1. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf 25
148
175
195
Find the weight of the sphere, in the open air, in pounds force. [pause] In this problem, a sphere, ---
dsphere
sphere
balance

2. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf 25
148
175
195
with a diameter of 1.5 feet, is submerged in water, which has a water temeprature of, ---
dsphere
sphere
balance

3. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf 25
148
175
195
60 degrees fahrenheit. [pause] The weight of the sphere when measured on a balance, under water,
dsphere
sphere
balance

4. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf 25
148
175
195
equals, 85, pounds force. The question asks to find weight of the sphere, ---
dsphere
sphere
balance

5. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
???
air
if weighted on a balance, in the open air. [pause] The difference between the scale reading, when the sphere is submerged, ---
dsphere
sphere
balance

6. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
???
air
and when the sphere is in the open air, equals the bouyant force, F b, ---
dsphere
sphere
balance

7. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
???
air
which acts in the upward direction on the submerged sphere. [pause] The other forces acting on the submerged sphere include ---
dsphere
sphere
balance Fb

8. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
the upward force exerted on the sphere by the scale, which we know equals, --- Fb
wsphere,w

9. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
85 pounds force. [pause] That last force in this vertical force equilbrium, is, a downward force, which is, the weight of the sphere, --- Fb
wsphere,w = 85 [lbf]

10. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
in air, which is what we’re trying to solve for. [pause] Summing the forces in the vertical direction, ---
wsphere,A Fb
wsphere,w = 85 [lbf]

11. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
ΣFy = 0 = Wsphere,w + Fb-Wsphere,A
we can isolate the weight of the sphere, in air, which equals, ---
wsphere,A Fb
wsphere,w = 85 [lbf]

12. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
ΣFy = 0 = Wsphere,w + Fb-Wsphere,A
the bouyant force, plus, the weight of the sphere, in water. We know the weight of the sphere in water, ---
Wsphere,A = Fb+Wsphere,w
wsphere,A Fb
wsphere,w = 85 [lbf]

13. Find: Wsphere,A [lbf] in air wsphere,w = 85 [lbf] in water Tw = 60 Fo
dsphere=1.5 [ft]
air
water
85 lbf
ΣFy = 0 = Wsphere,w + Fb-Wsphere,A
equals, 85 pounds force, so we’re left to solve for the bouyant force, F b, ---
Wsphere,A = Fb+Wsphere,w
wsphere,A Fb
wsphere,w = 85 [lbf]

14. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
unit weight
volume
water
of water
of sphere
85 lbf
which equals the unit weight of water, times the volume of the sphere. [pause] The unit weight of water, equals, ---
wsphere,A Fb
wsphere,w = 85 [lbf]

15. Find: Wsphere,A [lbf] in air 85 [lbf] dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
γwater = ρ * g * gc
water
unit
85 lbf
density
conversion
acceleration
the density of water, times, the gravitational acceleration, times, a unit converstion. The density of water, slightly depends on, ---
constant
wsphere,A Fb
wsphere,w = 85 [lbf]

16. Find: Wsphere,A [lbf] in air 85 [lbf] dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
γwater = ρ * g * gc
water
unit
85 lbf
density
conversion
acceleration
the temperature of the water. [pause] At 60 degrees fahreneheit, the water density, equals, ---
constant
wsphere,A Fb
wsphere,w = 85 [lbf]

17. Find: Wsphere,A [lbf] in air 85 [lbf] dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
γwater = ρ * g * gc
water
unit
85 lbf
conversion
[lbm/ft3]
acceleration
62.37 pounds mass per cubic foot. [pause] The acceleration constant, equals, ---
constant
wsphere,A Fb
wsphere,w = 85 [lbf]

18. Find: Wsphere,A [lbf] in air 85 [lbf] dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
γwater = ρ * g * gc
water
unit
85 lbf
conversion
[lbm/ft3]
[ft/s2]
32.17 feet per second squared. [pause] And the unit converstion, g c, ---
wsphere,A Fb
wsphere,w = 85 [lbf]

19. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
γwater = ρ * g * gc
water
85 lbf
[lbm/ft3]
[ft/s2]
equals 1 over 32.17, pounds force second squared, divided by pound mass foot. Therefore, the unit weight of water, ---
lbf* s2 1
wsphere,A
32.17
lbm*ft Fb
wsphere,w = 85 [lbf]

20. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
γwater = ρ * g * gc
water
85 lbf
[lbm/ft3]
[ft/s2]
computes to pounds force, per cubic foot. [pause] The volume of the sphere, V sphere, ---
lbf* s2 1
wsphere,A
32.17
lbm*ft
γwater = [lbf/ft3] Fb
wsphere,w

21. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
62.37 [lbf/ft3]
water
85 lbf 4
Vsphere= π * r3 * 3
equals, 4/3 times PI, times the radius of the sphere, cubed. [pause] The problem provides the sphere diamter, as, ---
wsphere,A Fb
wsphere,w

22. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
62.37 [lbf/ft3]
water
85 lbf 4
Vsphere= π * r3 * 3
1.5 feet, which means the sphere radius, equals, half that value, ---
wsphere,A Fb
wsphere,w

23. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
62.37 [lbf/ft3]
water
85 lbf 4
Vsphere= π * r3 * 3
0.75 feet, and the volume of the sphere computes to, ---
wsphere,A
r=0.75 [ft] Fb
wsphere,w

24. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
62.37 [lbf/ft3]
water
85 lbf 4
Vsphere= π * r3 * 3
1.767 cubic feet. [pause] Now we can solve for the bouyant force, ---
wsphere,A
r=0.75 [ft]
Vsphere= [ft3] Fb
wsphere,w

25. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
62.37 [lbf/ft3]
water
85 lbf 4
Vsphere= π * r3 * 3
F b, which equals,
wsphere,A
r=0.75 [ft]
Vsphere= [ft3] Fb
wsphere,w

26. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
[ft3]
water
62.37 [lbf/ft3]
85 lbf
Fb = [lbf]
pounds force. [pause] And when added to the weight of the sphere, underwater, ---
wsphere,A Fb
wsphere,w

27. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
[ft3]
water
62.37 [lbf/ft3]
85 lbf
Fb = [lbf]
the weight of the sphere, in air, equals, ---
wsphere,A Fb
wsphere,w

28. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
air
[ft3]
water
62.37 [lbf/ft3]
85 lbf
Fb = [lbf]
pounds force. [pause]
Wsphere,A = [lbf]
wsphere,A Fb
wsphere,w

29. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
[ft3] 25
148
175
195
62.37 [lbf/ft3]
Fb = [lbf]
When reviewing the possible solutions, ---
Wsphere,A = [lbf]

30. Find: Wsphere,A [lbf] in air 85 [lbf] Wsphere,A = Fb+Wsphere,w
dsphere=1.5 [ft]
Wsphere,A = Fb+Wsphere,w
Tw = 60 F o
Fb = γwater * Vsphere
[ft3] 25
148
175
195
62.37 [lbf/ft3]
Fb = [lbf]
the answer is D. [fin]
Wsphere,A = [lbf]
answerD

31. * 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