Solution Monday, May 16, 2016 Problem Points Score

1 RENSSELAER POLYTECHNIC INSTITUTETROY, NYFINAL EXAMINTRODUCTION TO ENGINEERING ANALYSIS(ENGR-1100)NAME:SolutionSection: _____RIN:_____Monday, May16, :You will be graded on 5 problems, 20 Points per 1, 2, and 3 are mandatoryand will be turning in your exam, please make sure you have circled the two problemsyou want to be gradedoutof problems 4, 5 and 1(20 Points )A frame with a smooth pulley supports an 800 N load as shown(a) Draw free body diagrams of members AC and BCD;(5)(b) Write out the equations of equilibrium for AC;(6)(c) Write out the equations of equilibrium for BCD;(6)(d)Determine the horizontal and vertical components of force that the pins at A, B, and C exert ontheir connecting members.(3) Solution :(a)FBD of member AC FBD of member BCD( pt each force)(b) 800 0800 00 AxyxxxyyyMCCFACFAC (2 pts each equation)(c) 800 800 0800 0800 0 ByxxxyyyMCFBCFBC (2 pts each equation)(d)0 0 0 0 0 0 0108000 1 0 00100 0 0 0 1 0108000 0 0 101800xyxyxyAABBCC 4200 ,40004200 ,32003400 ,4000xyxyxyANANBNBNCNBN ( pteachforce) Problem 2(20 Points )Consider the following linear the equations in a matrix form asAX=B.

2 (1pt) the cofactors of all the entries ofAand hence write down the cofactor matrix.(10pts)Each entry 1pt and the cofactor matrix 1pt11092)1012(2682)86(1892)46(212101)161 5(189333231232221131211 CCCCCCCCCH ence: the classical adjoint ofA, adj(A).(2pts)Equation 1pt and adjoint 1pt 122211221)( the inverse matrix ofAusing the adjoint method.(3pts)Correct det(A) 1pt,equation 1ptand correctA-11pt 122211221)(det11)2(2)1( the equations forx1using Cramer s rule.(4pts)CorrectA11pt, det(A1) 1pt, equation 1pt,correctx11pt 9detdet9)122(2)163(2)89(3324431223det111 AAAxP = 100 NT1T2 CxCyT1T2150 mmExEyCE25 Problem 3(20 Points )A band belt is used to control the speed of a flywheel. If the coefficient of friction between the beltand the flywheel is , determine the magnitude ofthe tension in each side of the cable andthemaximumcouplethat can be stopped if the flywheelis rotating:a)Clockwise(15)b)Counter clockwise(5)Note:You need to draw all requiredFBD sfor this problema)FBD of Band N400TT080xT80xT320x1000M2121 CFBD of Drum:1 pt for correct T1and T2and 1 pt for correct ) ( ).

3 T( )80(TeTT1221)2/3( )2/3( b)If the rotation is counter clockwise, the only difference is that T1and T2will be reversed but since bothare 80 mm away from C the values of T1andT2, and Cwill not change from part mm80 mm320 mmxy312122 Problem 4(20 Points )Four forces act on a small airplane in flight as shown: its weightW, the thrust provided by the engineFT, the lift provided by the wingsFL, and the drag resulting from moving through airFD.(1)Express all forces in Cartesian (x-y) vector form(8)(2)Find the vector (x-y) form of the resultant of the four forcesFR(6)(3)The angle of the resultantFRfrom positivex-axis(2)(4)The angle of the resultantFRfrom positive axis of the planex (2)(5)What is the status of the plane (select 1) at this moment? Explain(2)(a)Staticequilibrium;(b)Deacce lerating;(c)Accelerating;(d)Constant :You need to show all work to receive full (1)FT= 10(-cos10 i+ sin10 j) kN = + (3p)FL= 24(sin10 i+ cos10 j) kN = + (3p)W=-25jkN (1p)FD= 3ikN (1 p)(2)FR=FT+FL+W+FD= + + + + 3i(4p)= + (2 p)(3)The angle ofFRwith respect tox-axis is (2 p)=180 = (4)The angle ofFRwith respect tox -axis is: (2 p) = + 10 = (5)Ans: (c), because there is a net force in almost the same direction as the motion of the plane.

4 (2 p) Problem 5(20 Points )In the system shown:a)Express all the forces inthe FBD in Cartesian Vector Form(3)b)Express all the moments in the system in Cartesian Vector Form(3)c)Determine the resultant force of the system(3)d)Replace the forcesystemwith a force and a coupleat point A(8)e)Can the system be reduced to oneforce? Explain(4) N)k 60-j 30i ( ) ,,04(..1470F:forceN70Nk 50F:force50Na).12,.12, ) ,(.2,H0);0,(.16, );0,(.16, );0,(0,E0).12,(0,D0);.12,(.16, );.12,(.16, );.12,A(0,:pointsofsCoordinate21 pt for each force mNk 6j 12i 406,.12., 6i 80,12., )21 pt for each moment Nk 10j 30i 20 FFRc)21 2 pts for components and 1 pt for unit mNk 72j 84i )k 6j 18i (126326-12-16kji10010)k 6j 18i (12F xrCCCN)k 10-j 30i (20beforeassame theisRd)2AG21 2 ptsfor resultant, 1 ptfor unit, 3 pts for cross product, and 2 pts for theandforce thesinceforcesinglea toreducedbecansystem theyes0132-288-42010 x21CR)e pts if dot product is not 6(20 Points )A bar is loaded and supported as shown.

5 End A of the bar is supported with a ball-and-socket and endB with a cable and a link.(1)Draw complete and separate FBD of the bar in the box given(7)(2)Express the cable tensionTin Cartesian vector form(3)(3)Write equations-of-equilibrium for the bar(7)(4)Solve for reaction components at supportsAandB(if the result is along the negative axisdirections, use - ), as well as the magnitude of the cable tensionT(3) Solution :(1)The FBD is shown (each force component is 1 p)(2)==(,,)(,,) =( + )lb(3 p)(3)Equations of equilibrium for the bar are: =+ 750=0[1] (1 p) = [2] (1 p) =+ + 500=0[3] (1 p)Because all forces not passing A are concurrent at B, we can use resultant force components torepresent these forces:R= ( 750)i+ ( )j+ ( 500 +Bz)klbBecause they-component passes throughA, onlyx-andz-components contribute moment about A.

6 =(+ 500) 8 ( 750) 8=(2 pto this or any intermediatesteps),which is a vector equation and equivalent to the following 2 scalar equations:(+ 500) 8=0[4] (1 p)( 750) 8=0 [5] (1 p)(4) Solving equations [1] through [5] for the 5 unknowns:From [5],T= 2358 lb; (1 p)( p for each of the following solutions)from [4],Bz=-500 lbFrom [1],Ax lb;From [2],Ay= 1929lb;From [3],Az FBD here.