Vector Mechanics for engineers Statics 7th - Cap 03, Notas de estudo de Engenharia Civil

Baixe Vector Mechanics for engineers Statics 7th - Cap 03 e outras Notas de estudo em PDF para Engenharia Civil, somente na Docsity! PROBLEM 3.1 86 mm: A 13.2-N force P is applied to the lever which controls the auger of a “sr snowblower. Determine the moment of P about 4 when «a is equal to 30º. 122mm SOLUTION First note — P=I32N P, = Psina = (13.2 N)sin30º = 6.60N P, = Pcosa = (13.2 N)cos30º = 11.4315 N j Noting that the direction of the moment of each force component about 4 I22mm — is counterclockwise, Ma = XpaP, + Voa o = (0.086 m)(11.4315 N) + (0.122 m)(6.60 N) — Bb mm = 1.78831 Nm or M, = 1.788 Nm)  86 mm — asp es Vou mm qi PROBLEM 3.2 The force P is applied to the lever which controls the auger of a snowblower. Determine the magnitude and the direction of the smallest force P which has a 2.20- N-m counterclockwise moment about 4 SOLUTION S6ram —+ E pos P For P to be a minimum, it must be perpendicular to the line joining points A and B. a=0=tan! (2) - tm em) = 54.819º x 86 mm Then My = r4bPmin or Pain = My T4B — 220N-m [1000 mm 149.265 mm Im = 14.7389N Pam = 1474N Á 548º or Pi, = 14.74N X 352º 4 min  PROBLEM 3.5 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28º, determine the moment of the 4-lb force about point B by resolving the force into components along 4BC and in a direction perpendicular to 4BC. SOLUTION First resolve the 4-Ib force into components P and Q, where O = (4.0 Ib)sin28º = 1.87787 lb Then Mp = ryBQ = (6.5 in.)(1.87787 lb) = 12.2063 Ib-in or My = 1221 bin) 4  PROBLEM 3.6 It is known that a vertical force of 800 N is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P which creates the same moment about B if a = 10º, (c) the smallest force P which creates the same moment about B. 450 mm 100 mm SOLUTION = (a) Have Ms, = tepFy A quo” = (0.1m)(800 N) o45m = 80.0 N-m MZ N or My = 800N:m ) 4 A (b) By definition Ed Mp = rypPsinO E (6) Foco f ] less om where 9 =90º — (90º — 70º) - a =90º- 20º - 10º A IDE, = « = 60º 80.0 N-m = (0.45 m) Psin60º P=205.28N or P=205N «4 (c) For P to be minimum, it must be perpendicular to the line joining points 4 and B. Thus, P must be directed as shown. Thus Mp = din = "4BPnin or 80.0 N:m = (0.45 m) Prim Ban = 171. 778N or Pin =1778N < 20º4 min  PROBLEM 3.7 — A sign is suspended from two chains 4E and BF. Knowing that the E * tension in BF is 45 Ib, determine (a) the moment about 4 of the force ' exert by the chain at B, (b) the smallest force applied at C which creates the same moment about 4 SOLUTION o a ci Teeth (a) Have M, = Toa X Tor er, “ A ” Noting that the direction of the moment of cach force component | by lo about 4 is counterclockwise, o Ter, | pt Fê Ma = XTary + YThpx o. x Le css = (6.5 R')(45 Ib)sin 60º + (4.4 ft — 3.1 ft)(45 Ib) cos60º = 282.56 Ib-ft or M, = 283]b-ft) 4 (b) Have Ma = tcA (Em b ns For Fc to be minimum, it must be perpendicular to the | line joining points 4 and C. trsE a | e A “Mi dE) Y Í py c e gs where d="a (65%) +(44R) = 7.8492R 282.56 Ib-ft = (7.8492 fU)(Fc) 'min (Ein = 35.999 1b d=tan! 448) 34095º 65% 9 =90º — d = 90º — 34.095º = 55.905º or (Ec) qm = 36.01b < 55.9º 4  7.6 mm 152.4 mm PROBLEM 3.10 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-N force directed along its center line on the ball and socket at B, determine the moment of the force about 4. titia |52 Mm + Firstnote dog = V(344mm P + (152.4 mm) = 376.25 mm 344 mm : 152.4 mm cos0 = "2 0— sin0 = ="2—— 376.25 mm 376.25 mm and Fog = (Fep cos0)i — (Hop sin 0) 125N : : = 33625 ma mm)i + (152.4 mm)j] Now M, = aa * Eca where oa = (410 mm)i — (87.6 mm) j : 7, 125N : “ Then M,=[(410 mm)i - (87.6 mm)j|x si62s (04 — 152.45) = (30770 N-mm)k = (30.770 N:m)k or My =30.8N:m ) 4  PROBLEM 3.11 é , A winch puller 4B is used to straighten a fence post. Knowing that the H/ tension in cable BC is 260 lb, length a is 8 in., length b is 35 in., and "4 length d is 76 in., determine the moment about D of the force exerted by / the cable at C by resolving that force into horizontal and vertical “> —* components applied (a) at point C, (b) at point E. Lo -— (a Slope of line EC = an = 5 im. m. Then Tam = EC) = o|/ | ti Tr = E (250 lb) = 240 1b and Tam = E(2s0 1b) = 100 1b Then Mp = Tas:(35 in.) — Tum, (8 in.) = (240 Ib)(35 in.) — (100 1b)(8 in.) o = 7600 Ib-in. 246 - in) sa ]R or Mp = 7600 Ib-in. ) d Tu e BR bs ZA Tê (b) Have Mp = Tip) + Tam (x) Fe | Sm pad = (240 Ib)(0) + (100 1b) (76 in.) (5) = 7600 Ib-in. or Mp = 7600 Ib-in.) 4  A PROBLEM 3.12 Itis known that a force with a moment of 7840 Ib-in. about D is required 1 to straighten the fence post CD. If a=8 in, b=35 in., and d = 112 | in., determine the tension that must be developed in the cable of winch puller 4B to create the required moment about point D. SOLUTION Then and Have Slope of line EC = sin LA WM2in+8in 24 24 Tr =DT Abs = 5 4b ' Ta Tay = 35 14 Mp = Tan()) + Taty (x) MU 7 . <. 