FIRST PREBOARD SOLUTION (HYDRAULICS)
PROBLEM:
A pressure gage at elevation 4.8 m on the side of a
storage tank containing oil reads 34.7 kPa. Another
gage at elevation 2.2 m reads 57.5 kPa. Compute the
specific gravity of the liquid.
SOLUTION:
P    h 
PROBLEM:
A 3.7-m high by 1.5-m wide rectangular gate AB is
vertical and is hinged at a point 150 mm below its
center of gravity. The total depth of water is 6.1 m.
What horizontal force F (kN) must be applied at the
bottom of the gate for equilibrium.
SOLUTION:
57.5  34.7  sg  9.81 4.8  2.2 
sg = 0.89
PROBLEM:
A barrel containing water weighs 1260 N. What will
be the reading on the scale (N) if a 5 cm by 5 cm piece
of wood is held vertically in the water to a depth of
0.60 m?
SOLUTION:
For every acting force, there must be an equal and
opposite reacting force. The buoyant force exerted by
the water upward against the bottom of the piece of
wood is opposed by the 5 cm by 5 cm area of wood
acting downward on the water with equal magnitude.
This force will measure the increase in scale reading.
BF   w Vsub
2
BF  9810  0.05  0.60  


BF  14.715 N
new scale reading = 1260 + 14.715
new scale reading = 1274.715 N
PROBLEM:
Determine the volume (m3) of an object that weighs
200 N in water and 300 N in oil.
h  y  2.4 
e
SOLUTION:
Wliquid  Wair  BF
200  Wair  9810V  1
300  Wair  0.8  9810  V  2
Wair = 700 N
V = 0.051 m3
3.7
 4.25 m
2
I
Ay
1.5  3.7 
e
3
12
1.5  3.7  4.25
e  0.268 m
P   hA
P  9.81  4.25 1.5 3.7 
P  231.39 kN
jdlr
FIRST PREBOARD SOLUTION (HYDRAULICS)
Mhinge  0]
 3.7

P  e  .15  F 
 0.15 
 2

231.39  0.268  0.15   F  1.7 
F = 16.06 kN
PROBLEM:
A stone weighs 600 N, and when it was lowered into
a square tank 0.610 m on a side, the weight of the
stone in water was 323 N. How much did the water
rise in the tank, in mm?
SOLUTION:
PROBLEM:
Determine “y” in Figure so that the flashboard will
tumble only when the water reaches their top?
Wliquid  Wair  BF
323  600  9810V
Vstone  0.028 m3
Vstone  Vrise
0.028   0.61  y 
y = 0.076 m
2
SOLUTION:
y
4
= 1.33 m
3
PROBLEM:
A square pole (s.g.=0.68), 80 mm by 80 mm by 6 m
long, is suspended by a wire so that 4 m is
submerged in water and 2 m is above the surface.
What is the tension (N) in the wire?
PROBLEM:
The hydraulic jack in the figure is filled with oil at
897 kg/m3. Neglecting the weight of the two
pistons, what force F, in Newtons, on the handle is
required to support the 8900-N weight for this
design?
SOLUTION:
W  BF  T
0.68  9810  0.08  0.08  6   9810  0.08  0.08  4   T
T=5N
jdlr
FIRST PREBOARD SOLUTION (HYDRAULICS)
SOLUTION:
SOLUTION:
θ
A
F2
W
BF
F1
A
T
F2  A2 p2
8900 

 0.075  p2 
2
4
p2  2014547.902 Pa
p2  p1
F1  A1p1
F1 

MA  0]
W  2.5cos   BF  2cos 
2
 
W  2.5  9810    0.08  4 2 
4
W = 157.79 N
Fv  0]
 0.025  2014547.902 
2
4
F1  988.889 N
MA  0]
F  400  988.889  25
F = 61.8 N
PROBLEM:
The uniform 5-m-long round wooden rod in Figure
FM-2 is tied to the bottom by a string.
1. Determine the weight(N) of the rod.
2. Determine the tension (N) in the string.
3. Determine the specific gravity of the wood
W  T  BF
2
 
157.79  T  9810    0.08  4
 4
T = 39.45 N
WV
2
 
157.79  sg  9810     0.08  5
4
Sgwood = 0.64
PROBLEM:
A barge 12.6 m long with a flat bottom and
rectangular ends 7.5 m wide by 3.0 m high has a
draft of 1.8 m when fully loaded and floating in an
upright position. The center of gravity when fully
loaded is on the axis of symmetry and 30 cm above
the water surface.
1. What is the initial metacentric height (m)?
2. What is the metacentric height (m) when the angle
of heel is 12º
3. What is the resulting couple (kN-m)?
jdlr
FIRST PREBOARD SOLUTION (HYDRAULICS)
MBo 
B2
 1  0.5tan2  
12D
2
 7.5 

1  0.5tan2  0 

12  1.8
MBo 
B2
1  0.5 tan2  

12D
 7.5 
1  0.5 tan2  12 

12  1.8
2

MBo  2.66 m
MBo  2.60 m
MB0  GBo (M is above G)
GBo  0.90  0.30
GM  MBo  GBo
GBo  1.20 m
 2.66  1.20
GM = 1.46 m
MBo  GBo (M is above G)
GM  MBo  GBo
 2.60  1.20
GM = 1.40 m
RM  W  GM  sin 
RM  BF  GM  sin 
BF   f Vsubmerged
 9.81 7.5  1.8 12.6  
BF  1668.68 kN
RM  1668.68  1.46  sin  12
RM = 506.53 kN – m
jdlr