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2001 Solved Problems in Esas

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414 views1,002 pages

2001 Solved Problems in Esas

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miguel
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* Floot, LBF Building V. Gullas St. ty Telfax (032) 2598759 / (032) 2544384 Celphone: 0917-3279002 | Building (In front of PSBA) R.Papa St. floc, Manila Telfax (032)7966291 jelphone: 0917-3062164 Be ite R. THON oe oy ee Pence Natt dents’ Oui 4985.2 rot ginose Uielan tin Maas Tobe fs Reveewer, Besavila Egincerib Revey Gover 1806-1984 ‘Facally- UY College'bf Engtieering, 19851987 Preston, OV Enginaing & eNiocire Ati Assen ee ae - “ROMEO'A, ROJAS, 3 BSEE,€It 1991 (Cure Eauge), BSECE: TJ 2 Class industia Eletvian, TESDA, Mar SME (Suna Curt Laug) Ln, hae rie Noon aes oe a = as Foriner Reviewer: CEERS = CPRC, MERIT Review Center“ and BE Ren 2 | geet een IMPORTANT: Ary copy ofthis book not bearing the signtare of anyone of ithe authors or of the publisher or thi page shal be'coesdered as comin Sroman illegal source. Pe Re mH Copyright © 2004 by Benchmark Publishing ALL RIGHTS RESERVED. No part of this book may be reproduced by any means or stored. in a database or retrieval system without the prior writen permission from the publisher Printed in the Republic ofthe Philippines Published by Benchmark Publishing Distributed by: EXCEL REVIEW CENTER 2 Floor, LBF Building V. Gullas St Cebu City TeVfax (032) 253-8759 Celphone: 0971-3279002 4° Floor, CMFFI Building, R. Papa St ‘Sempaloc, Manila Tevtax (02) 7365291 Celphone: 0917-3062164 ISBN 971-92903.0-7 PREFACE Nothing is more frustrating for an ‘examinee than the feeling that he/she ‘could have dori better had there been some review materials for the most ) cronrna) ae 1 Analytical Condition: Force (vectors) in equilibrium must satisfy ‘the three glven conditions. Rf ae - > Friction isthe force that arises to oppose the motion or impending ‘motion of two bodies in contact Stati friction the force between two stationary surfaces In contact that 'Prevents motion between them. Ithas a certain maximum value called starting fiction Dynamic or kinetic friction occurs when there is relative (siding) motion ‘at the interface of the surfaces in contact Rolling Friction occurs when one surface rotates as it moves over the ‘another surface but does not slp or slide at the point of contact, Tee Fer speci effet ot fins the folowing witbe zed i= coef of tation a= cocticent of manic or Wnt icon and He? He 6 Chapter x ~ Engineering Mechanics (Statics) sal of cota icon (igs) tere sack de) ‘eel of Ficion ‘© ‘PARABOLIC CABLES ‘The cable is parabolic if the loading is uniformly distributed horizontally and A the spant-sg aorta than 100: (= L >10 EO . 7 For Symmetrical Suppors: (Parabolic Cable) © The Tension (T) at the supports: 300i Solved Problems in ESAS - Excel Review Center © The Length of the cabie: > Approximate Formula 4 + r A & S ¢ ah NS H €@ Tension at the Supports © Tension at the lowest point were "T= tension at he support H = intent atthe lovest int W= etenty ofthe Sead 4= 3 1L = span or distance between suppats "Chapter 1 - Engineering Mechanics (Statics) eormany ‘The cable is a catenary if the loading Is uniformly distributed along the length Of the cable and the span-to-sag ratio Is lesser than or equal to 10. \a. For symmetrical supports ( Caterary) | @ Tension at the supports (T) and the intensity at the iowest point (H) © Half Length of the Cable s=csim (5) te) 200% Solved Problems in ESAS - Excel Review Center (© Tension at the Supports yee econ (22) ya ooosh (2) c te szesinn(%) —s.