7840 Ibein. = a Tus(0) + 5 Tan(lI2 in) Tap = 250 1b or Tip = 2501b 4  PROBLEM 3.15 Form the vector products Bx C and B'xC, where B = B', and use the results obtained to prove the identity sina cos 8 = Ssin (a + 8) ++sin(a— 6). SOLUTION First note B = B(cos fi + sin Bj) c B' = B(cos fi - sin Bj) C=C(cosai + sinaj) By definition |BxC|= BCsin(a - 8) (1) |B'xC|= BCsin(o + 6) 2) Now BxC = B(cos fi + sin Bj) x C(cosai + sina) = BC(cos Bsina — sin Bcosa)k (3) BxC = B(cos fi - sin j) x C(cosci + sinaj) = BC(cos Bsina + sin Bcosa:)k (4) Equating magnitudes of Bx C from Equations (1) and (3), (5) sin(a — 8) = cos sina — sin Bcosa Similarly, equating magnitudes of B' x C from Equations (2) and (4), sin(a + 8) = cos Bsina + sin Bcosa (6) Adding Equations (5) and (6) sin(a — 8) + sin(a + 8) = 2cos Bsina sinacosB = asin(a +B)+ asin(a -B)4  PROBLEM 3.16 A line passes through the points (420 mm, -150 mm) and (-140 mm, 180 mm). Determine the perpendicular distance d from the line to the origin O of the system of coordinates. SOLUTION / Have d=Apxr B(-40, 180) , | “e 24 -x where As = Ea Asp [raid] (Cosrai potes) [Toa im mim A(420,-150) ' and Ega = (-140 mm — 420 mm)i + [180 mm — (150 mm) |j = —(560 mm)i + (330 mm) j fra] = 4(-560) + (330)? mm = 650 mm =(560 mmJi + (330mm)j 1 . . dg = OCA COM) 1 (sg,33 dam 650 mm gs (61 +33) Tou = (0-x,)i+ (0 — »4)j = —(420 mm)i + (150 mm) j Fas(cs6i — 33;) x [=(420 mm)i + (150 mm) j | = 84.0 mm d = 840 mm 4  PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 4i — 2; + 3k and -2i+6j-—5k, (b) 7i+ j - 4k and -6i — 3k + 2k. SOLUTION (a) (b) Have where Then and Have where Then and p- AxB IA x BI A=4i-2;+3k B=-2i+6j-5k i jk AxB=|[4 2 3=(10-18)i+(-6+20)j+(24-4)k = 2(-4i+ 7;+10k) 2 6-5 ja x Bl= "(4 + (7) + (10) = 2v165 2(=4i + 7; + 10k) 1 q; AA À = —==(4i + 75 + 10k NTE or ros i + 7] + 10k) «4 q - AxB JA x B| A=N+j-4k B=-6-3]+2k i jk AxB=|7 1 -4=(2-12)i+(24-14)j+(-21+6)k = 5(-2i+2;-3k) -6 3 2] IAxBj=5 (2) +(2) +(-3) = 5vi7 . p= Wit 23) o A= 1 —=(-2i+2;-3k) 4 5V17 a! A )  PROBLEM 3.19 CONTINUED i jk Mo =[15 3 -45|Nm=[(9-9)i+ (22.5 -22.5)j+(-15+15)k [Nm s23 or M5 =04 This answer is expected since r and F are proportional (* = 2) Therefore, vector F has a line of action passing through the origin at O.  PROBLEM 3.20 Determine the moment about the origin O of the force F = —(1.5 Ib)i + (3 Ib)j — (2 Ib)k which acts at a point 4. Assume that the position vector of 4 is(a)r=(25fi-(1j+(2 Mk (b)r=(45 fi -(9 Dj + (6 Ok, (0) r= (4 Fi = (1 Dj + (TR. SOLUTION (a) Have Mo =rxF where F=(1.5lb)i+(31b)j+ (2 1b)k r=(25f)i-(LM)j+(28)k i jk Then Mo =|25 -1 2|bf=[(2-6)i+(-3+5)j+(7.5-15)k|lbft 1.5 3 2 or Mo =—(4Ib-fi)i + (2 1b-ft)j+ (6 Ib-f)k «4 (b) Have Mo, =rxF where F=-(L5lb)i+(31b)j-(21b)k r=(45fi)i-(9f)j+(6f)k i jk Then M9=|45 -9 6]b-f=[(18-18)i+(-9+9)j+(13.5-13.5)k Jlo-f -15 3 2 or M, =04 This answer is expected since r and F are proportional ( = > ) Therefore, vector F has a line of action passing through the origin at O. (c) Have Mo, =rxF where F=(L5lb)i-(31b)j-(21b)k r=(48)i-(1)j+(7R)k i jk Then Mo =|4 1 7|bf=[(2-21i+(-10.5+8)j+(12-1.5)kJb-f -15 3 2) or Mo =—(19 Ib-fi)i — (2.5 Ib-ft) j + (10.5 Ib-fi)k «4  PROBLEM 3.21 Before the trunk of a large tree is felled, cables 4B and BC are attached as shown. Knowing that the tension in cables 4B and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force sim exerted on the tree by the cables at B. 72 ON Sm “Ss . 4 A 09m ” Sete SOLUTION Have Mo = Fo X F; where ro = (84m)j F; = Tap + Toc =(0.9 m)i - (8.4 m)j + (7.2 m)k Tip = Anal up = 7 7 Z (0.9) + (8.4) + (7.2) m (777N) —(SIm)i-(84m)j+(12m)k Tec = AncThc = (990 N) Co ley (84) +(12) m  PROBLEM 3.23 An 8-Ib force is applied to a wrench to tighten a showerhead. Knowing that the centerline of the wrench is parallel to the x axis, determine the moment of the force about 4. SOLUTION M,=IuxF ro = (85in)i—(20in)j+(5.Sin)k F, = —(8c0s45ºsin12º) lb F, = —(8sin45º)lb F, = —(8cos45ºcos12º) lb F=—(1.17613 1b)i — (5.6569 Ib) j — (5.5332 Ib)k / i j k y and M,=| 85 20 55 llbin 117613 -5.6569 -5.5332 = (42.179 Ib-in))i + (40.563 Ib-in.) j — (50.436 Ib-in.)k or M, = (42.2 Ib:in.)i + (40.6 Ib-in.) j — (50.4 Ib-in.)k «4  PROBLEM 3.24 , A wooden board 4B, which is used as a temporary prop to support a E small roof, exerts at point 4 of the roof a 228 N force directed along BA. Determine the moment about C of that force. 