# Chapter 2 : Equilibrium of Floating Bodies¶

## Example 2.1 Page No : 41¶

In [2]:
# Variables
l = 4.
w = 2.
sg = 0.75
z = 9810.
d = 0.5

# Calculations
v = l*w*d
wg = v*z*sg
s = 24000.
V = ((z*v)-wg)/s
V1 = (v*z-wg)/(s-z)

# Results
print "volume in m3 when block is completely in water",V,"m**3"
print "volume in m3 when block and concrete completely under water",round(V1,5),"m**3"

volume in m3 when block is completely in water 0.40875 m**3
volume in m3 when block and concrete completely under water 0.69133 m**3


## Example 2.2 Page No : 42¶

In [1]:
import math

# Variables
d = 1
s = 0.75
w = 9810

# Calculations
a = math.pi*d*d/4
h = d*0.5
p = w*h*s 			# intensity of pressure on at horizontal interface
v = p*a  			#vertical upward force
w1 = w*s*a*d/3 			# weight of oil in upper hemisphere
vf = v-w1  			# net vertical upward force

# Results
print "minimum weight of upper hemisphere in N",round(vf,4),"N"

minimum weight of upper hemisphere in N 963.0945 N


## Example 2.3 Page No : 42¶

In [4]:
# Variables
w = 90.

# Calculations
# By archemde's principle
# weight of water print alced  =  weight of sphere
z = 9810
v = w/z
d = (v*12/3.142)**0.33333

# Results
print "external diameter of hollow of sphere in m",round(d,4),"m"

external diameter of hollow of sphere in m 0.3272 m


## Example 2.4 Page No : 43¶

In [5]:
# Variables
s1 = 13.6
s2 = 7.8
s3 = 1.

# Calculations
# by archimede principle
# weight of body  =  weight of liquid print laced
# s2 = s1*x+s3*(1-x)
x = (s2-s3)/(s1-s3)

# Results
print "fraction of steel below surface of mercury",round(x,2)

fraction of steel below surface of mercury 0.54


## Example 2.5 Page No : 43¶

In [2]:
import math

# Variables
w = 9810.
do = 1.25
a = math.pi*do*do*0.25

# Calculations
f1 = w*a*1
f2 = w*a*3 			# buoyancy force of 3m lenght of pipe
di = 1.2
s = 9.8
wg = w*s*3*((1.25**2)-(1.2**2))*0.25*math.pi
fa = f2-wg

# Results
print "buoyancy force in N/m",round(f1,3),"N/m"
print "upward force on anchor",fa,"N"

buoyancy force in N/m 12038.681 N/m
upward force on anchor 8367.36499746 N


## Example 2.6 Page No : 44¶

In [8]:
# Variables
a = 0.25
s1 = 11.5
s2 = 1.
z = 9810.
v1 = a*a*a*0.5
wc = v1*z
h = 0.016

# Calculations
# by archimede's principle
v2 = (a*0.5+h)*a*a 			# volume of cube submergerd
v = (v2-v1)/(s1-s2)
wl = v*s1*z

# Results

weight of lead attached 10.744 N


## Example 2.7 Page No : 45¶

In [9]:
# Variables
s1 = 19.3
s2 = 9.
x = 14./24

# Calculations
wg = x*10
wc = (1-x)*10
vg = wg/s1
vc = wc/s2
vt = vg+vc

# Results
print "volume of 10gm,14 carat gold in cm3",round(vt,3),"cc"

volume of 10gm,14 carat gold in cm3 0.765 cc


## Example 2.8 Page No : 46¶

In [10]:
# Variables
h1 = 0.05
h2 = 0.015
s = 41./40
l = h1/(s-1)
w1 = 25

# Calculations
# applying bakance in vertical direction
w = w1*(l+h1)/(h2)

# Results
print "weight of ship in in N",round(w,3),"kN"

weight of ship in in N 3416.667 kN


## Example 2.9 Page No : 47¶

In [9]:
# Variables
w = 700.
w1 = 20000.
d = 0.5
h = 1.
wd = 250.
z = 9810.

# Calculations
f = z*3.142*d*d*2*0.25/3
n = (w*4+w1)/(f-250)
n1 = round(n)

# Results
print "number of drums",n1

number of drums 22.0


## Example 2.10 Page No : 47¶

In [11]:
# Variables
a = 0.12
l = 1.8
s = 0.7
z = 9810.
wp = s*a*a*l*z
v = a*a*(l-0.2)
w = v*z
t = w-wp
sp = 110000.

# Calculations
# applying equilibrium balance
w = t/(1-(9810/sp))

# Results
print "weight of lead in N",round(w,3),"N"

weight of lead in N 52.733 N


## Example 2.11 Page No : 48¶

In [12]:
# Variables
d = 4.
h = 4.
s = 0.6
s1 = 1.

# Calculations
h1 = s*h/s1
v = 3.142*d*d*0.25*h1
x = h1/2
cog = h/2
h2 = cog-x
a = 3.142*d*d*d*d/64
bm = a/v
mh = bm-h2

# Results
print "metacentric height in m,negative sign indicte that cylinder is in unstable equilibrium",round(mh,4),"m"

metacentric height in m,negative sign indicte that cylinder is in unstable equilibrium -0.3833 m


## Example 2.12 Page No : 49¶

In [13]:
# Variables
d = 4.
s1 = 0.6
s2 = 0.9
l = 1.

# Calculations
h = s1*l/s2
cob = h/2
cog = l/2
dcog = cog-cob
i = 3.142*d*d*d*d/64
v = 3.142*0.25*d*d*h
bm = i/v
bm = dcog
l = (6*1.5)**0.5

# Results
print "maximium lenght of cylinder in m",l,"m"

maximium lenght of cylinder in m 3.0 m


## Example 2.13 Page No : 51¶

In [4]:
import math

# Variables
s = 2.
w = 340.
v = 0.5*s*s*s
z = 9810.

# Calculations
w1 = z*4
gb = s/4-s/8
i = s*s*s*s/(12)
v = 4
bm = i/v
gm = bm+gb
p = w/(w1*gm)
theta = math.degrees(math.atan(p))

# Results
print "angle through which cube will tilt in minutes",round((theta*60),3)

# note : rounding off error.

angle through which cube will tilt in minutes 51.059


## Example 2.14 Page No : 51¶

In [9]:
import math

# Variables
l = 60.
b = 9.
w = 16.*1000000
w1 = 160.*1000
y = 6.
q = 3.
sp = 10104.

# Calculations
i = 0.75*l*b*b*b/12
v = w/sp
bm = i/v
mcd = 2-bm
cogd = gm+mcd

# Results
print "metacentric height %.3f m "%gm
print "position of centre of gravity below the water line %.3f m"%cogd

metacentric height 1.145 m
position of centre of gravity below the water line 1.419 m


## Example 2.15 Page No : 53¶

In [15]:
import math

# Variables
w = 450000.
y = 5.5
w1 = 80.*1000000
q = 3.

# Calculations
p = 12.5*1000
n = 120.
T = (p*60000)/(2*math.pi*n)
z = T/(w1*gm)
theta = math.degrees(math.atan(z))

# Results
print "angle of heel in degree %.4f"%theta

# note : rounding off error.

angle of heel in degree 1.2066

In [ ]: