summaryrefslogtreecommitdiff
path: root/Engineering_Mechanics_by_Tayal_A.K./chapter16_14.ipynb
blob: 7bd00420278bd50927862256c11ae7f122621c89 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
{
 "metadata": {
  "name": "chapter16.ipynb"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 16: Kinetics Of A Particle : Work And Energy"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-1,Page No:432"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "k=1000 # N/m # stiffness of spring\n",
      "x_1=0.1 # m # distance upto which the spring is stretched\n",
      "x_2=0.2 # m \n",
      "x_0=0 # initial position of spring\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Work required to stretch the spring by 10 cm from undeformed position is given as,\n",
      "U_10=-(k/2)*(x_1**2-x_0**2) # N-m \n",
      "# Work required to stretch from 10 cm to 20 cm is,\n",
      "U=-(0.5)*k*(x_2**2-x_1**2) # N-m\n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The work of the spring force is \",round(U_10),\"N-m\"\n",
      "print\"The work required to stretch the spring by 20 cm is \",round(U),\"N-m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The work of the spring force is  -5.0 N-m\n",
        "The work required to stretch the spring by 20 cm is  -15.0 N-m\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-3,Page No:436"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "M_A=100 # kg # mass of block A\n",
      "M_B=150 # kg # mass of block B\n",
      "mu=0.2 # coefficient of friction between the blocks and the surface\n",
      "x=1 # m # distance by which block A moves\n",
      "g=9.81 # m/s^2 # acc due to gravity\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Consider the respective F.B.D\n",
      "# Applying the principle of work and energy to the system of blocks A&B and on simplifying we get the value of v as,\n",
      "v=(((-mu*M_A*g)+(M_B*g))/(125))**0.5 # m/s \n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The velocity of block A is \",round(v,2),\"m/s\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The velocity of block A is  3.19 m/s\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-4,Page No:440"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "M=500*10**3 # kg # mass of the train\n",
      "u=0 # m/s # initial speed\n",
      "v=90*1000*3600**-1 # m/s # final speed\n",
      "t=50 # seconds\n",
      "F_r=15*10**3 # N # Frictioal resistance to motion\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Acceleration is given as,\n",
      "a=v*t**-1 # m/s^2\n",
      "# The total force required to accelerate the train is,\n",
      "F=M*a # N\n",
      "# The maximum power required is at, t=50s & v=25 m/s\n",
      "P=(F+F_r)*v*(10**-6) # MW\n",
      "# At any time after 50 seconds, the force required only to overcome the frictional resistance of 15*10^3 N is,\n",
      "P_req=F_r*v*(10**-3) # kW\n",
      "\n",
      "# Results\n",
      "\n",
      "print\"(a) The maximum power required is \",round(P,3),\"MW\"\n",
      "print\"(b) The power required to maintain a speed of 90 km/hr is \",round(P_req,2),\"kW\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) The maximum power required is  6.625 MW\n",
        "(b) The power required to maintain a speed of 90 km/hr is  375.0 kW\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-5,Page No:440"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "W=50 # N # Weight suspended on spring\n",
      "k=10 # N/cm # stiffness of the spring\n",
      "x_2=15 # cm # measured extensions\n",
      "h=10 # cm # height for position 2\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Consider the required F.B.D.\n",
      "\n",
      "# POSITION 1: The force exerted by the spring is,\n",
      "F_1=W # N\n",
      "\n",
      "# Extension of spring from undeformed position is x_1,\n",
      "x_1=F_1/k # cm\n",
      "\n",
      "# POSITION 2: When pulled by 10 cm to the floor. P.E of weight is,\n",
      "P_E_g=-W*h # N-cm # (P_E_g= P_E_gravity)\n",
      "\n",
      "# P.E of the spring with respect to position 1\n",
      "P_E_s=(0.5)*k*(x_2**2-x_1**2) # N-cm  # (P_E_s= P_E_ spring)\n",
      "\n",
      "# Total P.E of the system with respect to position 1\n",
      "P_E_t=P_E_g+P_E_s # N-cm # (P_E_t= P_E_total)\n",
      "\n",
      "# Total energy of the system,\n",
      "E_2=P_E_t # N-cm\n",
      "\n",
      "# Total energy of the system in position 3 w.r.t position 1 is:\n",
      "x=-(100)**0.5 # cm\n",
      "x=+(100)**0.5 # cm\n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The potential energy of the system is \",round(E_2),\"N-cm\"\n",
      "print\"The maximum height above the floor that the weight W will attain after release is \",round(x),\"cm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The potential energy of the system is  500.0 N-cm\n",
