{
"metadata": {
"name": "",
"signature": "sha256:b1d2399b6b4acd9a65f6af74a5e09d523ac3468105509cc56b7fc106dd581d3b"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2 : Equilibrium of Floating Bodies"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1 Page No : 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"l = 4.\n",
"w = 2.\n",
"sg = 0.75\n",
"z = 9810.\n",
"d = 0.5\n",
"\n",
"# Calculations \n",
"v = l*w*d\n",
"wg = v*z*sg\n",
"s = 24000.\n",
"V = ((z*v)-wg)/s\n",
"V1 = (v*z-wg)/(s-z)\n",
"\n",
"# Results \n",
"print \"volume in m3 when block is completely in water\",V,\"m**3\"\n",
"print \"volume in m3 when block and concrete completely under water\",round(V1,5),\"m**3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"volume in m3 when block is completely in water 0.40875 m**3\n",
"volume in m3 when block and concrete completely under water 0.69133 m**3\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2 Page No : 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Variables\n",
"d = 1\n",
"s = 0.75\n",
"w = 9810\n",
"\n",
"# Calculations \n",
"a = math.pi*d*d/4\n",
"h = d*0.5\n",
"p = w*h*s \t\t\t# intensity of pressure on at horizontal interface\n",
"v = p*a \t\t\t#vertical upward force\n",
"w1 = w*s*a*d/3 \t\t\t# weight of oil in upper hemisphere\n",
"vf = v-w1 \t\t\t# net vertical upward force\n",
"\n",
"# Results \n",
"print \"minimum weight of upper hemisphere in N\",round(vf,4),\"N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"minimum weight of upper hemisphere in N 963.0945 N\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3 Page No : 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"w = 90.\n",
"\n",
"# Calculations \n",
"# By archemde's principle\n",
"# weight of water print alced = weight of sphere\n",
"z = 9810\n",
"v = w/z\n",
"d = (v*12/3.142)**0.33333\n",
"\n",
"# Results \n",
"print \"external diameter of hollow of sphere in m\",round(d,4),\"m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"external diameter of hollow of sphere in m 0.3272 m\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4 Page No : 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"s1 = 13.6\n",
"s2 = 7.8\n",
"s3 = 1.\n",
"\n",
"# Calculations \n",
"# by archimede principle\n",
"# weight of body = weight of liquid print laced\n",
"# s2 = s1*x+s3*(1-x) \n",
"x = (s2-s3)/(s1-s3)\n",
"\n",
"# Results \n",
"print \"fraction of steel below surface of mercury\",round(x,2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"fraction of steel below surface of mercury 0.54\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.5 Page No : 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Variables\n",
"w = 9810.\n",
"do = 1.25\n",
"a = math.pi*do*do*0.25\n",
"\n",
"# Calculations \n",
"f1 = w*a*1\n",
"f2 = w*a*3 \t\t\t# buoyancy force of 3m lenght of pipe\n",
"di = 1.2\n",
"s = 9.8\n",
"wg = w*s*3*((1.25**2)-(1.2**2))*0.25*math.pi\n",
"fa = f2-wg\n",
"\n",
"# Results \n",
"print \"buoyancy force in N/m\",round(f1,3),\"N/m\"\n",
"print \"upward force on anchor\",fa,\"N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"buoyancy force in N/m 12038.681 N/m\n",
"upward force on anchor 8367.36499746 N\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.6 Page No : 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"a = 0.25\n",
"s1 = 11.5\n",
"s2 = 1.\n",
"z = 9810.\n",
"v1 = a*a*a*0.5\n",
"wc = v1*z\n",
"h = 0.016\n",
"\n",
"# Calculations \n",
"# by archimede's principle\n",
"v2 = (a*0.5+h)*a*a \t\t\t# volume of cube submergerd\n",
"v = (v2-v1)/(s1-s2)\n",
"wl = v*s1*z\n",
"\n",
"# Results \n",
"print \"weight of lead attached\",round(wl,3),\"N\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"weight of lead attached 10.744 N\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.7 Page No : 45"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"s1 = 19.3\n",
"s2 = 9.\n",
"x = 14./24\n",
"\n",
"# Calculations \n",
"wg = x*10\n",
"wc = (1-x)*10\n",
"vg = wg/s1\n",
"vc = wc/s2\n",
"vt = vg+vc\n",
"\n",
"# Results \n",
"print \"volume of 10gm,14 carat gold in cm3\",round(vt,3),\"cc\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"volume of 10gm,14 carat gold in cm3 0.765 cc\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.8 Page No : 46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"h1 = 0.05\n",
"h2 = 0.015\n",
"s = 41./40\n",