072 lo 0.96m ori SOLUTION Have Me = rue X Fpy where ryc = (0.96 m)i - (0.12 m) j + (0.72 m)k and Ega = Apa pa po “| (0 1mi + (1.8 m)j- (0.6m)k 1 doy+(8;+(06/ m (een) =—(12.0 N)i + (216 N)j- (72 N)k i j - Mç=|0.96 -0.12 02]Nm -120 216 -72 = —(146.88 N-m)i + (60.480 N.m) j + (205.92 N:m)k or Mç = (146.9 N-m)i + (60.5 N-m) j + (206 N:m)k «  PROBLEM 3.25 The ramp 4BCD is supported by cables at corners C and D. The tension in each of the cables is 360 Ib. Determine the moment about 4 of the force exerted by (a) the cable at D, (b) the cable at C. (a) Have Ma = Teu X Tp where re = (92in)j Toe = ApeTe . (24 ni (132 mi (120 ind (360 1b) (24) + (132) + (120)? in. = (48 Ib)i + (264 1b) j = (240 Ib)k ij ok M,=|0 a O |lb-in. = —(22,080 Ib-in.)i — (4416 Ib-in)k 48 264 -240 or M, = (1840 Ib-f)i — (368 Ib-f)k «4 (b) Have M, = Tux Teo where Fou — (108 in.)i + (92 in.) =(24 ini + (132 in)j = (120 in.) k Tog = Acoleg = 5 z 5 (360 Ib) (24) + (132) + (120)" in. = (48 Ib)i + (264 Ib) j — (240 Ib)k ij k . M,-|l08 92 O |lbiin. -48 264 -240 = (22,080 Ib-in.Ji + (25,920 Ib-in.) j + (32,928 Ib-in)k or M, = (1840 Ib-fi)i + (2160 Ib-ft)j + (2740 Ib-)k «4  PROBLEM 3.28 In Problem 3.21, determine the perpendicular distance from point O to cable BC. r 1 Problem 3.21: Before the trunk of a large tree is felled, cables 4B and Sam BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. a es de Sim << a . 4 IC 19 es 12m Have IMo!|= Tacd where d = perpendicular distance from O to line BC. Mo = po X Toc ro = 84mj (5.1m)i — (84m)j + (12m)k Tac = Anclbc = (990 N) (517 + (84 + (127 m = (510 N)i — (840 N) j + (120 N)k io od k M5=|0 84 O|=(1008N-m)i - (4284 N-m)k 510 840 120 and IMo|= ((1008)) + (4284)” = 44010 N-m <. 44010 Nm = (990 N)d d = 44454m or d=445m 4  PROBLEM 3.29 In Problem 3.24, determine the perpendicular distance from point D to a line drawn through points 4 and B. | Problem 3.24: A wooden board 4B, which is used as a temporary prop to support a small roof, exerts at point 4 of the roof a 228 N force directed D=b.lêm along B4. Determine the moment about C of that force. =| 18m IMpl= Fasd d = perpendicular distance from D to line 4B. Mp = Fyp X Fm rup = —(0.12m)j+ (0.72m)k Eu = Auf = (=(0.1m)i + (1.8m)j = (0.6m)k) 01) +(18) +(0.6 m =—(12.0N)i + (216N)j- (72N)k (228) i j Mp=| 0 012 072]Nm -120 216 -2 = —(146.88 N-m)i — (8.64 N-m)j - (1.44 N-m)k and IMpl= (146.88) + (8.64) + (1.44) = 147.141 Nm * 147.141 Nem = (228N)d d = 0.64536m or d =0.645m 4  PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from point C to a line drawn through points 4 and B. Problem 3.24: A wooden board 4B, which is used as a temporary prop to support a small roof, exerts at point 4 of the roof a 228 N force directed =. along B4. Determine the moment about C of that force. ori lcTonsa Have IMel= Fasd where d = perpendicular distance from C to line 4B. Mc = ryc X Fpy des ryc = (0.96 m)i — (0.12 m) j + (0.72m)k po SÊ =(0.1m)i + (1.8m)j - (0.6)k 1 Fou = Apa pa — Elotmbi + (L8m)i- (06)k) (0.tmji + (1 8m)j - (0.6) ) 28 n) (0.1) + (1.8 + (0.6) m =—(12.0N)i + (216 N)j- (72 N)k i j k Mç =|0.96 -0.12 0.72] Nim 120 216 N = (146.88 N-m)i — (60.48 N-m)j + (205.92 N-m)k and IMcl=((146.88) + (60.48)” + (205.92)” = 260.07 Nm “. 260.07 Nm = (228 N)d d=1.14064m or d=1.14Im 4  PROBLEM 3.32 In Problem 3.25, determine the perpendicular distance from point 4 to a line drawn through points C and G. Problem 3.25: The ramp 4BCD is supported by cables at corners C and D. The tension in each of the cables is 360 Ib. Determine the moment about 4 of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION Have where and [My] = Tod d = perpendicular distance from A to line CG. M, = "ou X Teo Tou = (108in)i + (92in)j Tog = Acolco = (24 in)i + (132 in.) = (120 in)k (36015) (24) + (132) + (120)? in = —(48 lb)i + (264 Ib) j = (240 Ib)k ij k - M,=/108 92 O |lbin. —48 264 -240 = —(22,080 Ib-in.)i + (25,920 Ib-in.) j + (32,928 lb-in.)k [Ml] = /(22,080)” + (25,920)? + 32,928)” = 47,367 Ib-in “. 