=esinn (2 where Te tensinat he apport = enn te vet sit ciel bce peretingh eile ero Y Se hal eng ofthe able ini dara sor dance between sips Gen Centrotd or center of gravity isthe point where the weight of the body is “concentrated, and at the point object wil not to rotate nor tend to rotate. Inertia i the natural tendency of an object to remain at rest when itis at rest fr inmotion, to continue moving at constant speed. “Moment or toque is the cross product of force and the perpendicular distance to which the force is applied. memento texgue 4 fore ‘=! moment rm or perpendiua distance “Moment of Inertia the natural tendency of the body to rotete of tend to rotate ‘Juste the distibution of area, volume or mass elements of the body. this also known as the second moment. «© Parallel Rais Thoerem CTransier As inert} “The moment of inertia ofthe body at a certain axis is equal to the sum of the moment of inertia with respect to the centroidalaxis parallel tot. and the product of the area and the square of the shortest distance between the two paral! oe SaaS sof ‘tance = caida oc nesta ats 2 cannon. MOMENT OF INEITIA. OF COMMON GEDMETBIC FIEURES (0 Triangle i 300% Solved Probiems in ESAS ~ Excel Review Center = Chapter 1 = Engineering Mechanics (Statics) © Rectangle | @ Thin.walled hollow Sphere @ circle ze. NA Toss For lus parlleazis theorem © Ellipse Yee (CL) mass momen OF INERTIA OF COMMON GEOMETRIC SHAPES (© Homogeneous Solid Sphere 300i Goived Problems in ESAS ~ Excel Review Center © Rectangular plate, axis through center, (Chapter 1 - Engineering Mechanics (Statics) Test I Problem: Two forces of 20 units and 30 units act at right angles. What is the magnitude of the resultant force?, A 4 B42 c. 24 D. 36 Problem: . ‘A rope is stretched between two rigid poles 40 feet apart: A load of 100, Tbs was placed at the midpoint of the rope that caused it to sag 5 feet. ‘What is the approximate tension in the rope in Ibs? A 190 lbs. B 160s €. 208 ibs D. 215ibs Problem: A box is being pulled by a force of 20 lb exerted in’ rope inclined 30° ‘with horizontal, What is the effective component of the force pulling the box? A 1889Ibs B 16.21 ibs ©. 17.32 bs D. 18.12 ibs Problem: [An electric post is supported by a guy wire which exerts a pull of 100 N fon the top of the post. if the angle between the guy wire and the ground | 60°, determine the vertical component of the force supporting the pole, A. 866N B 668N C. 766N D. 966N Problem: ‘The resultant of two forces in a plane is 400 Ibs at 120°. If one of the forces is 200 lbs at 20° what isthe other? ‘A. 427.77 Ibs at 144.28 degrees B. 435.77 lbs at 124.27 degrees 1-15 “or Sofved Problems in ESAS ~ Exeel Review Genter Chapizr 7 > Engindoving Mechanics eatics) ©. 407.27 tos at 194.82 degrees 10, EE Board April 2001 & 37727 bs at 144.38 degrees Determine the divergence of the vector. V= ii) + (oy) + k(xyz) at the I eet ) for) + ky) at the 6. Problem: Determine the resultant ofthe following coplanar forces: 90 tbs, 210% | A 9.00 4130 bs, 260°; 15 lbs, 30° and 58 ibs, 80". B. 11.00 13.00 ©. 7.00 A. 132.78 ibs, 55° B. 135.94 bs, 235° C. 143.78 bbs, 218° 11. Probles D. 127/80 bs, 225° What is the cfoss product A x B of the vectors, A=it4j+6k and 2 Problem: Oe oy +3) + 8K? The five forces shown act at t ran point O. What is the magnitude i ofthe resultant force? ow | B -iti+k A SEN | ©. 