MECH 314 Problems 4, Spring 2014

MECH 411 Introduction to Fluid Mechanics October 30, 2004 American University of Beirut, Fall 2004-2005 Handout # PS2 Selected Problems from Problem Set 2 Problem 2.97 B A R θ x y y = R sin θ dFFFF=-p rrrr dS ˆ r = r = r = r = cos θ xxxx + sin θ yyyy ˆ ˆ ˆ dS = R dθ rrrrˆ θ0 B A R θ x y y = R sin θ dFFFF=-p rrrr dS ˆ dFFFF=-p rrrr dS ˆ r = r = r = r = cos θ xxxx + sin θ yyyy ˆ ˆ ˆ dS = R dθ rrrrˆrrrrˆ θ0 The gauge pressure at y is pg(y) = p(y) − pa = ρ g (H − y) = ρ g (H − R sin θ) (1) The unit outward normal vector is ˆn = ˆr = cos θ ˆx + sin θ ˆy (2) The force due to pressure distribution is F = � dF = � S −pg(y) ˆn dS = � dF = −WR � π π/4 ρ g (H − R sin θ) (cos θ ˆx + sin θ ˆy) dθ = −ρgWR �R 4 − H √ 2 � ˆx + ρgWR � R �3π 8 + 1 4 � − H � 1 + 1 √ 2 �� ˆy (3) Noting that the position vector on the gate is r = x ˆx + y ˆy = R(cos θ ˆx + sin θ ˆy), the moment about the center of coordinates is MO = � r × dF = R∗ × F where R∗ = x∗ ˆx + y∗ ˆy is the center of pressure. Then satisfying the x and y components of the above equation yields � −x dFy = −x∗Fy � y dFx = y∗Fx 1 Notice that since dF is always along the radial coordinate, the moment about O is zero, i.e. x∗Fy = y∗ dFx. The center of pressure R∗ = x∗ ˆx + y∗ ˆy is determined as follows x∗ = � π π/4 x dFy Fy = 2 3 R � R √ 2 − 3 H � 4 √ 2H − 3 Rπ − 2 R + 8 H (4) y∗ = � π π/4 y dFx Fx = −1 3 R � R √ 2 − 3 H � 2 √ 2H − R (5) With R = 2 m, W = 4 m, H = 4 m, ρg = 10050 N/m3, we get Fx = 140.4 kN (6) Fy = −239.5 kN (7) x∗ = −0.385 m (8) y∗ = 0.6565 m (9) x y B FBx FBy A FA Fx Fy x* y* O θ0 x y B FBx FBy A FA Fx Fy x* y* O θ0 To get the reaction forces at A and B, we employ � F = 0 and � M = 0 about any chosen point. Then FAx + FBx = −Fx ⇒ FBx = −Fx + FA cos θ0 FAy + FBy = −Fy ⇒ FBy = −Fy − FA sin θ0 moments about B Fy(R + x∗) − Fxy∗ + FAR(1 + 2 cos θ0) sin θ0 = 0 Solving for FAx, FAy, FBx and FBy, we get FA = 140.3 kN FAx = −99.2 kN FAy = 99.2 kN FBx = −41 kN FBy = 140 kN 2 Problem 2.119 mg γ � � � � V Wθ F � � F � � θ W W/3 mg γ � � � � V Wθ F � � F � � θ W W/3 We assume that the angle θ is small so that the floating volume is Vfloat = (Wθ)(W) 2 (D) where D is the depth. The weight of the piece of wood must balance the buoyant force γwoodVwood + mg = γwaterVsubmerged ⇒ γwoodHWD + mg = γwater � HWD − W 2Dθ 2 � (10) where mg is the weight placed at the end of wooden beam. For equilibrium the sum of moments about the center of gravity of the wooden beam must be zero. We split the buoyant force into two components. Component Fb1 due to a rectangular volume of fluid and component Fb2 due to triangular volume of fluid. The component Fb1 due to the rectangular volume of water passes throught the center of gravity of the wooden beam. This is because the angle θ is small. Then the moment induced by the weight placed at the edge of the wooden beam must be balanced by Fb2 as shown in the Figure. Then mgW 2 = Fb2 �W 2 − W 3 � = γwater(D)(W)(Wθ) 2 �W 6 � ⇒ θ ≃ 6mg γwaterDW 2 Setting mg = 5 lbf, W = 9 ft, D = 4/12 ft, γwater = 62.4 lbf/ft3, we get θ = 0.0178 rad or 1.02o. 3 Substituting in equation (10), we get γwood = 42.4 lbf/ft3. The specific gravity of wood is then SG ≃ γwood/γwater = 0.68. 4 Problem 2.147 We start by applying Newton’s second law per unit volume a = g sinθ � θ θ � θ � g g cosθ � ▽p sˆ x y a = g sinθ � θ θ � θ � g g cosθ � ▽p sˆ x y � F = ρ a ⇒ −∇p + ρg = ρ a The free surface is characterized by uniform pressure, pa. Then the pressure gradient is normal to the free surface. If ˆs is the unit vector along the free surface, then ∇p · ˆs = 0, then g · ˆs = a · ˆs Note that g · ˆx = a · ˆx = a = g sin θ0 Then we must have ˆs = ˆx i.e. the water surface is aligned with the x coordinate. 5