# hw1 - Problem 2.58 The cables A B and 0 help support a...

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Unformatted text preview: Problem 2.58 The cables A, B, and 0 help support a pillar that forms part of the supports of a structure The magnitudes of the forces exerted by the cables are equal: IFAi— — iFgl— — §Fg|. The _magnjtude of the vectot: sum MM “yew of the three forces 13 200 kN What' IS |FA|? Solution: _ Use the angles and magnitudes to determine the vector components, take the sum, and solve for the unknown. The angles between each cable and the pillar are: Measure the angles counterclockwise form the m-axis. The force vec- tors acting along the cables are: FA : IFAIGCOS 303.70 +jsin 303.70) = 0.554SIFAIl — 0.8319iFAij : |F3|(ico:3323.1° +jsin323.1”) z 0.7997'F5Ei —- 0.6004313}; L3 = |FC|(i cos 333.4o +jsin 333.41") = 0.8944IFg|i—0.4472ingj The sum of the forces are, noting that each is equal in magnitude, is 2F: (2. 2489 FAEi — 1. 8795mm). The magnitude of the sum is given by the problem: 200 : in; (2.2439)2 + (1.8795)? x 2.931}FA!, from which IFAi = 68.24 kN ‘ i l i w l w L ‘ Problem 2.97 The 70-m—tall tower is supported by 1 three cables that exert forces F A B, F Ag, and F A D on it. The magnitude of each force is 2 kN. Express the total force exerted on the tower by the three cables in terms of scalar components. Solution: The coordinates of the points are A (0, 70, 0), B (40, The resultant force exerted on the tower by the cables is: O, 0), C (—40. 0, 40) D (—60, 0, "60) The position vectors corresponding to the cables are: x (750 , OJi + (o H 70)j + (—60 — 0)k = —60i — 70k — 60k = (—40 — 0)i + (0 — 70)j + (40 — 0)k = —4Di — 70j +40k " = (40 —0)i+ (o — 70)j+ (n —0)k :40i770j+0k _. Z 3 The unit vectors corresponding to these position vectors are: rAD —60_ 7o, 60 lgmpl 110 110 110 = “0.545511— 0.6364j — 0.5455k I'AC . 70, 40 “AC 21ml: R" gs” e : ~0.4444i -— 0.7778j + 0.4444k rAB 40 , 70 , . : = m _ — +0k=0.49631—0.8685' +Uk "‘43 mg: 80.61 80.5J J —l——.l_’—‘ “AD The forces are: FAB : [FABIUAB = 0.9926i — 1.737j +014: FAG = fFAc|uAC : —D.8888i i 1.5556j “1- 0.8888 FAD = iFADillAD = —1.0910i— 1.2728j — 1.09101: Problem 2.122 In Problem 2.121, determine the vec— tor component of T normal to the bar OD. Problem 2.123 The disk A is at the midpoint of the sloped surface. The string from A to B exerts a 02- lb force F on the disk. If you resolve F into vector components parallel and normal to the sloped surface, what is the component normal to the surface? _.,-_._- .1 4) if: ‘15s} Moi (3:9,, Solution: Considera line on t e sloped surface from Aperpendic— ular to the surface. (see the diagram above) By SIMILAR triangles we see that one such vector is r N : Sj + 2k. Let us ﬁnd the component ofF parallel to this line. The unit vector in the direction normaE to the surface is my 8j + 2k _ e : —-—- : 1i 2 0.970 +0243}: N Ier x/s? + 22 J The unit vector 6113 can be found by (we , mli + (as - aAh’ + (213- — ZANI (we — “)2 + (as *yaiz + (Zn - ZAP Point B is at (0, 6, 0} (ft) and A is at (5, I, 4) (ft). Substituting, we get 3:113 = 9A3 : “0.5154” 0.515j e 0.492k NOWF = iFieAB =(O-2)3AB _ F : 70.123i “t“ 0.123.; — 0.0984k (lb) The component of F normal to the surface is the component parallel to the unit vector eN. ’ ' FNORMAL : (F'QN)6N £(0.955)eN FNQRMAL = 0i + 0.0927j + 652351511. Problem 2.124 In Problem 2.123, what is the vector component of F parallel to the surface? Solution: From the solution to Problem 2.123, F E —0.123i + 0.123j - 0.0984k (lb) and FNQRMAL : 0i + 0.0927j Jr 0.023211: (lb) The component parallel to the surface and the component normal to the surface add to give F(F : FNORMAL + Fpmallel)- Solution: From the solution of Problem 2.12:, |T| = an N, and the com- ponent of T parallel to bar CD is Tparanel : “7.52 N. The component of T normal to bar CD is given by Triormal : V iTi T (Tperallel) : 49-4 N- Thus Fparallei = F — FNORMAL Substituting, we get Fpmncl = 41.12311 + 0.0304j — 0.121614: lb Problem 2.141 The cable BC shown in