214 7)-5« & stan 1 oe t D. 2i+7)+5k c. 1558N Ay_77 ON DL 158.1N xe 42, EE Board September 2001 i ‘The thres vectors described by 10 cm / at 120k degrees, k= 0, 1, 2 encompass the sides of an equlateral triangle. Determine the 3 the vector cross product: 0.5 [(10// at 0 deg) x (10 / at 120, EE Board March 1998 In the system shown, 2 5 ko block rests on @ horizontal table top and is attached with ezotal sting to 8 second A 6 reese snow. What he S80 8 ave fr the mass, & soo ete tel locket romath 5 433 stone ue 43. EE Board September 2001 A. seek Sener ipan at Tak degrees, k= 0 1,23, 4 encompass the 8 teekg res rear pentagon, Determine the magnitude ofthe vector 8 en eee suc {iol a 144 deg) x (10/2278 deQ) B. 20K A. 1904 9: SEBoard April 2008 8 are Shai Smersorl vector: 6. dese 5 tes (xy) + ye) + K(Bzx) B= ikyz) + (22x) + KCOHY) Determine the scalar product at the point (1.2.2) 14. Problem: A ue What is the angle between two vectors A and B if A = 4i + 12] + 6k and Raat B= 24i— 8) + 6k? c. 132 cme A. 168.45" 8 84.32" C8632 D. 2464" 2001 Solved Problems in ESAS - Excel Review Center 15. EE Board March 1998 Given the 3-dimesional vectors A =i by) + iy) + k Gen) B =i(yz) + j@z) + ky) Determine the magnitude of the vector sum |A+B| at coordinates (2.1, A 292 B 2088 C. 27.20 D. 2473 416, EE Board October 1997 ‘A 100 kg weight rests on a 30° incined plane. Neglecting friction how ‘much pull must one exert to bring the weight up the plane? A 86.67 kg B. 100k9 C.70.71kg D. 50kg 17. EE Board October 1992 ‘A simply supported beam is fve meters in length. It carries a uniformly distributed load including ts own weight of 300 Nim and a concentrated load of 100 N, 2 meters from the left end. Find the reactions if reaction Aat the left end and reaction B atthe right end A. Ra=BI0N, Re= 700N B. R,=820N, Ra=690N C. Rk 830 N, Ro= 680N D. Ra=B40N, Ro=670N 16. EE Board October 1993 ‘A beam is loaded as. shown below. Solve for Ri and Re A. 140840KN B 120830KN ©. 1208 40KN DL 1408 30KN Zz Chapter 1 ~ Engineering Mechanics (Statics) 19. EE Board October 1993 2 gman can exert a maximum pull of 1.000 N but wishes to fit @ new Stone doo fo his cave weighing 20000 NW ne uses ey loser must the fucrum be tothe stone tan te his kena fo POM MUCH A. 10 times nearer 8. 20 times farther ©. 40 times farther ©. 20 times nearer Problem: The three concurrent forces acting on the body a8 shown aren equitoium, 5! Which of the following most nearly —_ 308 gives the vertical component of the TTB KN fores? A. 788kN B. 702kN ©. 732kN D. 653KN 21. Problem: Two men are just able to lit @ 200 kg weight with a crowbar when the fulcrum for the lever is 0.3 m from the weight. The fiet man exerts he Strength at 0.9 m while the second ie at 1.5 m from the flaws respectively, the men interchanged positions, they can raise a 240 kg Weight. What force does the frst man exert? A 40kg B 50K Cc. 8Okg D. 55K9 22. Problem: Determine the reaction Reon a simply supported beam as shown, 510 ibs 520 ibs 530 lbs 540 ibs pom> 2001 Solved Problems in ESAS - Excel Review Center 23. Problem: fw ‘A'15048 cylindrical tank Is at rest : ‘as shown, Determine force P equired to move the tank up the higher-level surface a tome & ere af 24. Problem: The system as shown isin ‘equilioium, Determine the force exerted by cable AB. A. 88.67 Ibs B 86.97 bs C. 79.68 bbs D. 76.98 bs i 25. Problem: JRinpod whose legs are each 4 meters long supports load of +000 kg. | Fa foct of the tripod are at the vertices of a horizontal equisterel 4 tangle whose side is 3.5 meters, Determine the load on each leg A seastta seas |. meee otter 191 JR certain cable is. suspended between two supports at the same A ecation and 500 ff apart. The load is 500 lbs per horizontal foot 2003 Solved Problems in ESAS ~ Excel Review Center. 