APPMI21 Semester 2 2024 Class Test 3 / Klastoets 3 02/09/2024 ‘Time Limi Memorandum 60 Minutes Student Number ~ This class test contains 8 pages (including this cover page) and 4 problems, Check to see if any pages are missing. Enter| Problem | Points | Score all requested information on the top of this page, and put your initials on the top of every page, in case the pages become sep-|_1 6 arated. / Hierdie klastoets bevat 8 bladsye (insluitend hierdie voorblad) en 4 probleme. Kyk of enige bladsye ontbreek. Vul| 2 7 alle gevraagde inligting bo-aan hierdie bladsy in, en plas jou voorletters bo-aan elke bladsy, ingeval die bladsye los kom. 3 12 -You may not use your books, o any notes on this class test. / Jy mag nie jou boeke, of enige notas in hierdie klastoets gebruik.) 4 15 -You are required to show your complete work on each problem| on this test. / Dar word van jou verwag om jou volledige werk| Total: | 50 vir elke probleem in hierdie toets te wys. - Do not write in the table to the right. / Moenie in die tabel aan die regterkant skryf nie. The following rules apply: / Die volgende reéls ell * If you use a “fundamental theorem” you must indicate this and explain why the theorem may be applied. / As jy ’n “fundamentele stelling” gebruik moet dy dit aandui on verduidelik waarom die stelling toegepas kan word. * Organize your work, in a reasonably neat and coherent way, in the answer sheet pro- vided. Work scattered all over the page without a clear ordering will receive very little credit. / Organiser jou werk, op ‘n redelike netjiese en samehangende manier, in die antwoordblad wat voorsien word. Werk versprei oor die bladsy sonder ’n duidelike logiese uiteensetting, sal baie min krediet ontvang, © Mysterious or unsupported answers will not receive full credit. A correct an- swer, unsupported by calculations, explanation, or algebraic work will receive no credit; an incorrect answer supported by substantially correct calculations and explanations might still receive partial credit. / Geheimsinnige of ongesteunde antwoorde sal nie volledige krediet ontvang nie, 'n Korrekte antwoord, nie ondersteun deur berekeninge, verduideliking, of algebraiiese werk sal geen krediet ontvang nie; ‘n ver- keerde antwoord ondersteun deur wesenlik korrekte berekeninge en verduidelikings mag dalk nog gedeeltelike krediet ontvang * You have 10 min free reading time. / Jy het 10 min ekstra leestyd.APPMI21 Class Test 3 / Klastoets 3 - Page 2 of 8 02/09/2024 1. (6 points) (a) (2 points) Define coplanar force system Defnieer saamvlakkige kragstelsel. Solution: Coplanat force system is a group of forces whose lines of actions belong Vf to the same plane. (b) (4 points) Correct the following statements: / Korrigeer die volgende stellings: Solution i. The resultant of a force system is defined to be the simplest system that can ve replace the original system without changing its external effect on a rigid (/ body. Die resultant van 'n kragstelsel word gedefinioer as die eenvoudligste stelsel wat, die oorspronklike stelsel kan vervang, ii, The wrench is a collinear and unidirectional force-couple. SL Die skroef is 'n nie-kollineére en eenrigting krag-koppel. 2. (17 points) The force system acting on a structural member consists of the couple C and the forces F,,F2 and Fs. Given C = 200.N.m, Fi, = 100.V, Fy = 90.V and Fy = 120N. Die kragstelsel wat op 'n struktuur gedeelte inwerk, bestaan uit die koppel C en die krag F,,F2 en Fs. Gegee C = 200 .N.m, Fy = 100.V, F;=90.V en Fy = 120, Figure 1: Question 2 / Vraag 2 (a) (7 points) Determine the equivalent force-couple system with the force acting at point G. Bepaal die ekwivalente krag-koppelstelsel met die krag wat by punt G inwerk.APPMI21 Class Test 3 / Klastoets 3 - Page 3 of 8 02/09/2024 Solution: Reference: Sample problems 3.1 & 3.10 Because of the three-dimensional nature of this problem, we will use vector algebra to solve it. The first step is to express the three forces and the couple in vector form: Fy = 10045 = wo 2 = 100 A=) + 5k |AB| = 51.221 — 76.82) + 38.41kN m J F2=90iN, Fs = 120jN C= 20056 = wo = 200 ( = 160i - 120kN-m When we move the forces F, and F, to point G, they are concurrent with F3. Adding the three forces, we get for the resultant force: R=F,+F2+F3 (—51.22i — 76.82j + 38.41k) + 901 + 120) wa 38.781 + 43.18] + 38.41kN ‘The couple of transfer that arises from moving