Find: SG air Wcyl,A=350 [N] fluid Wcyl,S=200 [N] r h 0.78 0.89 1.01 1.12 Find the specific gravity of the fluid, SG. [pause] In this problem, --- r= 20 [cm] submerged cylinder h= 12 [cm]
Find: SG air Wcyl,A=350 [N] fluid Wcyl,S=200 [N] r h 0.78 0.89 1.01 1.12 a cylinder is submerged in a fluid. The cylinder measures --- r= 20 [cm] submerged cylinder h= 12 [cm]
Find: SG air Wcyl,A=350 [N] fluid Wcyl,S=200 [N] r h 0.78 0.89 1.01 1.12 20 centimeters in the radius, and 12 centimeters ---- r= 20 [cm] submerged cylinder h= 12 [cm]
Find: SG air Wcyl,A=350 [N] fluid Wcyl,S=200 [N] r h 0.78 0.89 1.01 1.12 in height. When submerged, the cylinder exhibits an apparent weight of ---- r= 20 [cm] submerged cylinder h= 12 [cm]
Find: SG air Wcyl,A=350 [N] fluid Wcyl,S=200 [N] r h 0.78 0.89 1.01 1.12 200 Newtons. When the cylinder is in open air, it’s weight is, --- r= 20 [cm] submerged cylinder h= 12 [cm]
Find: SG air Wcyl,A=350 [N] fluid Wcyl,S=200 [N] r h 0.78 0.89 1.01 1.12 350 Newtons. [pause] To solve this problem, we’ll first write out, ---- r= 20 [cm] submerged cylinder h= 12 [cm]
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 r= 20 [cm] h= 12 [cm] air r fluid the sum of the forces in the vertical direction, which equals zero, because, --- h
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 r= 20 [cm] h= 12 [cm] air r fluid the acceleration of the cylinder equals 0. [pause] A bouyant force, F b, ---- h
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] h= 12 [cm] air r fluid acts upward on the cylinder, as well as the normal force experienced by --- h Fb
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] h= 12 [cm] air r fluid the bottom of the tank, which is equal to the weight of the submerged cylinder. Lastly, the self-weight of the cylinder, --- h Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] h= 12 [cm] air r fluid acts in the downward direction. [pause] The bouyant force is a function of, --- Wcyl,A h Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] h= 12 [cm] air r fluid the specific gravity, therefore, we’ll solve for the bouyant force, ---- Wcyl,A h Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] air r fluid which equals, the weight of the cylinder in air, minus, the weight of the cylinder, when submerged. [pause] After plugging in --- Wcyl,A h Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] air r fluid these weights, we compute the bouyant force, equals, --- Wcyl,A h Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] air r fluid 150 Newtons. [pause] The bouyant force, equals, --- Wcyl,A h Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] air r fluid Fb = ρ * g * Vcyl rho times g, times V cylinder. In this equation, V cylinder represents, ---- Wcyl,A h Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] air r fluid Fb = ρ * g * Vcyl the volume of the cylinder, which is a function of the radius, ---- Wcyl,A volume of h cylinder Fb Wcyl,S
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl and height, of the cylinder. [pause] After plugging in these dimensions, ---- volume of cylinder
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl the volume of the cylinder, equals, ---- volume of cylinder
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] 15,080, centimeters cubed. [pause] If we convert the units to meters cubed, ---- volume of cylinder
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] the volume of the cylinder equals, --- m 1 volume of * cm 100 cylinder
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] 0.01508 meters cubed. [pause] The gravitational acceleration constant, --- m 1 volume of * cm 100 cylinder Vcyl = 0.01508 [m3]
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] g, equals, 9.81 meters per second squared, and the variable rho, --- m 1 g= 9.81 [m/s2] * cm 100 Vcyl = 0.01508 [m3]
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] the density of the fluid, equals, the specific gravity of the fluid, --- m 1 g= 9.81 [m/s2] * cm 100 Vcyl = 0.01508 [m3]
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = π * r 2 *h Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] times, the density of water, where the density of water, equals, --- m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw Vcyl = 0.01508 [m3]
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = 0.01508 [m3] Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] 999 kilograms per meters cubed. [pause] After making these subtitutions, --- m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw ρw= 999 [kg/m3]
Find: SG Wcyl,A=350 [N] Wcyl,S=200 [N] ΣFy = 0 = Fb + Wcyl,S - Wcyl,A r= 20 [cm] Fb = Wcyl,A - Wcyl,S h= 12 [cm] Fb = 150 [N] Vcyl = 0.01508 [m3] Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] our equation for the bouyant force, reduces down to, --- m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw ρw= 999 [kg/m3]
Find: SG = 150 [N] SG * 999 [kg/m3] * 9.81 [m/s2] * 0.01508 [m3] Fb = 150 [N] Vcyl = 0.01508 [m3] Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] one equation and one unknown variable, and that unknown variable, is, ---- m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw ρw= 999 [kg/m3]
Find: SG = 150 [N] SG * 999 [kg/m3] * 9.81 [m/s2] * 0.01508 [m3] Fb = 150 [N] Vcyl = 0.01508 [m3] Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] the specific gravity of the fluid, SG, which equals, --- m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw ρw= 999 [kg/m3]
Find: SG SG = 1.015 SG * 999 [kg/m3] * 9.81 [m/s2] * 0.01508 [m3] = 150 [N] Fb = 150 [N] Vcyl = 0.01508 [m3] Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] 1.015. [pause] m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw ρw= 999 [kg/m3]
Find: SG 0.78 0.89 1.01 1.12 SG = 1.015 SG * 999 [kg/m3] * 9.81 [m/s2] * 0.01508 [m3] = 150 [N] Fb = 150 [N] Vcyl = 0.01508 [m3] Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] when reviewing the possible solutions, --- m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw ρw= 999 [kg/m3]
Find: SG = 150 [N] 0.78 0.89 answerC SG = 1.015 1.01 1.12 SG * 999 [kg/m3] * 9.81 [m/s2] * 0.01508 [m3] = 150 [N] Fb = 150 [N] Vcyl = 0.01508 [m3] Fb = ρ * g * Vcyl Vcyl = 15,080 [cm3] the answer is C. [fin] m 1 g= 9.81 [m/s2] * cm 100 ρ= SG * ρw ρw= 999 [kg/m3]
* 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