41. EE Board April 1997 Four turns of rope around a horizontal post will hold @ 1000 1b weight twith a pull of 10 los. Find the coefficient of fiction between the rope and the post. A 018 B 018 © 022 D. 030 42, Problem: ‘Two pulleys 2t in diameter are connected by a belt so that, each has: the belt wrap around half ofits circumference. The coefficient of friction between the belt and the pulley is 0.20. Ifthe tension in tight side is 300 tbe determine the tension in the slack side when the belt is about to slip. A 145.23 ibs B. 155.34 ibs ©. 160.05 lbs D. 163.85 bs 43. Problem: [A object weighing 400 N is held by a rope that passes over a horizontal ‘rum. The coefficient of friction between the rope and the drum is 0.25. if the angle of contact is 150°, determine the force that will raise the object. A. 750N B 760N Cc. 770N D. 780N 44, Problem: Determine the abscissa of the ‘centroid for the shaded area as shown, 25 35 45 48 goe> 45. Problem: Determine the moment of inertia with respect to its vertical Centoial is of the shaded 352.65 355.69 362.19 367.12 pow> 46. Problem: Determine the moment of inertia of the T-section as shown about its centroidal axis parallel to the bate. A. 12098.12 B 1009431 c. 1109212 DB. 13001.42 EE Board October 1999 A. trapezoid has two equal slanting sides, a 6 om base and 23 cm top parallel to and 5 om above the base. Determine the moment of inertia of the ‘trapezoid area relative to the top side, in om A 218.75 B. 240.63 c. 284.69 D. 198.86 48, EE Roard March 2998 Jers has a demeer of 20cm, Determine the moment of inertia of tho cedar area oltve tothe ts perpendicular fo the tea Tu the center of the circle in cm*. vd 14,280 45708 q279 19007 pom> 2001 Solved Problems in ESAS EE Board March 1998 ‘An isosceles triangle has a 10 fem base and @ 10 om altitude. Determine the moment of inertia ofthe triangular area relative to a line parallel to the base and {through the upper vertex in om? A 2.750omt & 3.025 emt ©. 21500 em* D. 2273.on* Problem: Given the truss shown. Find the force on member AC. 5000 we A. 12,000 KN. B. 13,000kN ©. 42,500 kN D. 13,500 kN (Chapter 1 ~ Engincering Mechanics (Statics) SOLUTION TO TEST I a R= (20) + (30) R Pa R = 96.05 units 3 0=75.963" mr 100=Te0s0+ Toos® 100 o T0088 400 2 c08 75.963" T=206 tbs TE rvssexe =Feos8 Fence = 2000830" Fores = 17.32 [0s T Fy =Fsing Fy =100sineo" Fy =866N | ood Solved Probleis in ESAS - Exel Review Canter Chapter 1~ Engincering Mechanics (Statics) HEB isin casne aw rh Lhe FZ =F? +R? -2F Roos 100" F? = (200}? + (400/* -2(200(400)c08100" Siti.s6 ® 47727 be “7878 T 3 Breet Using sine aw sin 100" sina | Refer tothe vector alagram ‘a77.27 400 A a=5562 ' = 180+ ” & 0; (4-207) =180" | 0 =236 0, +(6562" -20")=180" u 0, = 144.38" ij TCs | eden cox s0N yess fess | where: “BF, = 20 + 30 008 30° + 40 cos 60° - 60 sino" ea EF, = 35.98 EF, = 30 sin 30° + 40 sin 60° + 50 + 60 cos 30 “= 151.60 i a BF, = 151.6 R = (35.98)? +(151.60)"_ R =1558N ! Fe=190 Fe=i30 o i YF = 1800820" + 55cos80" ~90c0830" - 190c0880" Sp.=-77975 Diy = 5sing0" + S5sin80" -90sin30" -1908in80" SR =-111.98 R= (a5) (26) een : ‘The maximum force that can produce the ‘fiction will be; ((): [err ere + (191.367 eum 136.94 Ibs ae f= (0.50)(5 kg X9.60 mis*) f= 245N (90,7) sin37?