F; and F, is the moment about G of Fy + Fy = (51.224 ~ 76.82) + 38.41k) + 904 = 38.781 — 76.82j + 38.41kN 4 ‘The resultant couple is the sui of the couple of transfer and the couple C, resulting C® = ren x (Fi + F2)+C i i k -4 30 | +1601 - 120k V/ 38.78 —T6.82 38.41] 901 + 116] + 360k (b) (5 points) Determine the wrench that is equivalent to the force system. Bepaal die skroef wat ekwivalent is aan die kragstelsel.APPMI21 Class Test 3 / Klastoets 3 - Page 4 of 8 02/09/2024 Solution: Force-Couple System Reduction As shown in part (a) the original force system can be reduced to the force-couple system: the force R,, acting at the origin O, and the couple Cx, where R= 38.81 + 43.25 + 38.4kN C¥ = 390i + 116j + 360k ‘We begin by determining the axis of the wrench, defined by the unit vector A in the direction of R: ee R88) + (43.27 + (88.4? 557i + 0.6217 + 0.552k uU R 38.81 + 43.2) + 38.4k vA ‘The component of C¥ in the direction of A can now be obtained: c= (C8 .d)d Y ‘The magnitude of this vector is Ch =CF.d vA = (3901 + 116] + 360k) - (0.5571 + 0.621j + 0.552) / = 488m which gives CE = CPA = 488(0.557i + 0.621j + 0.552k) vA = 272i + 303} + 269k Nm (©) (5 points) Find the coordinates of the point where the axis of the wrench crosses the xy-plane. Bepaal die koérdinate van die punt waar die as van die skroef die xy-vlak sny. Solution The wrench consists of the force-couple system: R = 38.81 + 43.2) + 384kN Cf = 2724 + 303) + 269k N-m To find the coordinates of the point where the axis of the wrench intersects theAPPM121 Class Test 3 / Klastoets 3 - Page 5 of 8 02/09/2024 ry-plane, we must find C%, the component of CG that is normal to A. We have cR=c®-cf = (890i + 116j + 360k) ~ (2724 + 303) + 269k) = 118i — 187] + 91kN-m i+ yj be the vector from the origin O to A, the point where the wrench, intersects the 7y-plane. We have rxR=OF o> ijk J rc uv 0 18i — 187j + 91k ass 432 384 After expanding the determinant, we get 38.4yl~ 38-40) + 48.22 ~ 88.8))k = 1181 — 187} + 91k Equating the coefficients of i and j yields 3.07m v, 87m 4 The third equation, obtained by equating the coefficients of k, is not independent of the preceding two equations, as can be easily verified. 384y= 118 y -384r= 187 orAPPM121 Class Test 3 / Klastoets 3 - Page 6 of 8 02/09/2024 Figure 2: Question 3 / Vraag 3 3, (12 points) Determine the resultant of the four forces and one couple which act on the plate as shown in Figure 2. Bepaal die resultant van die vier kragte en een koppel wat op die plat inwerk soos aangetoon in Figuur 2 Solution + R.= DF, = 40-+80c0s30° — 6000s 45° = 66.9.V Whe 44 Ry = SF, = 50+ 80sin30° + 60sin 45° = 132.4N Yi R= VO FI = 148.3N 4 -a2 1% ‘To determine the line of action of R. / Vv 6 = arctan (328 + M, = 140 ~ 50(5) + 60.cos 45°(4) — 60sin 45°(7) —237N.m M,= Ri d= 4 = Bi =16m 4% The resultant R may be applied at any point on the line which makes 63.2° angle with xcaxis and is tangent at point A to a circle of 1.6 m radius with center O.APPMI21 Class Test 3 / Klastoets 3 - Page 7 of 8 02/09/2024 Figure 3: Question 4 / Vraag 4 4. (15 points) The transmission tower OA is being hoisted into position by the cables AB and AC. The resultant of the cable tensions P and Q, along with the 2400-N weight of the tower, is a force R. acting at point O. Determine P, Q, and R. Die uitsaaitoring OA word opgehys in posisie in deur die kabels AB en AC. Die resultant, van die spanning in die kabels P en Q en die 2400 N gewig van die toring, is 'n krag R. wat by punt O inwerk. Bepaal P, Q en R. Solution From the figure we have A(0, 40, 69), B(—28, 40, 0), C(32, -36, 0) —28i — 807 — 69k P= Prag = P(e ae ( V28? + 80? + 697 ) = P(-0.2569% — 0.732) — 0.63138) N 324 — 764 — 60k = Oran = Q ( 3% = TE = 69k cae a VaR + 70? + OF W = ~2400k N (intersects the y axis at y=20 m) Because R acts at point O, > Mo =0, we have ) = Q(0.29764 — 0.70685 — 0.6417k) N Y (roa x P) + (roa x Q) + (20j x W) = i i k i j k ij ok 0 40 69 }+Q) 0 40 69 |+|0 20 0 0.2562 07327 -0.6313| — |-0.2976 —07068 -0.6417| |o 0 —2400) Expanding the determinants lead and equating like components: 25.26P +23.1Q = 48000 Q)\ \v, —17.68P + 20.53Q 0 (2) \4 wa 10.25P —119Q = 0 (3)APPM121 Class Test 3 / Klgstocts 3 - Page 8 of 8 02/09/2024 v ‘ Solving (1) and (2) gives P = 1063.N; Q=915N v R=P+Qiw J 1063(—0.2562i — 0.732) — 0.6313k) + 915(0.2976i — 0.7068j — 0.6417k) — 2400k = ~1425j ~ 3660k.N V/