# Engineering Mechanics

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ACE Engineering Publications Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally| Kolkata Chapter- 1 Force and Moment Systems   Engineering   Mechanics    Solutions    for    Vol    –   I     _   Classroom   Practice   Questions   01. Ans: (b) Sol: Assume F 1  = 2F 2  (F 1 >F 2 ) F 1x  = 2F 2  R =  cosF4FF 222221  260 =  cosF4FF4 222222  260 2  =  cosF4F5 2222  ------ (1) R  1  =  cosFF2FF 2x1222x1  180 =    180cosF.F.2FF4 222222  180 2  =  cosF4F5 2222  ------ (2) 260 2  =  cosF4F5 2222  180 2  =  cosF4F5 2222   2222 10F 180260       F 2  = 100N, 260 2  = 5(100) 2 +4(100) 2 cos         = 63.89 Where   angle between two forces. 02. Ans: (b) Sol:  Let the angle between the forces be   Where, R is the resultant of the two forces. If Q is doubled i.e., 2Q then resultant (R   ) is  perpendicular to P.  cosQ2P sinQ290tan   P + 2Q cos   = 0 P = –2Q cos   ------(i) Also,  cosPQ2QPR  22  R = Q [using eq.(i)] 03. Ans: (b) Sol: Since moment of F about point A is zero.   F passes through point A, R = 260 (180–   )   F 1  F 1x  R  1  =180 F 2   6m F x  F A 0 3m x B Y F   R Q P   R    2Q P    : 2 : ME – GATE _ Vol – I _ Solutions ACE Engineering Publications Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally| Kolkata m N180M F0    m N90M FB    0M FA    0F3F180M yxF0    F x  = 60N ……. (1) 906F3FM yxFB    60  3–6F y  = -90   F y  = 6270 F y  = 45N   F = 2y2x FF    = 22 4560    = 75 04. Ans: (a) Sol:    160w0 wdxdw w =  160 dxx90 = 90 160121 121x    = 90     1602/3 x32 = 60 (16) 3/2  w = 3840 N The moment due to average force should be equal to the variable force R   d =  dw   x 3840  d = x.dx.x90 160   = 90  1505.1 dxx 3840d = 90 1605.2 5.2x      d = 9.6 m 05. Ans: (c) Sol: Moment about ‘O’   M 0  = 100sin 60  3 = 300  23 = 1503 = 259.8  260 N 06. Ans: (a) Sol: F R   =  F y  F R   = 100+150–25+200 (upward force Positive downward force negative) R = 425 N For equilibrium  M A  = 0 (since R = resultant) Let R is acting at a distance of ‘d’ 425  d = 150  0.9+25  2.1–200  2.85   d = 1.535m (from A) 100 N 150 N 25 N 200 N A C D B 1.2m 0.9m 0.75m 16m dw 360 N/m x dx    : 3 : Engineering Mechanics ACE Engineering Publications Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally| Kolkata Chapter- 2 Equilibrium of Force System   120   B   75   45   6060   F BC F AB 200   Fig: Free body diagram at ‘B’   F CD 75   105   60   F BC Fig: Free body diagram at ‘C’   4575   P   T AB  T AC  60 o   30 o   600N   A   01. Ans: (d) Sol: Resolve the forces along the inclined surface  F x  = 0 Pcos45 –Wsin30 = 0 P = 45cos30sin300    P = 212.13 N 02. Ans: (a) Sol: T AB  cos60   = T AC  cos30   T AB  = 3 T AC  T AB  sin60   + T AC  sin30   = 600 N 600T21T23 ACAC       T AB  = 520 N ; T AC  = 300 N 03. Ans: (c) Sol: For Equilibrium of Point ‘B’ )120sin(200)4560sin( F)7560sin( F BCAB   F BC  = 223.07 N From Sine rule at “C”. 105sinP)7560sin( F)4575sin( F BCCD   P = 135sin105sin07.223    P = 304.71 N A   B   C   D   P   200   75 o 45 o 45 o 60 o Y   X    N   W 30 o   P   300   45 o   30 o   Wcos30   Wsin30

Sep 22, 2019

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Sep 22, 2019
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