47,367 Ib-in. = (360 Ib)d d =131.575in or d=10.96ft «4  PROBLEM 3.33 line drawn through points D and E. In Problem 3.25, determine the perpendicular distance from point B to a Problem 3.25: The ramp 4BCD is supported by cables at corners C and D. The tension in each of the cables is 360 Ib. Determine the moment about 4 of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION Have where and [Ms] = Toed d = perpendicular distance from B to line DE M, = Fio X Tor Fe = —(108in.)i + (92in.) j (24 inJi + (132 in.)j = (120 in )k - E (360 Ib) (24) + (132) + (120) in. Top = AprTo; = = (48]b)i + (264 1b) j — (240 Ib)k io jk - Mp =|[-108 92 O |lbin. 48 264 -240 = —(22,080 Ib-in.)i — (25,920 Ib-in)j — (32,928 Ib-in.)k [Ms] = 4(22,080)" + (25,920)" + (32,928)? = 47,367 lb-in. <. 47,367 Ib-in. = (360 Ib)d d = 131.575in. or d=10.96f 4  PROBLEM 3.34 Determine the value of a which minimizes the perpendicular distance from point C to a section of pipeline that passes through points 4 and B. Assuming a force F acts along 4B, Mel =|rac x F|= F(d) where d = perpendicular distance from C to line AB F= Am = (8 mis (7 mi - Cmt p (8) + (7) + (9) m = F(0.57437)i + (0.50257) j — (0.64616)k ryc = (Im)i - (2.8m)j- (a -3m)k i i k “Me=| 1 -28 3-a |F 0.57437 0.50257 —0.64616 = [(0.30154 + 0.50257a)i + (2.3693 — 0.57437a) j + 2.1108k]F Since Mel=qrucxP) or JrucxF)= (dry “. (0.30154 + 0.50257a)” + (2.3693 - 0.57437a)) + (2.1108) = d? Setting <a?) = 0 to find a to minimize d 2(0.50257)(0.30154 + 0.50257a) + 2(-0.57437)(2.3693 — 0.57437a) = 0 Solving a=2.076lm or a=208m 4  PROBLEM 3.37 Consider the volleyball net shown. Determine the angle formed by guy wires 4B and 4C. SOLUTION First note AB = [ru] = ((-185m) + (-24m) + (06m =3.15m AC = frc;4) = (Om)! + (-24m) + (18m) =30m and rma = —(1.95m)i — (2.40 m)j+ (0.6m)k reu = (2.40 m)j + (1.80m)k By definition Toa" Teia = [Estou |coso or (-1.95i-2.40;+0.6k)-(-2.40j + 1.80k) = (3.15)(3.0)coso (=1.95)(0) + (-2.40)(-2.40) + (0.6) (1.8) = 9.45coso = cos0 = 0.72381 and 0 = 43.630º or 0=436º 4  PROBLEM 3.38 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD. SOLUTION First note 4€ = reu) = (24) + (187 m = 3m AD = [tou] = (12) + (24) + (03) m=27m and reu = —(24m)j+ (1.8m)k roa = (1L2m)i- (24m)j+ (03m)k By definition Feia * Fpya = IFcralfroua)coso or (-2.4j+1.8k)-(1.2i-2.4;+0.3k) = (3)(2.7)coso (0)(1.2) + (-2.4)(-2.4) + (1.8)(0.3) = 8.1coso and cos9 = a = 0.77778 0 = 38.942º or 0=389º4  PROBLEM 3.39 Steel framing members 4B, BC, and CD are joined at B and C and are braced using cables EF and EG. Knowing that E is at the midpoint of BC and that the tension in cable EF is 330 N, determine (a) the angle between EF and member BC, (b) the projection on BC of the force exerted by cable EF at point E. SOLUTION (a) By definition Asc Apr = (1)(I)coso where Ac -ómi-(tSmi(2mk 15 as 12%) Vais) +(45P+(12Jm 205 (7m)i - (6m)j+ (6 m)k a Agr = — (HP +(om 1 - (l6i-45j-12k) (=i-6)+ 6h) 20.5 11.0 (16)(=7) + (-4.5)(-6) + (-12)(6) = (20.5)(11.0)coso (-7i- 6; + 6k) = cos9 and 0= cos! (52) = 134.125º 225.5 or 0=1341º4 (b) By definition (Ter)pc = Tor cosO = (330 N)cos134.125º =-229.26N or (Tr) ac = —230N 4  PROBLEM 3.42 Slider P can move along rod 04. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular. Dimensions ir mm SOLUTION The requirement that member OA and the elastic cord PC be perpendicular implies that Aos: Ape = 0 or Aos rep = 0 — (0.24m)i + (0.24 m)j - (0.12m)k (0.24) + (0.247 + (0.12 m 2 3 3 3 where Ao4 Letting the coordinates of P be P(x,y,z), we have rop = [(0.18 - x)i + (0.30 — y)j+ (0.24 — 2)k Jm Si -38) -[(0.18 = x)i + (0.30 - 5)j+ (0.24 - =)k |= 0 (1) dop (9:19: Since rpo = Aosdop = (a +2)-k), 2 2 + Then x= 3 dor. y= 3%or, z= 3 %or (2) Substituting the expressions for x, y, and z from Equation (2) into Equation (1), Lo. : 2 : 2 : 1 sli +2j- wi(ous - Edop + (030 - Eos | + (024 + dos e) =0 or 3dop = 0.36 + 0.60 — 0.24 = 0.72 = dop = 024m or dop = 240 mm 4  PROBLEM 3.43 Determine the volume of the parallelepiped of Figure 3.25 when E ()P=+7in)i-(Lin)j+(2in)k Q=(0 inji-(2in)j+ (4 in)k, and RR S=5 ini + (6 in)j = (Link, (b) P= (Linji+(2in)j- (Link, LL Q=8inJi-(lin)j+(9 ink, and S = (2 inJi+(3 in)j+(Lin)k. » E 1 SL ii SOLUTION Volume of a parallelepiped is found using the mixed triple product. (a) Vol=P-(QxS) 7-2 =[3 -2 Ain” =(-14+168+20-3+36-20)inº -s 6-1 =187in” or Volume = 187 in? (b) Vol=P-(QxS) 124 =|-8 -1 9Jin?=(-1-27+36+16+24-2)inº 231 = 46in? or Volume = 46 in? «  PROBLEM 3.44 Given the vectors P=4i -2;+P.k, Q=i+3j-5k and S = 6 + 2j — k, determine the value of P. for which the three vectors are coplanar. SOLUTION For the vectors to all be in the same plane, the mixed triple product is zero P-(QxS)-0 42 .0=|1 3 -S=-12+40-60-2+P.