Prob- lem 2.140 exerts a BOO-lb force F on the hook at B. (a) Determine rAB x F and rAC x F. (b) Use the deﬁnition of the cross product to explain why the result of (a) are equal. Solution: (a) From Problem 2.140, the unit vector 9130 = 0i — 0.6j + 0.3k, and rAB = —12i+ 6j — 12k Thus F ﬂ |F|eBo E 0i — 180j + 240k, and the cross product is i j k I'AB X F x 712 6 —12 m —7‘20i + 28305 + 2160k(ft-lb) 0 —180 240 The vector TAG : (4 e 16)i+ 0j + (8 —12)k = —12i+ Oj — 4k. Thus the cross product is i j k rag x F r #12 0 M4 = —720i + 2880j + 2160k (ﬂuib) 0 —180 240 (b) The deﬁnition of the cross product is r x F = ErHFlsinBe. Since the two cross products above are equal, #343 IiFI sin 913 = |rA0§|Fisin92a Note that rAc = FAR + 1‘30 from Proh— leni 2.1l6. hence mm; X F m I'AB X F + r50 x F = |rABHF|sin91e + |ch|EF|sin08 : IrABHFlsinﬂle, since 1'30 and F are parallel. Thus the two results are equal. H Problem 2.142 The rope AB exerts a SO-N force T on the collar at A. Let rCA be the position vector from point C to point A. Determine the cross product roA x T. Solution: The vector from O to D is r01) = (:L‘D -— mcﬁ + {3,113 — yc}j + (23 w 2:0)k. The magnitude ofthe vector Ircnl = V (we — \$002 + (to) — yo)? + (213 — ZCP- i The components of the unit vector along CD are given by 11.03,; = i (mp —mc)/jrcpl,ucpy : (yp iyc)/IrgD|,etc. Numerical . ; 1 values are [ram = 0.439 m, ucpx = 41.456, Hcpy = 41.684, . I i and etc-Dz : 0.570. The coordinates of point A are given by wA : ! rec + iI'C'Ai'U-Cbma 3,1,; = my + |rcA|uch, etc. The coordinates of point A are (0.309, 0.162, 0.114) in. The vector I‘CA is given by ray; =(:1:A ~mg)i+(yA—yg)j+(zA—zc}k. ThevectorrCA is rCA : (-0.091)i + (70.137}j + (0.114)k m. The vector from A to B and the corresponding unit vector are found in the same manner as from C to D above. The results are ITABI = 0.458 m, MAE” = —0.674, HA3!) : 0.735, and UABz : 0.079. The force T is given ‘ by T = iTiUAB- The result is T = -—33.7i + 36.7j + 3.931: N. The cross product rCA x T can now be calculated. i j k I'CA X T m —0.091 —0.138 0.114 —33.7 35.7 3.93 : (W4.65)i + (»3.53)j + (—7.98)k N-m Problem 2.158 Suppose that the forces F A and F 3 shown in Problem 2.163 have the same magnitude and FA .FB z 600 N2. What are FA and F3? Solution: From Problem 2.163, the forces are: The dot product: FA - F3 : |FA|fF3|(U.6233) : 600 N2, from _ i FA =EFAE{icos40“ sin50“+jsin40“+kcos40“c0550°) W I lF i 600 3103N A i B : W x . ’ n {F A!(0.58681 + 0.6428j + 0.4924k) 0.6233 FB = |FB |(—icos 60-a sin 30° +jsin 60° + kcos 60“ cos 30°) and : IFBIHJ 25i+0 Sﬁﬁj +0 433k) FA = 18.21i + 19.95j + 15.28k (N) i F5 : ”7.7m + 26.87j + 13.441: (N) H Problem 2.159 The rope CE exerts awspoﬁ forcejl‘ on the door ABC'D. Determine theé ctoi: componenft‘fw of T in the direction parallel to the liné"f point B. '“poihtit'tn (0,0.2.0) m ' (0.4,0.25,—0.I) to i i ‘ .. . A-(0.5,0,0)m x i " (0353,02) :‘n Solution: Two vectors are needed, to}; and I‘AB. The end points of these vectors are given in the ﬁgure. Thus, rc E : (\$13 — wgﬂ + (yg — ycﬁ + (21; —— zc)k and a similarform holds for 1343. Calculating these vectors, we get 1-3;; = 0.4i + 0.05j — 0.11: m and 1343 : —0.15i+ Clj + 0.2k In. E (0’4’0'25’411) m The unit vector along CE is 80E = 0.963i + 0.12Gj — 0.241k C and the force T, is T : ITIBCE. Hence, T : 500(0.963i + (09-10)!“ A (05 0mm 5 , 0.120j —— 0.241k) = 482i + 60.23 -— 120k N. The unit vector along ' ’ ' x . J i' AB is given by mm a —-O.6i + Oj + 0.8k and the component of - 5 T parallel to AB is given by TAB i T I GAB‘ Thus, TAB : B {482)(—0.6) + (60.2)(0) + (—120}(0.8) = —385.2 N (035,0,02) In Problem 2.160 in Problem 2.169, let r 30 be the po— sition vector from point B to point 0. Determine the cross product r30 x ’1‘. i Solution: The vector from B to C is i 5 r30 = (30 W meli + (:90 e yell + (20 " ZB)k z —o.35i + 0.2j w 0.2k m. The vector T is T : 482i + 60.2j — 1201: N. The cross product of these vectors is given by i j k ' 1 mm x '1‘ : —o.35 0.2 “0.2 : ——12.oi m l38j m 117kNm i 482 