_ eet m= 1.89 ko Awi(ay) +292) B2) peilyz)+i(22)+K(399) eB =(xy Kye) #( 292X220 +3219) At (12.9) x= Y= 2 Z=S jpoB=(1N 2K2K3)+(2K2HSH AAA *(BNSH AsBa138 Divergence = 2x-+¥ ‘at @2A) 7X2 SY EBZ pivergence = 212) -3 + (2X2) Divergence = 9 KAKS ANZ) ee ee Chapter 1 ~ Engineering Mechanics (States) i AxB=|1 2 ik 46 3 5 axB= (201+12)+3k)-(6K+5)+180) AxB=216Ti-6k fii kl axe-|1 4 8| Naa 2.3 5) Axio tead as A.crove 8" ‘change the | A AxBe| 1 Thus: B= 102120" Re-write the gh A=101+0)+0k xB =(204+12) 3k) ~ (B+ 51+18)) AxB=247/-5 given vectors in rectangular frm using ealoltor soz0"=10+j0 5 +j8.66 wen vectors into thee-dimensionat (j,k) format B= -5i+ 8.65)+08 ik | 0 0 0 5 606 0| Axb-[0+0+866K]-[0+0+0) AxB=06.6k os|AxB|=0.5(668) 05|A x8|= 43.3 units 2003 Solved Problems in ESAS - Excel Review Center aah Rewrite the given vectors into three-dimensional (jj k) format: A=-8,091+5.877)+ 0K B=-8.081-5.877)+0« i ik | AxB=|-809 5877 0 809 -5877 0 Ax B=[0+0+ 45.548k]-[-47.545k +0+0] AxB=95.09 Thus: 25|A x8] =2.5(95.09) 2.5|A xB|=287.725 unts Ale 12)+ 6K B= 24\-8)+ 6k nla +02 OF alens lel= Vea scorer jel=26 eB =4(24)+(12\-8)+6(6) AeB=26 .. Chapter 7 ~ Engineering Mechanics (Suatie}) AB =|Aljoosd-> formula 36 = 14(28)0080 136 o=84.328° (ay 4 2y) +) ( 2y2 + 22K) + k( 32x4 Sey) where wye2,zet A+B = i(642) + j(446) +49 +18) A+B =81 + 10j+ 27k asa) = VleF t0F Way = 20.88 DFraines =9 P=Wsino P=100sin30" 400(2) +1500(2.5)-Ra(6)=0 Ry =700N m0 Rq(6)~100(3)~1500(2.5)=0 30d Solved Probicma in ESAS - Excel Review Genter ‘Ghapter # ~ Engineering Mechanics (Statics) Referto the closed polygon (viangl} Using cosine law: (62.2) =(64.5) +(77.8'-2(64.5\77.8)c08a a= 50,702" 0 = 90°--(80"+a) 0 = 90°30" 50.782" 0=9.208" Let. F =vertical component ofthe 77.8 KN force F=77.8cos9.208° F=76.797 kN XMa=0 Ry(4)~120(3) -20(6)- 40(2)=0 R= 140 KN Tam =O rm 300(0:3}= (0.9) 8.5) g og {os IM, =0 B=60-0.6A>Eq 1 | 40(2) +120(1)-R_(4)~20(2) =0 y= 40 KN Maen = 3400.3) = B(0.9) + A(1.5) 102=0.98+1.5A-+Eq.2 ' Mascon =9 W0%2)-F(x;) = 0 ‘Substitute Eq.1 in Eq.2: W0%2)= FOG) . . (20,000) =(1000), 102=0.9(60-0.68) +458 % 102=54-0.54A+1.58 X= 20% A=50kg wie [<__—_ nln s ‘hus, the erm must be paced 20 snes nana athe steven to ib han Vein Wile the resultant load of che triangular loscing while W. Is the reoultant, of the rectangular losding as chown. 2001 Solved Problems in ESAS - Excel Review Center ‘Chapter 1 - Engineering Mechanics (Statics) a4 x= $196) m=t2in 1 =H) x= 18 in 100 tbs T,00830" ~T,cos4s" =0 T, =0.8165 T, > Equation 4 DR -0 T,sin4s’ + T,sin30°—100=0 0.707 T, +0.5 T, =100-+ Equation 2 Waxy + Wes 540(12)+ 720108) Ry 540 bs Substitute E91 in E92: O.707T, + (0.5)(0.8165T,) = 100 4.11821, = 100 7, =6967 ibs Using the Pythagorean theorem x= —# x=6.928 ft =M. Wx -P(12) 150(6.928)~-12P z00x Solved Problems in ESAS - Excel Review Center ER =0 ‘3(Re0830.33) = 1000 R=386.10 kg Note: Since the loading is horzontaly dstrbuted, the osble is parabolic Using the approximate formula ef _32t ae 5 ~s00+ 9007 _ 3200" (500) 5(600)° S=L+ > formula W=500 Ibs/tt (ey +H? formula (2 ws we ME formal ed a Substtute Eq in Ea. (at) -() a) Ved (a2 2 f415 Pu 6000 tye. ele 6000? =0. oseast +39,0625x10°U* (=U + 144001? ~6000°(25600) ‘Chapter 1 - Engineering Mechanics (Statics) Using the quadratic formula: 144008 {t4400¥ — 4(4y-25600(60007 2H) Be 1920054 2 052827 976.128 m 28 Using the approximate formula: sat +8h 2 a ae ae s2at 3140) ao 