Engineering Chapter 2 sample notes, Assignments of Law

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Chapter 2

Course: ME 304 Spring 2021

Fluid Mechanics

Instructor: Samaneh Farokhirad

Hydrostatic force on a plane surface

dF

o x

y

𝜃 𝑦

𝑑𝐴

𝑦!

ℎ

Geometric properties of some common shapes

Hydrostatic force on a plane surface What is the point of application of the force ?? (^) 𝐹 = 𝑝%𝐴 Single Force F

Calculate the moment about the origin 𝑑𝐹 = 𝜌𝑔ℎdA 𝑑𝐹^ =^ 𝜌𝑔𝑦𝑠𝑖𝑛𝜃𝑑𝐴^ Moment=Force×arm 𝑑𝑀 = 𝑦𝑑𝐹 = 𝜌𝑔𝑦'𝑠𝑖𝑛𝜃𝑑𝐴 Moment of Inertia for a certain area of surface: (^) 𝐼( = (^) ∫$ 𝑦'𝑑𝐴 w.r.t x-axis

Geometric properties of some common shapes

Hydrostatic force on a plane surface 𝐼) = 𝐼)& + 𝐴𝑦&^ ( 𝑀 = 𝜌𝑔𝑠𝑖𝑛𝜃𝐼) 𝑀 = 𝜌𝑔𝑠𝑖𝑛𝜃(𝐼)& + 𝐴𝑦&^ () Moment that the distributed pressure force exerts about the origin of coordinates F

Torque of the distributed force=toque produced by the concentrated equivalent force 𝑀 = 𝜌𝑔𝑠𝑖𝑛𝜃 𝐼)& + 𝐴𝑦&^ (^ = 𝐹×𝑦* (^) y of the resultant force

𝐼(+: product of inertia w.r.t x, y

Hydrostatic force on a plane surface 𝑦* = 𝑦& + 𝐼)&^ F 𝑦& 𝐴

Y-coordinate of the point of application of the force

X-coordinate of the point of application of the force 𝑑𝑀 = 𝑑𝐹×𝑥 = 𝜌𝑔𝑦𝑠𝑖𝑛𝜃𝑑𝐴×𝑥 𝑀 = 𝜌𝑔𝑠𝑖𝑛𝜃 ' $ 𝑥𝑦𝑑𝐴 𝐹×𝑥* = 𝜌𝑔𝑠𝑖𝑛𝜃𝐼), 𝐹 = 𝑝+ 𝐴 = 𝜌𝑔𝑦& 𝑠𝑖𝑛𝜃𝐴

𝑥* = (^) 𝑦𝐼),& 𝐴 𝐼), = 𝐼),& + 𝐴𝑥& 𝑦&^ 𝑥*^ =^

𝐼),& 𝑦& 𝐴 +^ 𝑥&

Hydrostatic force on a curved surface 𝐹)

𝜃, 𝜃'

Compute the components of the force separately (^) 𝐹^7 *

Horizontal component: Total force for the blue area: 𝑑𝐹 = 𝑝𝑑𝐴 𝑑𝐹- = 𝑝𝑑𝐴cosθ 𝑑𝐴cosθ 𝑑𝐴 𝜃 (^) 𝑑𝐴𝑐𝑜𝑠𝜃

𝐹^7 * = 𝐹^7 - 𝚤+̂ 𝐹^7. 𝚥̂ 𝑑𝐹 𝑑𝐹. 𝑑𝐹-

x^ 𝚥̂

Hydrostatic force on a curved surface (^) 𝐹 (^7) * = 𝐹 (^7) - 𝚤̂+ 𝐹 (^7). 𝚥̂ Vertical Component: 𝑑𝐹^

𝚥̂ h 𝑑𝐴

𝑑𝐹^ 𝑑𝐹" 𝑑𝐹. = 𝑝𝑑𝐴𝑠𝑖𝑛𝜃 𝑑𝐹. = 𝑑𝑊 (^) Weight 𝐹. = ' 𝑑𝐹. = ' 𝑑𝑊 = 𝑊 =Weight of the fluid above the surface Conclusion: The vertical force (𝐹.) is equal to the weight of the fluid above the surface

𝑑𝐴 𝜃 𝑑𝐴𝑠𝑖𝑛𝜃

Hydrostatic force on a curved surface (Summary)