60.2 —120 . Problem 3.35 The collar A slides on the smooth ver- tical bar. The masses mg : 20 kg and m3 = 10 kg, When it : 0.1 In, the spring is unstretched. When the system is in equilibrium, h = 0.3 In. Determine the spring constant 14:. Solution: The triangles formed by the rope segments and the hor- izontal line level with A can be used to determine the lengths Lu and L3. The equations are ‘ Ln = #0125)2 + (0.1}2 and L3 = #0125)? + (0.3)2. The stretch in the spring when in equilibrium is given by 5 = L 5 -— Lu. Carrying out the calculations, we get Lu, = 0.269 m, L5 2 0.391 m, and 6 : 0.121 m. The angle, 9, between the rope at A and the horizontal when the system is in equilibrium is given by tame = 0.3/0.25, or 3 = 50.2°. From the free body diagram for mass A, we get two equilibrium equations. They are ZFI =—NA+Tcosﬁ':0 and EFF" : TsinﬁimAg:0. We have two equations in two unknowns and ban solve. We get N A = 163.5 N and T = 255.4 N. Now we go to the free body diagram for B, where the equation of equilibrium is T — m B y — Fat? 2 O. This equation has onty one unknown. Solving, we get k: = 1297 Nlm Problem 3.48 The SO-Ib cylinder rests on two smooth surfaces. (3) Draw the free-body diagram of the cylinder. (b) If a : 30°, what are the magnitudes of the forces exerted on the cylinder by the left and right surfaces? Solution: Isolate the cylinder. (a) The free body diagram of the isolated cylinder is shown. (b) The forces acting are the weight and the normai forces exerted by the surfaces. The angle between the normal force on the right and the :1: axis is (90 + E). The normal force is NR : |NR|(i 003(90 + 19) +jsin(90 + ,6» NR = |NRI(—isin,6 +j (205,6). The angle between the positive :1: axis and the left hand force is normal (90 7 or); the normal force is NI, : INLKisinoz +jcos CE). The weight is W = 0i — L“?! j. The equilibrium conditions are 2F =W+NR+NL =0. Substitute and collect like terms, ZFE = {—iNﬂgsmn+ INL|slnoz)i = 0, SEE : (|NR|cosﬁ+ lelcoso: - |W|}j : 0. Solve: lNRl z (sing) lNLI: em |W| sinﬁ ) sin(a + ,9) I and iNLi = ( For |W| = 50 lb, and 0: = 30°, [3 2 45°, the normal forces are |NL| =36.6lb, [NR| 325.911: Problem 3.49 For the 50—11) cylinder in Problem 3.48, obtain an equation for the force exerted on the cylinder by the left surface in terms of the angle 0: in two ways: (it) using a coordinate system with the y axis vertical, (b) using a coordinate system with the y axis parallel to the right surface. 1 Solution: The solution for Part (a) is given in Problem 3.48 (See free body diagram). \ i i l () <W—> Part (b): The isolated cylinder with the coordinate system is shown. The angle between the right hand normal force and the positive :1: axis is 180°. The normal force: N g = —|NR|i + DJ. The angle between the left hand normal force and the positive a: is 180 w (or + ,8). The normal force is NL : INL§(—icos{o: + [3) +jsi11(a + ,6». l _ , . . . Substitute and collect like terms, The angle between the weight vector and the posttive 3 arms is 43. The weight vector is W = iW|(i (308,3 — j sin 13). 2 Fa; = (—|NR| — lNLl cos(or + ,6) + lW] cos ,8)i 2 0, The equilibrium conditions are Ea}, = (|NL| sin(a + ,9) — |W| sinﬁ)j : 0. ZF:W+NR+NL:O. Solve: INLi =( |W| sinﬁl ) sin(rx + ,6) Problem 3.56 The system is in equilibrium. What are the coordinates of A? Solution: Determine from geometry the coordinates as, y. Isolate the cable juncture A. Since the frictionless pulleys do not change the magnitude of cable tension, and since each cable is anded with the same weight, arbitrarily set this weight to unity, [W| = 1. The angle between the cabie AB and the positive :1: axis is 0:; the tension in AB is ITABl = iCOSCt +jsina. The angle between A0 and the positive .1: axis is (180° - It3); the tension is TAG : ITacchosﬁ +jsin a), The weight is lW| : Oi 7 ji. The equiiibrium conditions are 2F =TAB+TA0+WEQ Substitute and collect like terms, ZR; = (0056! —- cosmi = 0, SF. = (sina+sinﬂ— 115:0. From the ﬁrst equation cos a = cos ,3. On the realistic assumption