(2+18) -6 2 1 so that p= am 20 or P = 1.700 4  PROBLEM 3.46 The 0.732 x 1.2-m -m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the $40 mm coordinate axes of the force exerted by the cord at C. SOLUTION E PP First note == (07327 - (0132) m 0,132m PE =020m I A | o 2 2 2 tz —» Then der = J(0.840)) + (0.720) + (0.720) m =132m — Tec and Te; = go ee) CE — (0.840 m)i + (0.720 m) j = (0.720 m)k . 132m (s4N) = —(36.363 NJi + (29.454 N) j - (29.454 N)k M, = ga X Ter rea = (0.360 m)i + (0.852m)j i i k M,=|0360 0852 0 |Nm [34.363 29.454 29,454] = —(25.095 N-m)i + (10.6034 N-m)j + (39.881 N.m)k . M, =-25.1Nm, M, = 10.60 Nim, M, = 39.9N:m 4  PROBLEM 3.47 A fence consists of wooden posts and a steel cable fastened to each post and anchored in the ground at 4 and D. Knowing that the sum of the moments about the z axis of the forces exerted by the cable on the posts at Band C is -66 N-m, determine the magnitude Tcp when Ta, = 56 N. SOLUTION Based on |M.|= ke-[(r9), x Tas] + [(rc), x Tep] where M. = —(66N-m)k (5), = (e), = (Um)i Tas = ApaTha . (1.5m)i = (Im)j+(3m)k 3.5m (s56N) = (24 N)i - (16 N)j + (48N)k Tcp = AcpTcp (2m)i= (Im)j-(2m)k To 3.0m Inss) = (66 Nem) = he (im) [(24 NJi = (16N)j+ (48 Nx) et (tmix Steo(a=5- 20) o dou -2n, “Tops (66 -24)N or Top =630N 4  PROBLEM 3.48 Im A fence consists of wooden posts and a steel cable fastened to each post and anchored in the ground at 4 and D. Knowing that the sum of the moments about the y axis of the forces exerted by the cable on the posts at B and Cis 212 N-m, determine the magnitude of Tp, when “Top = 33N SOLUTION Based on [(ro). x Tg + (re). x Te] where M, = (212N:m)j (15). = (8m)k (rc). = (2m)k Toa = ApaTha º (1.5 m)i = (Lm)j = (3m)k Tu 3.5m = SEM (15i = j+ 3h) Tep = AcpTcp º a) =(22i-11j-22k)N + (212Nm)=j- le m)k x E Si-j+ 9) +j-[(2m)kx(22i-11;-22k)N] 8(1.5) or 212 = 3.5 Toa + 2(22) RC: 2418.6667 or Ta = 490N 4  PROBLEM 3.51 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that M, =230lb-in, M, = 200 Ib-in., and M, = —35 Ib-in., determine the magnitude of P and the values of 4 and 6. Basedon M, = (Pcosg)[(9 in)sind |- (Psind)[(9 in)coso] (1) M, = (Pcosg)(S in.) (2) M, = (Psing)(Sin) (3) Then eos % ai or tang = = 0175 4 =9.9262º -200 or 4=993º4 Substituting é into Equation (2) 200 Ib-in. = —(P cos9.9262º)(5 in) P = 40.608Ib or P=406b+4 Then, from Equation (1) 230 Ib-in. = [(40.608 Ib)cos9.9262º |[ (9 in.)sing ] — [(40.608 Ib)sin9.9262º |[(9 in.) coso] or 0.98503sind — 0.172380cos0 = 0.62932 Solving numerically, 0=489º 4  PROBLEM 3.52 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that M, = —180 Ib-in.and M, = —30 Ib-in., determine the moment M, of P about the x axis when 8 = 60º. Basedon M, = (Pcosg)[(9 in)sino |- (Psing)[(9 in)coso | (1) M, = —(Pcosg)(S in.) (2) M, = —(Psing)(S in.) (3) Then Equation (3). M. —(Psing)(5) Equation (2) M, —(Pcosó)(5) -30 or — =tan -180 , bp =94623º From Equation (3), —30 Ib-in. = —(Psin9.4623º) (5 in.) -. P=36497]b From Equation (1), M, = (36.497 1b) (9 in.) (c0s9.4623º sin 60º — sin9.4623º cos60º) = 253.60 Ib-in. or M, = 254 Ibiin 4  PROBLEM 3.53 The triangular plate 4BC is supported by ball-and-socket joints at B and D and is held in the position shown by cables 4E and CF. If the force exerted by cable 4E at 4 is 220 lb, determine the moment of that force about the line joining points D and B. Mpp = ApB' (rao x Tu) (48in)i = (14in)j - = 0.96i - 0.285 Ao 50in. ' 8 rp =(4in)j+ (8in)k — [(B6in)i= (24in)j+ (8in)k] ( Tag = Aagl4p = 44in 220 lb) = (180 1b)i — (120 1b)j + (40 Ib)k 0.960 -0.280 O = Mpp=| 0 4 Bllbin 180 120 40 = (0.960)[(-4)(40) - (8)(-120) | + (-0.280)[ 8(180) - 0] = 364.8 Ib-in. or Mp = 365 Ib-in. d  PROBLEM 3.56 A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by « cable EG at E is 61.5 N, determine the moment of that force about the line joining points D and 7. SOLUTION Have Mp =Apy* E x Teo] (1.6 m)i — (0.4 m)j h 10 TON where tor 04/17 m =L(4-)) 7 Fou = —(10.9m + 0.8m)k = (11.7 m)k Trg = ArgTic (2mi-(36mi- (Im 5 N) 23m =S[(12N)i- (3.6 N)j- (117N)k] 410 SN(117m) D = . My > o 0 q O4tm — Rm DI j : Eca “17 12 36 117 = (14.1883 N-m)ffo = ((-1)(-3.6)]+ [(-n)(-1)(1.2) - 0) = 187.286 Nm or Mpy = —187.3N:m 4  PROBLEM 3.57 ! A rectangular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge 04. ad DA. se ã 5 ; =x, SOLUTION (0a) +| Have Mo = ho” (rei x P) o SGT x x where o” | ; (0Ajz From triangle OBC a 04) =— (04),=4 Z a(l a 04). = (04), tan30º = 2| — | = 8 (04), = (04) tmnsor = 8) 6 Since (04) = (04): + (04)) + (04? 