Title HPGE 15 Solutions.pdf ✅

15 Hydraulics: Buoyancy Solutions SITUATION 1. A 5000 kg iceberg floats in seawater. If the density of the iceberg is 890 kg/m3 and the density of seawater is 1030 kg/m3 . ▣ 1. What is the volume of the ice above the surface? [SOLUTION] Determine the volume of the iceberg VT = m ρice VT = 5000 kg 890 kg m3 VT = 5.62 m3 Determine the volume of ice submerged Wice = Ww 5000 kg = (1030 kg m3 ) Vsub Vsub = 4.85 m3 Compute for the volume of ice above the surface: Vsurface = VT − Vsub Vsurface = 5.62 m3 − 4.85 m3 Vsurface = 0.764 m3 ▣ 2. What vertical force will fully submerge the iceberg? [SOLUTION] F = γwVsurface F = (1030 kg m3 × 9.81 m s 2 ) (0.764 m3 ) F = 7.716 kN SITUATION 2. A block of wood floats in water with 5 cm projecting above the water surface. When placed in glycerin of specific gravity of 1.35, the block projects 7.5 cm above the liquid.
▣ 3. Determine the total length of wood. [SOLUTION] BFwater = BFglycerin γwVw = γgVg γwAblock(h − 5 cm) = 1.35γwAblock(h − 7.5 cm) h = 14.64 cm ▣ 4. Determine its specific gravity. [SOLUTION] Wblock = BFwater SGblockh = (1)(h − 5cm) SGblock = 0.659 SITUATION 3. A stone weighs 280 N in alcohol (sg = 0.80) and 220 N in glycerin (sg = 1.25). ▣ 5. What is the volume of the stone? [SOLUTION] Wstone − BFal = 280 N Wstone − BGgl = 220 N Wstone − 0.8 (9810 N m3 ) Vstone = 280 N Wstone − 1.25 (9810 N m3 ) Vstone = 220 N From the two equations, Wstone = 386.67 N Vstone = 0.01359 m3 ▣ 6. What is its specific weight? [SOLUTION] γ = W V γ = 386.67 N 0.01359 m3 γ = 28.45 kN/m3

0.184 N = (0.3 × 9810 N m3 ) (0.01 m × 0.05 m × 0.05 m) + T T = 0.110 N ▣ 9. Determine the specific weight of the metal part. [SOLUTION] T + BF = W 0.110N + (9810 N m3 ) (6600 mm3 × ( 1 m 1000 mm) 3 ) = γ (6600 mm3 × ( 1 m 1000 mm) 3 ) γ = 26.53 kN/m3