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Principles of Engineering Mechanics (2nd Edition) Episode 4 pot

Principles of Engineering Mechanics (2nd Edition) Episode 4 pot

Principles of Engineering Mechanics (2nd Edition) Episode 4 pot

... moment of external forces = C moment of (mass x acceleration) = moment of the rate of change of = rate of change of the moment of or momentum momentum We may n~~ke use of ... methods The magnitude of 4 is C’C 4. 7 BC 0.19 4 = - = - = 24. 7 rads2 To determine the sense of 4 we note that the normal component of urn is c’c in the sense of c’ to c; thus ... [-rsinOwo+I~in4%~]i VG = [-r~in8w~+asin4o~~]i- [bc0s4%~1j (jBc = sec4 - sinOwoZ - sin40Bc2 k ~c = -[~~~s~w~~+I(cos~~~~-sin~~~~]i a~ = - [rcosowo2 + a (cos4%c2 - sin4(jBc )I...
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Principles of Engineering Mechanics (2nd Edition) Episode 5 pot

Principles of Engineering Mechanics (2nd Edition) Episode 5 pot

... Moment of inertia of a body about an axis 77 Figure 6.8 = (pLdrrdO)r2 hence for the whole body 2lr a pLr3 drd9 IGz= Io Io = 1,“ pLr3dr2.rr = pL2.rra4 14 = 4. rrpLa4 The mass of ... 6.16 The track of the wheels of a vehicle is 1 .4 m and the centre of gravity G of the loaded vehicle is located 92 Energy k. e. = 4M (O x B ) - (W x B) + 4ZG w2 = ~MB~~~ ... in Fig. 6 .43 has a total mass of 760 kg. Each rear wheel moment of inertia of 6 kg m2. The moment of inertia of the front whee1s may be neg1ected. 6.21 See Fig. 6 .41 , A lift...
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Principles of Engineering Mechanics (2nd Edition) Episode 7 pot

Principles of Engineering Mechanics (2nd Edition) Episode 7 pot

... centre of mass G falls through a vertical distance ~ (4. 675)~ = 45 .0 J ho = (AG)(sin 0, - sin45") b) The free-body diagram on the left of Fig. 8.19 is for a fixed quantity of fluid. ... velocity at t = 6.728 s is given by v = +[O+ 11.63 +4( 2.1 24+ 6.117+9. 842 ) + 2 (4. 159 + 8.008)] = 36.1 mls Example 8.7 A sphere of mass ml is moving at a speed u1 in a direction ... of gravity G is located as shown. The mass of the truck is 1OOOkg and its moment of inertia about G is 650 kgm'. The mass of the wheels may be neglected. a) If the angle of...
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Principles of Engineering Mechanics (2nd Edition) Episode 6 doc

Principles of Engineering Mechanics (2nd Edition) Episode 6 doc

... BC and CD as shown in Fig.7 .42 . Each has a length of 0.5m and a mass of 2.0kg. A torsion spring (not shown) at A has a stiffness of 40 Ndrad, one end of the spring being fixed and ... consequence of equations 8.8 to 8.10 is the relationship which exists between the velocity of approach and the velocity of recession of the points of contact. Velocity of approach ... oscillations. Figure 7 .44 Determine the natural frequency of each system for 7.13 An electric locomotive develops a constant power output of 4MW while hauling a train up a gradient of slope a,.cSin...
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Principles of Engineering Mechanics (2nd Edition) Episode 9 ppsx

Principles of Engineering Mechanics (2nd Edition) Episode 9 ppsx

... -3.01 -45 .00 which makes the curve pass through the critical 2/r -6.99 -63 .43 4/ 7 - 12.30 -75.96 1 o/r -20. 04 - 84. 29 Once scales for the graphs have been chosen, all of the graphs of ... three 1/(107) -0. 04 -5.71 particular values of K are sketched in Figs 1/ (4~ ) -0.26 - 14. 04 10.31(a), (b) and (c). 1 /(2 7) -0.97 -26.57 In Fig. 10.31(b) a value of K has been chosen ... S)z = CDw (iv) 4, = k2W (VI The flow into the tank is and the net inflow is equal to the area A times the rate of change of height h,: 41 - 40 = ADho (4 Equations (i) to...
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Principles of Engineering Mechanics (2nd Edition) Episode 10 docx