2 2 or é -(5) (08); (55) 2 a 2 (04), = “2 =a g Then r = Siafi est 40 = 33 23 1 2 1 a Dad an Aos 7 + RE + NE P=AgcP — (asin309)i - (acos30º)k (p = Ai —Bk) Tcio = ai  PROBLEM 3.57 CONTINUED os — 0 0 2 0 3 Lopo) 2 É 23 (o(£) 1  PROBLEM 3.59 The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is partially supported by members EF and GH. Knowing that the compressive force exerted by member EF on the walkway at HF is 5400 lb, determine the moment of that force about edge AD. SOLUTION Having Ma = hp" (reu x Ter) 2Ai+(3R)) 1 where An= (ami+ GA =——(8i+ 5) º ley-qgya dês ren = (Ti (3R)j (enc 7ni+[3n + (Mi (6h n T ay E (8 n 5400 Ib) Tor = AprTor = = 600[ (1 Ib)i + (4 Ib)j + (8 Ib)k | 8 10 600 600 - Map =—==|7 3 Oflb-ft===(-192 — 56) lb-ft no = es, Uno Jos ( 12-59) = -18,456.4 Ib-ft or M,p = 18.46 kip-f «  PROBLEM 3.60 The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is partially supported by members EF and GH. Knowing that the compressive force exerted by member GH on the walkway at H is 4800 lb, determine the moment of that force about edge 4D. SOLUTION x Having Map =Ap* (raia x Ton) Cami+0Mi 1. where Ap = SEE = (8 +) a (oa) (ap n dos Faia = (20 Ri = (6 8) = 2[(10 Ri = (38) n (16-20 )i+[68+ (16)(3 1) 5 +( (8) + (87 + (8) 1 = 1600[=(Llb)i + (2 1b) j+ (2 1b)k | Tor = AgnTon = De (agoo lb) (1600 lb)(2 t) O soa Map = +" Hho -3 = (48-20 AD 65 65 ( ) -1 2 2 = 26,989 Ib-ft or M,p = 27.0 kip-ft 4  PROBLEM 3.61 Two forces F) and F; in space have the same magnitude F. Prove that the moment of F; about the line of action of F; is equal to the moment of F> about the line of action of F,. SOLUTION First note that FR=AM Let M, = moment of F, about the line of action of M, and M, = moment of F about the line of action of M, Now, by definition M, = Arma x E) = Arara x do) M = Ao (run x F) = Ao(ras x Ah Since H=6=F and LyB = EmA M, = An(rava x Ao) F M, = Ao (rea x A)F Using Equation (3.39) An(rona x 1) = Ao (-rma x A) so that M, = Arma x M)F Mp = Mi 4  PROBLEM 3.63 y In Problem 3.54, determine the perpendicular distance between cable CF . and the line joining points D and B. Sin A Padii Problem 3.54: The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables 4E and CF. If the force exerted by cable CF at C is 132 Ib, determine the moment of that force about the line joining points D and B. Í 36 in, 14 in. SOLUTION Have (Mp) = Apa "(Key X Ter) where = 0.96i — 0.285 . rem = (8in)j - (16 in)k á Tor = Acrlcr a — (24 ini = (36 in)j- (8 mk (132 b) 44 in. = (72 lb)i — (108 1b)j — (24 Ib)k 0.96 -0.28 O Mpa=| 0 8 1éllbin 72 -108 -24 = —1520.64 Ib-in. Only the perpendicular component of Tc contributes to the moment of Tcy about line DB. The parallel component of Tc; will be used to obtain the perpendicular component.  PROBLEM 3.63 CONTINUED Have (Ter era = App: Ter = (0.96; — 0.285) -[ (72 1b)i — (108 1b) j — (24 Ib)k] = [(0.96)(72) + (-0.28)(-108) + (0)(-24) |b = 99.36 lb Since Tor = (Tor pero * (Ter )pasaia (er pero = (Tor? - (Ler acata = (132) - (99.36 = 86.900 lb Then Mpp = (er peso (a) —1520.64 Ib-in. = (86.900 Ib)d d = 17.4988 in. or d=1750in. 4  PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between cable EF and the line joining points D and 1. Problem 3.55: A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EF at E is 66 N, determine the moment of that force about the line joining points D and 1. SOLUTION Have Mpy = Apr “(reu x Ter) (Womi-(04m)j 1,4. he Ay = S—D + = -(4- ver 2 04/17 m 17 (4-3) Fey = (54m)k 12m)i-(3.6m)j+(54m)k ta = Must - V2MB= DS MIA(SAM ç) .om =6[(2N)i-(6N)j+(9N)k] 410 My - CNC o il=1mn879Nm A? 2 -6 9 Only the perpendicular component of T,; contributes to the moment of T;r about line DI. The parallel component of T;p will be used to find » the perpendicular component. +. q paia atm dsm Have Ao da == Fen (Ter parana = Apr Ter Mom = ao(s = D)-[(12N)i- (36 N)j+ (54 N)k] =-L(48 + 36)N N ale 5 1  PROBLEM 3.65 CONTINUED Since Tap = (Tr) pero + (Teo) parate 2 2 (Tro per = lho) — (Teo) ara 2 = 60.651N Then Mpy = (Tec) pero (a) 187.286 N-m = (60.651 N)(d) d =3.0880m or d=309m«4  PROBLEM 3.66 In Problem 3.41, determine the perpendicular distance between post BC and the line connecting points O and 4 Problem 3.41: Slider P can move along rod O4. