34 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A 0.3[m] = f(0.4[m]) 0.3 [m] depth P = ρ * g * y FR= * g * ρ *[ 1/3 * y3 - 1/3 * y3 ] for y1, and, y2. We’ll also substitute in the values for --- 2 1 y2=0.4 [m] FR= * g * ρ *[ 1/3 * y3 ] y1=0 [m]

35 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A ρ = 2,400 [kg/m3] 0.3 [m] g=9.81 [m/s2] FR= * g * ρ *[ 1/3 * y3 - 1/3 * y3 ] the acceleration, and the density, and the reaction force, F R, --- 2 1 y2=0.4 [m] FR= * g * ρ *[ 1/3 * y3 ] y1=0 [m]

36 Find: FR [kN] 0[m] = f(0[m]) w=0.75 * y dy y FR= P * A 0.4 [m]
A=w * dy w A= 0.75 * y * dy FR= ∫P * A ρ = 2,400 [kg/m3] 0.3 [m] g=9.81 [m/s2] FR= * g * ρ *[ 1/3 * y3 - 1/3 * y3 ] equals, kiloNewtons. [pause] 2 1 FR= [kN] y2=0.4 [m] FR= * g * ρ *[ 1/3 * y3 ] y1=0 [m]

37 Find: FR [kN] w=0.75 * y 156 377 FR= P * A 752 A=w * dy 1,234
A= 0.75 * y * dy FR= ∫P * A ρ = 2,400 [kg/m3] g=9.81 [m/s2] FR= * g * ρ *[ 1/3 * y3 - 1/3 * y3 ] When reviewing the possible solutions, --- 2 1 FR= [kN] y2=0.4 [m] FR= * g * ρ *[ 1/3 * y3 ] y1=0 [m]

38 Find: FR [kN] w=0.75 * y 156 377 FR= P * A 752 A=w * dy 1,234
A= 0.75 * y * dy FR= ∫P * A ρ = 2,400 [kg/m3] g=9.81 [m/s2] answerB FR= * g * ρ *[ 1/3 * y3 - 1/3 * y3 ] the answer is B. [fin] 2 1 FR= [kN] y2=0.4 [m] FR= * g * ρ *[ 1/3 * y3 ] y1=0 [m]

39 * ΔP13=-ρA* g * (hAB-h1)+… depth to centroid kg*cm3 1,000 g * m3
a monometer contains 5 different fluids, which include, --- kg*cm3 1,000 * g * m3

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