"l = h1/(s-1)\n",
"w1 = 25\n",
"\n",
"# Calculations \n",
"# applying bakance in vertical direction\n",
"w = w1*(l+h1)/(h2)\n",
"\n",
"# Results \n",
"print \"weight of ship in in N\",round(w,3),\"kN\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"weight of ship in in N 3416.667 kN\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.9 Page No : 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"w = 700.\n",
"w1 = 20000.\n",
"d = 0.5\n",
"h = 1.\n",
"wd = 250.\n",
"z = 9810.\n",
"\n",
"# Calculations \n",
"f = z*3.142*d*d*2*0.25/3\n",
"n = (w*4+w1)/(f-250)\n",
"n1 = round(n)\n",
"\n",
"# Results \n",
"print \"number of drums\",n1\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"number of drums 22.0\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.10 Page No : 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"a = 0.12\n",
"l = 1.8\n",
"s = 0.7\n",
"z = 9810.\n",
"wp = s*a*a*l*z\n",
"v = a*a*(l-0.2)\n",
"w = v*z\n",
"t = w-wp\n",
"sp = 110000.\n",
"\n",
"# Calculations \n",
"# applying equilibrium balance\n",
"w = t/(1-(9810/sp)) \n",
"\n",
"# Results \n",
"print \"weight of lead in N\",round(w,3),\"N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"weight of lead in N 52.733 N\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.11 Page No : 48"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"d = 4.\n",
"h = 4.\n",
"s = 0.6\n",
"s1 = 1.\n",
"\n",
"# Calculations \n",
"h1 = s*h/s1\n",
"v = 3.142*d*d*0.25*h1\n",
"x = h1/2\n",
"cog = h/2\n",
"h2 = cog-x\n",
"a = 3.142*d*d*d*d/64\n",
"bm = a/v\n",
"mh = bm-h2\n",
"\n",
"# Results \n",
"print \"metacentric height in m,negative sign indicte that cylinder is in unstable equilibrium\",round(mh,4),\"m\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"metacentric height in m,negative sign indicte that cylinder is in unstable equilibrium -0.3833 m\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.12 Page No : 49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"d = 4.\n",
"s1 = 0.6\n",
"s2 = 0.9\n",
"l = 1.\n",
"\n",
"# Calculations \n",
"h = s1*l/s2\n",
"cob = h/2\n",
"cog = l/2\n",
"dcog = cog-cob\n",
"i = 3.142*d*d*d*d/64\n",
"v = 3.142*0.25*d*d*h\n",
"bm = i/v\n",
"bm = dcog\n",
"l = (6*1.5)**0.5\n",
"\n",
"# Results \n",
"print \"maximium lenght of cylinder in m\",l,\"m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximium lenght of cylinder in m 3.0 m\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.13 Page No : 51"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"# Variables\n",
"s = 2.\n",
"w = 340.\n",
"v = 0.5*s*s*s\n",
"z = 9810.\n",
"\n",
"# Calculations \n",
"w1 = z*4\n",
"gb = s/4-s/8\n",
"i = s*s*s*s/(12)\n",
"v = 4\n",
"bm = i/v\n",
"gm = bm+gb\n",
"p = w/(w1*gm)\n",
"theta = math.degrees(math.atan(p))\n",
"\n",
"# Results \n",
"print \"angle through which cube will tilt in minutes\",round((theta*60),3)\n",
"\n",
"\n",
"# note : rounding off error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"angle through which cube will tilt in minutes 51.059\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.14 Page No : 51"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"# Variables\n",
"l = 60.\n",
"b = 9.\n",
"w = 16.*1000000\n",
"w1 = 160.*1000\n",
"y = 6.\n",
"q = 3.\n",
"sp = 10104.\n",
"\n",
"# Calculations \n",
"i = 0.75*l*b*b*b/12\n",
"v = w/sp\n",
"bm = i/v\n",
"gm = (w1*y)/(w*(math.tan(math.radians(q))))\n",
"mcd = 2-bm\n",
"cogd = gm+mcd\n",
"\n",
"# Results \n",
"print \"metacentric height %.3f m \"%gm\n",
"print \"position of centre of gravity below the water line %.3f m\"%cogd\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"metacentric height 1.145 m \n",
"position of centre of gravity below the water line 1.419 m\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.15 Page No : 53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"# Variables\n",
"w = 450000.\n",
"y = 5.5\n",
"w1 = 80.*1000000\n",
"q = 3.\n",
"\n",
"# Calculations \n",
"gm = (w*y)/(w1*math.tan(math.radians(q)))\n",
"p = 12.5*1000\n",
"n = 120.\n",
"T = (p*60000)/(2*math.pi*n)\n",
"z = T/(w1*gm)\n",
"theta = math.degrees(math.atan(z))\n",
"\n",
"# Results \n",
"print \"angle of heel in degree %.4f\"%theta\n",
"\n",
"# note : rounding off error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"angle of heel in degree 1.2066\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}