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 0.12 m and the tension in the cord is 30 N, Dimensions in mm determine (a) the angle between the elastic cord and the rod 04, (b) the projection on OA of the force exerted by cord PC at point P. SOLUTION Assume post BC is represented by a force of magnitude Fac where Fac = Faci Have Mos = dos? (mo x Fsc) where dos = (0.24 m)i + (0.24 m)j - (0.12m)k — 25,2; 1 0.36m 3 303 rao = (0.18 m)i + (0.24 m)k 221 Mos = 5 Fipe 0.18 0 0.24] = Lhe (048 - 0 18) = -0.22Fpe 010 Only the perpendicular component of Fac contributes to the moment of Fac about line O4. The parallel component will be found first so that the perpendicular component of Fsç can be determined. 2.2, 1 : Foc(paratte) = Ao4" Fac = (3 + 317 24) Fred 2 = the Since Fac = (Fara + (Fo) pesp 2 2 2 2 2F, (Fac) = ao (Faca = (ac) - (2) = 0.74536F ye Then |Mo4l = (Esc )pero (2) 0.22Fpc = (0.74536Fpc)d d = 0.29516m or d = 295 mm 4  360 mm De um ka Ar | | B “4 PROBLEM 3.67 In Problem 3.45, determine the perpendicular distance between cord DE and the y axis. Problem 3.45: The 0.732 x 1.2-m lid ABCD of a storage bin is hinged along side 4B and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D 852 mim SOLUTION 4º f—— 0:132m se se Ei na zom0m E Bar 132m First note == (0.732) = (0.132 m =0.7220m Have M, = (ro x To) where roa = (0.132; + 0.720k) m Top = AprTor (0.360 m)i + (0.732 m)j — ( 1.08m 0.720 Mk (4 N) = (18 NJi + (36 N)j - (36 N)k 0 1 0 x - M,=/0 0.132 0.720) = 12.96 N:m I8 36 36 Only the perpendicular component of Tp; contributes to the moment of Tbg about the y-axis. The parallel component will be found first so that the perpendicular component of Tp; can be determined Top(parates) =: Toe =36N  PROBLEM 3.69 A couple M of magnitude 10 Ib-ft is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners 4 and D, (b) at corners B and C, > (c) anywhere on the block. Po (a) Have or (b) Have (0) Have M = Pd 10 Ib-Rt = P(10 O 12in. . P=12b or Ban = 12.00 1b 4 doc = (BE) + (EC) =(10in) + (6in)) = 11.6619 in. M = Pd 10 1b-ft = P(uL6619 im) 1h ) 12in. P=10.28991b or P=10.291b4 2 2 dic = (AD) + (DC) = (10 in) + (16 in)” = 289 in. M = Pdic 10 b-fi = P(aV8o io) ) in, P = 6.3600 lb or P=636lb«4  | x PROBLEM 3.70 Two 60-mm-diameter pegs are mounted on a steel plate at 4 and C, and o i two rods are attached to the plate at B and D. A cord is passed around the ad pegs and pulled as shown, while the rods exert on the plate 10-N forces as indicated. (a) Determine the resulting couple acting on the plate when T= e 36 N. (b) If only the cord is used, in what direction should it be pulled to " create the same couple with the minimum tension in the cord? | — an (c) Determine the value of that minimum tension. SOLUTION (a) Have M=x(Fd) = (36 N)(0.345 m) — (10 N)(0.380 m) =8.62N:m M=862Nm )4 p DOLOm (6) SON A E DER Have M=Td=8.62N-m For T'to be minimum, d must be maximum. « Toin must be perpendicular to line AC tang = 0880m 433333 0.285m and 0 = 53.130º or 0=531º4 Have M = Toindmax where M =8.62N:m duas = to 380) + (0.285 + 2(0 030)] m=0.535m 8.62 Nim = Tin (0.535 m) Tain = 16.1121N or Tin = 16.11N 4  PROBLEM 3.71 The steel plate shown will support six 50-mm-diameter idler rollers mounted on the plate as shown. Two flat belts pass around the rollers, and rollers 4 and D will be adjusted so that the tension in cach belt is 45 N. ! Determine (a) the resultant couple acting on the plate if a = 0.2 m, (b) the “ value of a so that the resultant couple acting on the plate is 54 Nm clockwise SOLUTION (a) Note when a = 0.2 m, rc; is perpendicular to the inclined 45 N forces. Have M=x(rd) =—(45N)fa + 0.2m + 2(0.025m)] - (45 N)[2av2 + 20.025 m)] For a =0.2m, M = —(45 N)(0.450 m + 0.61569m) = 47.956 N-m or M=480N-m ) 4 M=540Nm ) M = Moment of couple due to horizontal forces at 4 and D 45 t + Moment of force-couple systems at C and F about C. 54.0 Nim = 45 N[a + 0.2m + 2(0.025 m) | +[Mc + My + E(a+02m)+ E, (2a) ] where Mc = —(45 N)(0.025 m) = —1.125 Nm Mp = Mç = 1.125 Nm  PROBLEM 3.73 Knowing that P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. M=M,+M, M, = jp X Pc rem = (0.96 m)i — (0.40 m)j Po = (100 N)k iodo ok M, =/9.96 -040 O |=(40N-m)i + (96 N-m)j 0 0 100) Also, M, = pa X Por roya = (0.20 m)j — (0.55 m)k Pp = AgpPp (0.48 m)i + (0.55 m)k a (0.48) + (0.55) m = —(96 N)i + (110 N)k (146N) ij k M,=|0 0.20 -0.55] N:m -96 0 110 = (22.0 N.mji + (52.8 N-m)j + (19.2 N-m)k  PROBLEM 3.73 CONTINUED and M =[(40 N.mji + (96 N-m)j]+ [(22.0 N-m)i + (52.8 Nem) j+ (19.2 Nm)k | = (62.0 N-m)i + (148.8 N-m)j + (19.2 N-m)k Mi = [22 +M2+M2 = (62.0) + (148.8) + (192)? = 162.339 Nm or M =1623N-m 4 M 620i+148.8;+19.2k “MO 162.339 = 0.38192i + 0.91660j + 0.118271k cos0, = 0.38192 + 0,=67.547º or 0,=67.5º4 cosd, = 0.91660 - 0, =23.566º or 0,=236º4 cos0, = 0.118271 “ 0, = 83.208º or 0,=832º 4  PROBLEM 3.74 Knowing that P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. M=M,+M, M, = "oc X Ei race = (10 ini E = (4Ib)k My =(10in)ix (4 Ib)k = (40 Ib-in.) j Also, M; = ryp x Ep ryr =(Sin)i+(3in)j Ep = AgpFp º =(Sin)i+ (3in)j+(7in)k As + (3) + (77 in -1b 83 (71b) SR) ain) DM qi nm Im. Me Hs 3 oj= - 783 83 -5 3 7 (21i + 35 + 0k) = 0.76835(21i + 35j) Ib-in.  