Horizontal component: 𝐹- = (^) ∫ 𝑝𝑑𝐴. = Force acting on the vertical projection of the surface

Vertical component:

𝑑𝐹^

The vertical force ( 𝐹. ) is equal to the weight of the fluid above the surface 𝐹. = 𝑊 x

x

Example 1 The 4 - m-diameter circular gate is located in an inclined wall of a large reservoir shaft along containing its horizontal water diameter, (γ = 9. (^80) and𝑘𝑁/ 𝑚the#) .water The gatedepth is ismounted 10 m at on the a shaft. (b) shaft Determine to open the the gate moment. that would have to be applied to the

A 𝑀$ = 0 𝑀 = 𝐹#(𝑦# − 𝑦$) 𝑀 = 1230 × 10 %𝑁 11 .6𝑚 − (^) 𝑠𝑖𝑛60°10𝑚 = 1. 07 × 10 '𝑁. 𝑚

Example 2 Consider two volumes of water, one salty and one fresh water, separated by a wall. Given that the depth of the salty water is 7m, (a) find the depth of fresh water required to give a zero resultant force on the wall. (b) Will the moment due to fluid forces be zero when the resultant is zero? Why?

7m (^) h? Salty Water𝐹^9 Fresh Water^ 𝐹^8

𝐹L = 𝐹M 𝐹^ =^ 𝑝%𝐴 𝛾Lℎ&L𝐴L = 𝛾Mℎ&M𝐴M 𝛾L = 10. 1 𝑘𝑁/𝑚N^ 𝛾M = 9. 8 𝑘𝑁/𝑚N ( 10. 1 𝑘𝑁/𝑚N)(^72 𝑚)( 7 × 1 𝑚() = ( 9. 8 𝑘𝑁/𝑚N)(ℎ 2 𝑚)(ℎ× 1 𝑚() ℎ = 7. 11 𝑚

(a) 𝑥

Example 3 This figure shows a cross section of a submerged tunnel used by automobiles to travel under a river. Find the magnitude and location of the resultant hydrostatic force on the circular roof of the tunnel. The tunnel is 4 mi long. A 𝐹( = 0 A 𝐹) = 0 𝐹" = 𝑊 = 𝛾*+,-.∀

No horizontal force due to the symmetry

= 62. (^4) 𝑓𝑙𝑏𝑡% 4𝑚𝑖 (^5280) 𝑚𝑖𝑓𝑡 2 20𝑓𝑡 70𝑓𝑡 − 𝜋 2 20𝑓𝑡 &^ = 2. 86 × 10 /^ 𝑙𝑏

= 𝛾*+,-.𝑙 2𝑅ℎ − 𝜋 2 𝑅&

ℎ = 𝑙 =50𝑓𝑡 4 𝑚𝑖 + 𝑅

Exercise # 3: calculate the exact location of the resultant force

2R

h

Determine the magnitude and direction of the force that must be applied to the bottom of the gate shown in the figure to keep the gate closed. ℎ$ = 𝑦$ 𝐹# = (^1000) 𝑚𝑘𝑔%^9 .81𝑚 𝑠& 1 .7𝑚 2𝑚× 0 .8𝑚 = 26700 𝑁

𝑦$ = 𝐻 + 0 .8𝑚 2

𝑦# = 1 .7𝑚 + ( 1121 .7𝑚^ 2𝑚)(2𝑚^0 ×.8𝑚 0 .8𝑚^ %)

Example 4

𝐹# = 𝛾ℎ$𝐴

𝑦# = 𝑦$ + (^) 𝑦𝐼($$𝐴

A 𝑀! = 0 Summing moment about the hinge A 𝑀! = 𝐹 0 ℎ − 𝐹#(𝑦# − 𝐻) = 0 𝐹 0 = 𝐹#(𝑦# ℎ^ − 𝐻) = 26700𝑁^0.^1 8𝑚.^73 −^1.^3 𝑚 𝐹 0 =^ 14400𝑁

= 2 .1𝑚 − 0 .8𝑚 + 0 .4𝑚 = 1 .7𝑚

= 1 .73𝑚

𝐹$

𝑦$ 𝑦% 𝐹$ 𝐹%