Principles of Engineering Mechanics (2nd Edition) Episode 10 docx

... phase margin of 45 " this value of I G(jw) 1 should be unity so C needs to be reduced by a factor of 4. 4 64. This gives C = 10 14. 4 64 = 2. 24. Example 10.7 Figure 10.51(a) shows ... 1 .42 7 rads the phase angle C$ = -135.02". Using the value of C = 10 as in a) above we find that the corresponding value of 'I G(jw) 1 is 4. 4 64. For a phase margin of 45 " ... (recos4-2re4sin4+2L8cos~)ee + (rc$+ 2L&+ re2sin4cos4)e4 (11.20) 11.6 Motion relative to translating axes been reduced. In many problems it is often easier to express the motion of...
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Principles of Engineering Mechanics (2nd Edition) Episode 12 pps

Principles of Engineering Mechanics (2nd Edition) Episode 12 pps

... the yz plane (1 2. 64) The proofs of these two theorems are similar to those given for moments of inertia in section 6.3. Using the definition of second moment of area equation 12.62 ... of cross-section as shown in Fig. 12. 24. The applied in a plane normal to the axis of the shaft. elemental shear force is In the case of beams the loading is transverse to the axis of ... of area and denoted by I. Similar to moment of inertia, the second moment of area is often written as I=Ak2 where A is the cross-section area and k is known as the radius of...
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Principles of Engineering Mechanics (2nd Edition) Episode 14 pps

Principles of Engineering Mechanics (2nd Edition) Episode 14 pps

... 1.385 ds2, 0 .43 6 ds2 4. 1 87.0 N m anticlockwise 4. 2 4. 3 21.0 kN, 3 .49 kN, 14. 4 kN 5.9 a) 0.5 rads anticlockwise, 4. 4 a) (-2lOi+505Oj) N, 4. 5 a) 190N h 52", 4. 7 -30kN, ... anticlockwise 8.16 a) (-11.311' +4. 69j) ds2, 8.17 20 .44 kN 8.1 (2 348 1'- 540 j+ 3924k) N 8.7 a) 11.4kN, 8.11 (23.4k) rads', b) (- 54. 4i-tO.23) N, (9.371'+ 12.5j) N ... 4. 13 FA = (-2i+ 247 .5j) N, b) d[Roxl/m]i, c) d/[3~,x,/(2m)]i, FB = (-252.5j- 11.9k) N, Fc = (5j+ 1.9k) N d) d[2Rox1/(3m)]i 4. 14 42 04N 3 .4 (61i+ 19j) ds 4. 15 7000 kg, 69.4...
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Principles of Engineering Mechanics Second Edition pdf

Principles of Engineering Mechanics Second Edition pdf

... Rcose = 49 41~0 ~41 " = 3729.0 m y = Rsin8 = 49 41sin41" = 3 241 .0 m so A is located at point (3729, 3 241 ,800.2) m. For point B, r= 7037m, 8 = 73 .4& quot;, 4 = 1.3"; ... XY z (44 - 1 jk - AxB= A, A, A, (4. 3) Fx Fy Fz Bx By Bz 4. 5 Couple From Fig. 4. 4 we note that by the definition of the vector product of two vectors, equation 4. 1 may be ... Fig. 4. 14 is constructed of bars which are assumed to be connected by frictionless pins at their ends. Figure 4. 12 FIX + Fh + F3x = 0 Fly + FZy + F3y = 0 (4. 14a) (4. 14b) (4. 14c)...
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Principles of Engineering Mechanics Second Edition H. R. Harrison potx

Principles of Engineering Mechanics Second Edition H. R. Harrison potx

... acceleration of B, aB , can be scaled from the figure. + 1.5(3)(-0. 342 3+ 0. 940 j) % = VA +%/A % = 4i + (0.34i+ 4. 01j) thus = (4. 34i +4. 91.) m/s2 The speed of B is the magnitude of VB ... = 49 41.0 m x = Rcose = 49 41~0 ~41 " = 3729.0 m y = Rsin8 = 49 41sin41" = 3 241 .0 m so A is located at point (3729, 3 241 ,800.2) m. For point B, r= 7037m, 8 = 73 .4& quot;, ... Part 3 Engineering Mechanics syllabuses of degree courses in engineering. The emphasis of the book is on the principles of mechanics and examples are drawn from a wide range of engineering...
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