PROBLEM 3.76 Knowing that P = 210 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Have M=M+M,+Mp i j k where M,="emxPo=|0.96 -040 O |=(40N-m)i+ (96 N-m)j 0 0 100 | k M,=r54xP =| 0 0.20 -0.55|= (22.0 N-mi + (52.8 N-m)j+ (19.2 N-m)k -96 O no (Sec Solution to Problem 3.73.) i i k M, = r;4 XP; = |0.48 0.20 1.10] = (231 Nem)i + (100.8 Nm)k 0 210 0 M =[(40 +22 + 231)i + (96 + 52.8)j + (19.2 + 100.8)k ] Nem = (293 N-mji + (148.8 Nm) j + (120 N:m)k IMj=[M7+M2 + M? = (293) + (148.8) + (120)? = 349.84 Nem or M = 350 N:m M 29% +1488+120k 6 935525, 042533] 4 0.34301k “Mo 349.84 cos6, = 083152 0,=33121º or 9,=33.1º4 coso, = 042533 0,=64828º or 0, = 648º 4 cosg. = 034301 0,=69.940º or 9. = 699º 4  PROBLEM 3.77 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. Knowing that the holes are perpendicular to the surfaces of a Man To: É o the workpiece, replace the couples applied to the drills with a single ” equivalent couple, specifying its magnitude and the direction of its axis. 5 (9 a - vh ã A SOLUTION Nave M=M+M,+M; mo | E where M, = (1.1 b-ft)(cos25ºj + sin 25ºk) M, = (Li b-f)j 2sº & M, = (1.3 1b-ft)(c0s20ºj — sin 20º) M = (-0.99694 — 1.1 - 1.22160) j + (-0.46488 + 0.44463)k Mal 3 =—(3.3185 Ib-ft)j — (0.020254 Ib-ft)k 20º [7 e and qMj=[M2+M2+M? = (0) + (33185) + (0020254)? Z = 33186 Ib-ft or M =332]b-ft «q A Mo. (0)i — 3.3185j — 0.020254k “IM 3.3186 = —0.99997; — 0.0061032k cos0, = 0 « 0,=90º or 0, = 90.0º 4 cosd, = —0.99997 - 0,=179555º or 0, =179. 64 cos0, = —0.0061032 0.=90.349º or 0,=903º 4  PROBLEM 3.78 The tension in the cable attached to the end C of an adjustable boom 4BC is 1000 N. Replace the force exerted by the cable at C with an equivalent force-couple system (a) at 4, (b) at B. , Tcos soº (a) Based on LF: F,=T=1000N or F, = 1000N = 20º 4 ZM, M,= (Fsinsoe)(d,) = (1000 N)sin50º(2.25 m) = 1723.60 Nm or M, =1724Nm ) 4 (b) Based on ZF: F,=T=1000N or F; = 1000N = 20º 4 LMp: Mp = (Tsins0º)(dy) = (1000 N)sin50º(1.25 m) = 957.56N:m or M, =958N:m ) 4  PROBLEM 3.80 CONTINUED LF, Psin60º = = Pp + Po (700 N)sin60º = 87.5 N + Pp, Pyy = 518.72 N 2 2 Po = (Pos) + (Poy) = (350) + (518.72) = 625.76 N B, 9 = (22) - (252) = 55.991º x 350 or P,=626N < 560º 4  PROBLEM 3.81 A landscaper tries to plumb a tree by applying a 240-N force as shown. Two helpers then attempt to plumb the same tree, with one pulling at B and the other pushing with a parallel force at C. Determine these two forces so that they are equivalent to the single 240-N force shown in the md figure. o. . . e” SS. pa Based on ZF,: —(240 N)cos30º = —Fy cosa — Fo cosa or (Fo + Fo)cosa = —(240 N)cos30º (1) ZF,: (240 N)sin30º = Fasina + Fesina or (FE; + Fe)sina = (240 N)sin30º (2) From Equation (2), tana = tan30º Equation (1) a=30º Based on ZM: [(240 N)cos(30º - 20º) |(0.25 m) = (Fp cos10º)(0.60 m) From Equation (1), Fy = 100 N or F; = 100.0N H= 30º « —(100 N + Hc )cos30º = —240c0530º Fe =140N or F; = 1400N H 30º €  PROBLEM Two helpers 3.82 A landscaper tries to plumb a tree by applying a 240-N force as shown. (a) Replace that force with an equivalent force-couple system at C. (b) attempt to plumb the same tree, with one applying a horizontal force at C and the other pulling at B. Determine these two forces if they are to be equivalent to the single force of part a. SOLUTION (a) Based on 2Mc (b) Based on or ZH: —(240 N)cos30º = —H.. cos30º Fe =240N or F; =240N DM 30º 4 : [(240N)cosl0º |(d,)= Me d,=025m Mc = 59.088 Nm or Mc =59.1Nm ) 4 XF,: (240 N)sin30º Fasina Fasina = 120 (1) EMp: 59.088 Nem — [(240 N)cos10º |(dc.) = —Fr (de c0s20º) 59.088 Nm — [(240 N) cos10º |(0.60 m) = —F.[ (0.60 m) cos20º | 0.56382Fc = 82.724 Fc = 146.722 N or E. =1467N-— 4 ( . and ZF,: —(240 N)cos30º = 146.722 N — Fgcosa Facosa = 61.124 (2) From a - a =1.96323 a = 63.007º or a =630º4 From Equation (1), p= o = 134.670 N sin 63.007º or F; =134.7N > 63.0º 4