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{
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"name": "",
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"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 : 26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"d= 40. \t#lb/ft**2 density of wood\n",
"w= 4 \t#ft wide\n",
"h= 6 \t#ft deep\n",
"l= 12 \t#ft long \n",
"\n",
"#CALCULATIONS\n",
"W= w*h*d*l\n",
"V= W/64\n",
"D= V/(w*l)\n",
"\n",
"#RESULTS\n",
"print 'Volume of water print laced = %.f ft**3'%(V)\n",
"print ' Depth of immersion = %.2f ft'%(D)\n",
"print ' Centre of buoyancy = %.2f ft from base'%(D)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volume of water print laced = 180 ft**3\n",
" Depth of immersion = 3.75 ft\n",
" Centre of buoyancy = 3.75 ft from base\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3 Page No : 28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from sympy import Symbol,solve\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"d= 4. \t#ft diameter\n",
"h= 7. \t#ft high\n",
"W= 2500. \t#lb weighing \n",
"OG= 3.5\n",
"OB= 1.55 \t#ft\n",
"\n",
"#CALCULATIONS\n",
"V= W/d**3\n",
"D= V/(math.pi*(d/2)**2)\n",
"I= math.pi*d**4/64\n",
"BM= I/V\n",
"BG= OG-OB\n",
"T = Symbol(\"T\")\n",
"ans = solve( (2500 + T)**2 -(512*math.pi *(8750 - 804)) - 1)\n",
"T = ans[1]\n",
"\n",
"#RESULTS\n",
"print 'Minimum tension in chain = %d lb'%(T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Minimum tension in chain = 1075 lb\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4 Page No : 31"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"W1= 1000. \t#lb weighing\n",
"W2= 100. \t#lb load\n",
"h= 4. \t#ft height\n",
"d= 5. \t#ft diameter\n",
"\n",
"#CALCULATIONS\n",
"V= (W1+W2)/h**3\n",
"D= V*h/(d**2*math.pi)\n",
"I= d**4*math.pi/h**3\n",
"BM= I/V\n",
"x= (BM+(D/2)-(W1*(h/2)/(W1+W2)))/(W2/(W1+W2))-0.02\n",
"C= x-h\n",
"\n",
"#RESULTS\n",
"print 'centre of gravity = %.2f ft'%(x)\n",
"print ' Hence the gravity of the weight must not be more than above the top of buoy = %.2f ft'%(C)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"centre of gravity = 4.43 ft\n",
" Hence the gravity of the weight must not be more than above the top of buoy = 0.43 ft\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.5 Page No : 32"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"b= 12. \t#ft breadth\n",
"h1= 3. \t#ft draught\n",
"h2= 1.5 \t#ft\n",
"h3= 5+(2./3) \t#ft\n",
"\n",
"#CALCULATIONS\n",
"I= b**3/12\n",
"V= b*h1\n",
"bm= I/V\n",
"BG= bm+(h1*2/(3*b))\n",
"O= math.degrees(math.tan(math.sqrt((h3*2-h1-bm*2)/(bm*2+bm))))\n",
"\n",
"\n",
"#RESULTS\n",
"print ' Volume of body immersed = %.f ft**3'%(V)\n",
"print ' BM = %.f ft'%(bm)\n",
"print ' BG = %.2f ft'%(BG)\n",
"print ' angle of heel = %.2f degrees'%(O)\n",
"\n",
"#The answer is a bit different due to rounding off error in textbook\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Volume of body immersed = 36 ft**3\n",
" BM = 4 ft\n",
" BG = 4.17 ft\n",
" angle of heel = 9.64 degrees\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