that both angles are in the same quadrant, then or = ,6. From the second equation sin 0: = (é) or or : 30°. With the angles known, geometry can be used to determine the coordinates 2:, y. The origin of the :12, y coordinate system is at the pulley B, so that the coordinate . :1: of the point A is positive. Deﬁne the positive distance 5 as shown, so that h+e b—m Similarly, ( Reduce to obtain :0 =b— hector —£cotcu. Substitute into the ﬁrst equation to obtain a: r (in) (1).. hector). Multiply this equation by tan a and use 5 = m tan 0: to obtain 5 = (toga) {hm heota}. The sign of the coordinate y is determined as follows: Since the co- ordinate m is positive, the condition (I) —— hoot. a) > U is required; with this inequality satisﬁed (as it must be. or the problem is invalid), e is also positive, as required. But the angle a: is in the ﬁrst quadrant, so that the point A is below the pulley B. Thus 3; = —e and the coordinates of the point A are: 1 m:(%)(behcotoz), y:—§(btana—h). 3:300 E Z i . l l i E l : Problem 3.76 The cable AB keeps the 8-kg collar A in place on the smooth bar CD. The y axis points upward. What is the tension in the cable? Solution: The coordinates of points 0' and D are C (0.4, 0.3, 0), and D (0.2, 0, 0.25). The unit vector from 0 toward D is given by 9019 = 9013314- BCDyj + eaDzk 2 —U.4561 m 0.68% + 0.570k. The location of point A is given by an; = mg + dCAeCDzs with similar equations for yA and ZA- From the ﬁgure, dCA = 0.2 m. From this, we ﬁnd the coordinates of A are A (0.309, 0.162, 0.114). From the ﬁgure, the coordinates of B are B (0, 0.5, 0.15). The unit vector from A toward B is then given by 3A3 = eABmi + BAByj + 614sz : 70.57414» 0.735j + 0.079k. The tension force in the cable can now be written as TAB : —U.574TABi + 0.735TABj + 0.079TABR. From the free bod)r diagram, the equilibrium equations are: FNa: + TABBABm = 0, FNy + TABBABy " my : 0. and FNz +TAB‘3AB2 = 0. We have three equation in four unknowns. We get another equation from the condition that the bar CD is smooth. This means that the normal force has no component parallel to CD. Mathematically, this can be stated as F N - 60D x 0. Expanding this, we get FNweCDm + FNyecny + FNzeCDz E 0- We now have four equations in our four unknowns. Substituting in the numbers and solving, we get ‘ ’B’S'Tr'v' To Swim Fm. = 38.9 N. FM : 36.1 N, and FNZ : —4.53 N, Problem 3.77 In Problem 3.76, determine the magni- tude of the normal force exerted on the collar A by the smooth bar. 80111151011: The solution to Problem 3.76 above provides the mag- nitudes of the components of the normal force exerted on the collar at A. IFNi : V (1?pr + (FNy)2 ‘i" (FNz)2o Substituting in the values found in Problem 3.77, we get gFN: = 53.2 N. Problem 3.84 The mass of the suspended object A is 111,; and the masses of the pulleys are negligible. De- termine the force T necessary for the system to be in equilibrium. Solution: Break the system into four free body diagrams as shown. Carefully label the forces to ensure that the tension in any single cord is uniform. The equations of equilibrium for the four ob- jects, starting with the leftmost pulley and moving clockwise, are: 373T=0, RA3S=0, F—SRr-O, and 2T+2S+2R—mAg:0. We want to eliminate S, R, and F from our result and ﬁnd T in terms of WA and 9. From the ﬁrst two equations, we get S = ST, and R : BS 2 91". Substituting these into the last equilibrium equation results in 2T «1— 2(3T} + 2(9T) = mAg. Solving, we get T : mAg/26 . .9 Note: We did not have to solve for F to ﬁnd the appropriate value of T. The ﬁnal equation would give us the value of F in terms of mm and 9. We would get F = 27mA 9/26. If we then drew a free body diagram of the entire assembly, the equation of equilibrium would be F - T — m A g = 0. Substituting in the known values for T and I", we see that this equation is also satisﬁed. Checking the equilibrium solution by using the “extra" free body diagram is often a good procedure. ...
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