        "The maximum height above the floor that the weight W will attain after release is  10.0 cm\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-6,Page No:442"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "m=5 # kg # mass of the ball\n",
      "k=500 # N/m # stiffness of the spring\n",
      "h=10 # cm # height of drop\n",
      "g=9.81 # m/s^2 # acc due to gravity\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Consider the respective F.B.D.\n",
      "# In eq'n 1 substitute the respective values and simplify it further. In this eq'n of 2nd degree a=1 b=-0.1962 & c=-0.01962. Thus the roots of the eq'n is given as,\n",
      "a=1 \n",
      "b=-0.1962\n",
      "c=-0.01962\n",
      "delta=((-b+(((b**2)-(4*a*c))**0.5))/(2*a))*(10**2) # cm # We consider the +ve value of delta\n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The maximum deflection of the spring is \",round(delta,2),\"cm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum deflection of the spring is  26.91 cm\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-7,Page No:444"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "m=5 # kg # mass of the ball\n",
      "k=500 # N/m # stiffness of the spring\n",
      "h=0.1 # m # height of vertical fall\n",
      "g=9.81 # m/s^2 # acc due to gravity\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Consider the respective F.B.D\n",
      "# On equating the total energies at position 1 & 2 we get eq'n of delta as,\n",
      "delta=((2*m*g*h)/(k))**0.5 # m \n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The maximum compression of the spring is \",round(delta,3),\"m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum compression of the spring is  0.14 m\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-9,Page No:445"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "m=5 # kg # mass of the collar\n",
      "k=500 # N/m # stiffness of the spring\n",
      "AB=0.15 # m # Refer the F.B.D for AB\n",
      "AC=0.2 # m # Refer the F.B.D for AC\n",
      "g=9.81 # m/s^2 # acc due to gravity\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Consider the respective F.B.D\n",
      "\n",
      "# POSITION 1: \n",
      "P_E_1=m*g*(AB)+0 \n",
      "K_E_1=0\n",
      "E_1=P_E_1+K_E_1 #\n",
      "\n",
      "# POSITION 2 : Length of the spring in position 2\n",
      "CB=(AB**2+AC**2)**0.5 # m \n",
      "# x is the extension in the spring\n",
      "x=CB-AC # m\n",
      "# On substuting and Equating equations of total energies for position1 & position 2 we get the value of v as,\n",
      "v=(((E_1-((0.5)*k*x**2))*2)/m)**0.5 # m/s\n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The velocity of the collar will be \",round(v,2),\"m/s\"\n",
      "# The answer given in the text book (v=16.4 m/s) is wrong.\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The velocity of the collar will be  1.64 m/s\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-10,Page No:446"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "m=5 # kg # mass of the block\n",
      "theta=30 # degree # inclination of the plane\n",
      "x=0.5 # m # distance travelled by the block\n",
      "k=1500 # N/m # stiffness of the spring\n",
      "mu=0.2 # coefficient of friction between the block and the surface\n",
      "g=9.81 # m/s^2 # acc due to gravity\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Consider the F.B.D of the block\n",
      "# Applying the principle of work and energy between the positions 1 & 2 and on further simplification we get the generic eq'n for delta as, 750*delta^2-16.03*delta-8.015=0. From this eq'n e have values of a.b & c as,\n",
      "a=750\n",
      "b=-16.03\n",
      "c=-8.015\n",
      "# Thus the roots of the eq'n are given as,\n",
      "delta=(-b+((b**2-(4*a*c))**0.5))/(2*a) # m\n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The maximum compression of the spring is \",round(delta,3),\"m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum compression of the spring is  0.115 m\n"
       ]
      }
     ],
     "prompt_number": 28
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16.16-11,Page No:448"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Initilization of variables\n",
      "\n",
      "M=10 # kg # Here M=M_1=M_2\n",
      "g=9.81 # m/s^2 # acc due to gravity\n",
      "\n",
      "# Calculations\n",
      "\n",
      "# Consider the respective F.B.D\n",
      "# Applying the principle of conservation of energy and by equating the total energies at position 1 & position 2 we get v as,\n",
      "v=((M*g*4)/(25))**0.5 # m/s\n",
      "\n",
      "# Results\n",
      "\n",
      "print\"The velocity of mass M_2 is \",round(v,2),\"m/s\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The velocity of mass M_2 is  3.96 m/s\n"
       ]
      }
     ],
     "prompt_number": 29
    }
   ],
   "metadata": {}
  }
 ]
}