5.4=0.08667¢" — 54,000 = 656.676" —o* * 666.674" + 54,000 45.4 = 40+. Since the load is evenly tributed along the span ‘the cable is parabolic Using the quadratic formula: @ 667 + (666.679 - a(164,000) 2th = Sener 38 Take minus sign for minimum value: of =94.355 d=9.714m 0d Solved Problems in ESAS - Excel Review Center waist we 16 kgim wl ince the load of 20x? 214) + Equation 1 wa, formula Hae ot = 20(20= x1)? : quate £1 to E42 q x2 _ 20(20- “Hay 212) 2(20 - x, 300 - 80x, + 2x," 100-B0x, +x? ->Equation 2 | | | Using Substitute x; in Eq. ‘Chapter 1 - Engineering Mechanics (Statics) the quadratic formula 202 (E807 41800), 2 20+56.568 2 1.716 m 20(11.7167 24) H=343.16 KN 1 (Wa)? + > formula =v royfon rf +99810F Tem 5.53 N T=04 (60-+6)-» Equationt 6400 +4600 +0? = 400? + ¢? Substitute cin Eq. . yeStect (80 +07 = (400)? +0? 160¢ = 153,600 0-960 4(80 +960) 16 kg d= 180-150 Nove: The maximum tonsion of the cable is d=30m alnays the tension at the supports. yiastic (S00}? = (3007? +c? =400m Substitute cin Eat x=elnS2t x=400in300+ 800 x=277.258m L=2x 1 =2(277:258) L=554.52m = 2y4 =225m Wye 2y2 200m y? =SP+e? (225 = S? +0? > Equation! (2007 =S,? +02 + Equation2 Subtract E92 from Eq.1 (225)*- (200) =8?-8? $2 = 10625+8,? + Equation3 8,48, =300 S,=300-S, 3s} =(300-8,7 ? = 300" -6008, +8,? -> Equations Equate £9.3 to Ea 10625+8,2 = 300" 6005, +8,? _ 300? ~ 10625 500 8, =132.28m & Substitute So in £2: N= Woos30" + Psin30* N=250c0s30" + Psin30" 2 N=216.506+0.5 P DA-0 P0830? = Wsin30" +F Pcos30° = Wsin30 + uN 0.866 P = 250(0.5)+0.2(216.506-+0.5) 01866 P=125+43.3+0.1P P= 210.71 kg YM =0 N¢18¢0870") -(0.25N)(18sin70")~ 1200870) - 10 18-1)007 150-4220 268.00 -1108 45 +6155 1 8281477525 16,669 «0 Equation 1 Subsite Nina Dn Refer to the figure: 4,928(300) -1477.525+61.563x=0 xa146 ft Using Pythagorean theorem. W(0.5x)=N,(4) 12(0.5(3)=N,(4) Np = 44.145 N xe FiaNe F,=46.145.N Nzz Since the required force isto tend block vo move up the plane, use the cooficlent of kinetic fiction. F=Wsine+F F=Wsine +0.30N F=Wsino +0.30(We0s6) F = 250(sin30") +0.30(250)(c0830") F = 189.95 Ibs 200% Solved Problems in ESAS - Excel Review Center Chapter a= Es =a ‘Substitute Ne in Equation 3: P=74.24(c0830")+0.4(1214.32) uma (56)(92.2) n=0.01t Consider the FBD of block A: 2x rad um B= 4 tums x: B=8n 0881) ion Trio — Substitute values: 00830" =F; 1000 . gues 0.8667 = 0.25%, > Equation + e <— Fi=0.25N) 100 = en") DA =0 Int00 = (6n) int “M(ie) N+ Tsin30" =, h N, +0.5T = 309.81) N= 2043 -0.5T -» Equation 2 Tatts w Substitute E92in 1 ne “Tw angl ofwrapin iter { geet =025(294:3-057) 8 pilyo en a be 0.8667 =73.575-0.125T reerence o 120° oF x ta rains ‘Sea se Consider the FBD of block 8: Dro P=To0s30" + P'=74.24(c0830")+0.4N, + Equation 3 DR-9 N= W,+Tsin3o" 20(9.81)+74.24(8in90") 121432N A Fret formate Bexrad Substitute values: 1-46 x25 B=150" x 180" 62 rad Maze Using Varignon’s theorem: Since he force requied is Substitute values: sod to raise the object, then A Te=400.N it must be the tighter side or ‘he bigge foros. Aras = Apts + Rake +Aaks AT _ ganze) og ]= [era] +[2cay}io)+[6r2)](@) 400 ) T= 4000025) 1, =770.087 N vais 20s ra » ‘Az = atea ofa half circle Ay =9(15) Aras = (2) +4(2) +62) Au #82 By inspection: A, #1255 By inspection: x45 4 as 45) x= 2.122 3n Using Varignon's theorem: Anat = AX Ake (138-12 59 (8-195 5)=(12.5n\2.122) x=5.47 By inspection: d, =5.47-2.122 6, =3.348, Using transfer axis moment formula: Wyo = (h+ Aid?) ~(L,+Az6,7) eee) 1 (mts) +(135y0.977 page Sm). asf | J 2 Yen3 