1. 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]

2. 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]

3. 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]

4. 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]

5. 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]

6. 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]

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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]
meters cubed. [pause] The gravitational acceleration constant, --- m 1
volume of * cm
100
cylinder
Vcyl = [m3]

23. 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 = [m3]

24. 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 = [m3]

25. 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 = [m3]

26. 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 = [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]

27. 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 = [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]

28. Find: SG = 150 [N] SG * 999 [kg/m3] * 9.81 [m/s2] * 0.01508 [m3]
Fb = 150 [N]
Vcyl = [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]

29. Find: SG = 150 [N] SG * 999 [kg/m3] * 9.81 [m/s2] * 0.01508 [m3]
Fb = 150 [N]
Vcyl = [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]

30. Find: SG
SG = 1.015
SG * 999 [kg/m3] * 9.81 [m/s2] * [m3]
= 150 [N]
Fb = 150 [N]
Vcyl = [m3]
Fb = ρ * g * Vcyl
Vcyl = 15,080 [cm3]
[pause] m 1
g= 9.81 [m/s2] * cm
100
ρ= SG * ρw
ρw= 999 [kg/m3]

31. Find: SG
0.78
0.89
1.01
1.12
SG = 1.015
SG * 999 [kg/m3] * 9.81 [m/s2] * [m3]
= 150 [N]
Fb = 150 [N]
Vcyl = [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]

32. Find: SG = 150 [N] 0.78 0.89 answerC SG = 1.015 1.01 1.12
SG * 999 [kg/m3] * 9.81 [m/s2] * [m3]
= 150 [N]
Fb = 150 [N]
Vcyl = [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]

33. * 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