Using Varignon’s theorem: raat =Anys + Any (80+ 297) =(80)118) + (90,3) y=9.118 By inspection: 200% Solved Problems in ESAS - Excel Review Center | 4,=16-9.118 | 4, = 6.882 418-3 418 Using transfer axis moment formula: baationae) loons bal Dah? BE nat) (Ba Ad? ) (<9 +(coy6.882y || (2802. oy0.118 hea yo = 10,084.31 no ( By inspection: Chapter 1 ~ Engineering Mechanics (Statics) Using ante ans moment formula b= (hr Ae) +(e) + As0) ' Ce na) ‘(ase (dma? ( (2a yy =218.75 em* bate asa? j 36 100) Joa 58? «a re( 22 91237} s2[ 48" sar0{ $y] zat seat 2 y= 2208, a 15,708 cm* br? _ 101109 4 4 500 om* “The reaction at point A: (Rx) Mm, = 0 5000/24) = (Rax2) R= 9760 ‘By Method of Joint (@ joint 4) vise nen 3750 750 8 = ta (2.989 12572.3 kN aamsom grirase c (12572.3) cos 17.38° 2000 KN a _ CHAPTER 2 Engineering Mechanics - Dynamics Dynamies is the branch of mechanics that deals with bodies in motion. “Motion i a progressive change of position of the body. Kinematics is the branch of dynamics which describes the motion of bodies ‘without reference to the forces that either cause the motion or are generated a8 a result of motion. Kinematics is often referred as -eometry of motion. Recilinear motion type of motion in which the body moves in a straight line ‘or is moving in the direction peralel to its displacement. (@ RECTIUMEAR WOTIOK (Kinematics in one-dimension } 60 var mation. Koti wi anata pando elo ae atc: V abe ven set ory {2 Uniformiy accelerated motion. A motion with constant change in. velocity ‘or of uniform acceleration oe oo ie ‘Yi te xg rial veocty S isthe distance raversed (dsplacement {isthe tne Unt asthe acceleration, we (+) when the body accelerating, (i> Ve} tse () when the body declan, (Vi Social Cases Atte maximum height : 4s (+) when ging down |) when goog Up (agains the gray) | 4,0 (when ts cropped. and theo tha drops, [spot moving ths mst asa be the cae reef) ‘al vlty the time of tig = 98tmis = 981 emis! = 3221 ‘Phas "isthe tine fight om the maert of release tothe mania eight At the maximum range (R) © PROIECTIE (Kinematics in Two-Dimensions ) Projectile. A body which after being given an intial velocity (with an inital angle of release), i allowed to travel under the action of gravity alone. ‘Pra: "the ttl tie of Hight fam the moment the recon, ‘2 The General Equation ot Prelectle release othe masinum polo impact in ¥ ‘ust be postive forthe project ts above point of reease ust be negate forthe projectile its el point of recase any =0 then Chapter 2 ~ Engineering Mechanics (Dynamics) Uniform Motion 2001 Solved Problems in ESAS - Excel Review Genter __ ‘The Range of PofecteIncined at sn angle: (ciove or below) cum Brat snBeos (a xe) Boos! Uniform Angular Acceleration pe Ae she cost ain SS DO 7; te above equations canbe dried by using the gener equate of project and sbatsing X= ReesB :y=Ranp, ‘@ ROTATIONAL KINEMATICS: Angular Displacement (@) is the angle through which a rigid object rotates ‘bout a fied axis. S/ Unit: radlan (rag) Linear and Angular Retations Sere Vere Radian the SI unit of displacement defined as the circular arc length ($) traveled by point of a rotating body divided by the distance (r) ofthe point’ from the axis Phe, derad = 360 degrees = 1 reveuton vere anulr displacement in rans ular velocty (rade) gular accleaton (rede) fas feat ROG Kinetics is a branch of dynamics which deals with the study of bodies in ‘motion particularly the force involved in the motion, Average angular velocity owe) is the angular displacement of ah object ‘ivided by the time elapse. S! Unit: rad/s Instantaneous angular velocity is the angular velocity that exists at a ‘given instant. 7 Uni: rad/s Angular acceleration (2) isthe change in angular velocity divided by the time elapsed. Sf Units radi? Newton's First Law ( aw of nerzit) AA body at rest will remain at rest or in motion at that constant velocity, unless acted upon. by some unbalanced, external force, 2 Tangential velocity (Vi) is the linear Velocity ofa tangent point on a rotating body . It represents how fast the point is moving along the are ofthe circle. ST Unit: mis Tangential acceleration (a) the linear acceleration of a tangent point on a Fotating body. ST Unit: mis" Uniform circular motion is the movement of a point particle at constant tangential speed in circular path. 2-4 001 Solved Problems in ESAS ~ Excel Review Gunter Chapter a - Engineering Mechanics (Dynamics) Newton's Second Law (Lav of accelrston), ere: ‘The acceleration (a) of a body is directly proportional to the net force (Fa) ‘acting on land inversely propotional tos mass (m) EF = reversed eect force dla oppete nthe recon ofthe cern, > wag Foot ot <> an Fs BF oom, CCURVILINEAR TRANSLATION: Comtrpetat and Centrtugal Forces > The Normal Force(Centipetal force) Newton's Third Law (Lav Action and fesctn) For every force that acts on one body there is a second force equal in ‘magnitude and opposite in direction. FilSFe se Fa Newton's Law of Universal Gravitation ‘Every particle in the universe attracts every other particle with 8 force which is directly proportional to the product of the masses of two particles ‘and inversely proportional to the square of the distance between the center of tha mass, a meine gene wee: RECTILINEAR TRANSLATION: D’Alembert’s Principle The resultant of the external forces and the kinetic Teaction (reversed effective fore) acting on a body equals to 260. ‘Chapter 2 ~ Engineering Mechanics (Dynamics) © morose om woMeToM Impatse (1) isthe product of the average force and the time interval during which the force acts. ea Momentum (p) is the product of the mass and velocity. lesen af the cord W= weit ral force finde ie 1 tne or oe comple relation ice formas ta pace by. BANKING OF HIGHWAY CURVES: > deal angle of banking(no friction) Callison refers. to the mutual action of the molecules, atoms , and ete. w low ‘when they encounter one another. ‘Momentum is. conserved in ail collisions or explosions. In the conservation of — momentum; the final total momentum is equal to the initial total we momentum. The essential effect of colision is to redistribute the {otal momentum of the colliding objects. Al! collisions-conserve Pyaar ‘momentum, but not all of tem conserve Kinetic energy 28 Well ‘and about t skid: (wth fiction) (th Collision falls into three categories: “A Blastic. IN B. Inelastic and w PX ©. Completes Ielsic colisions taste Colision i acolision vtch conserves Kinetic energy Inelastic Collision 's @colision which. does not conserve kinetic energy. mee — Some kinetic energy i converted into heat energy, sound energy, and s0 forth. Completely Inelastic Collision is the colision in which the objects stick together afterward In such collisions the KE loss is maximum, cont tin Coil of Retitaton (0) ste roto. between the relative. speeds of Spedtion wo coliding objects fer and before they cade Teper oy = © Controidal Rotation Nome rant HE, = rotor! ete energy (2 = rlaonal peed

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