RENSSELER POLYTEHNI INSTITUTE TROY, NY FINL EXM INTRODUTION TO ENGINEERING NLYSIS ENGR-00) NME: Solution Section: RIN: Monda, Ma 6, 06 Problem Points Score 0 0 0 0 5 0 6 0 Total 00 N.B.: You will be graded on 5 problems, 0 points per problem. Problems,, and are mandator and will be graded. Before turning in our eam, please make sure ou have circled the two problems ou want to be graded out of problems, 5 and 6.
Problem 0 points) frame with a smooth pulle supports an 800 N load as shown a) Draw free bod diagrams of members and BD; 5) b) Write out the equations of equilibrium for ; 6) c) Write out the equations of equilibrium for BD; 6) d) Determine the horizontal and vertical components of force that the pins at, B, and eert on their connecting members. ) Solution: a) FBD of member FBD of member BD 0.5 pt each force) b) c) M 0. 0. 8000.5 0 F 800 0 F 0 M B 0. 800.05 8000.05 0 F B 800 0 F B 800 0 pts each equation) pts each equation) d) 0 0 0 0 0. 0. 0 0 0 0 0 800 0 0 0 0 B 0 0 0 0 0 0 0. B 800 0 0 0 0 800 0 0 0 0 800 00 N, 000N B 00 N, B 00N 00 N, B 000N 0.5 pt each force)
Problem 0 points) onsider the following linear equations. 5. Write the equations in a matri form as X=B. pt) 5. Determine the cofactors of all the entries of and hence write down the cofactor matri. 0pts) Each entr pt and the cofactor matri pt 0 0) 6 8 8) 6 8 ) 6 0 6) 5 8 Hence:. Evaluate the classical adjoint of, adj). pts) Equation pt and adjoint pt ) T adj. Find the inverse matri of using the adjoint method. pts) orrect ) pt, equation pt and correct - pt ) ) ) adj 5. Solve the equations for using ramer s rule. pts) orrect pt, ) pt, equation pt, correct pt ) 6) 8)
Problem 0 points) band belt is used to control the speed of a flwheel. If the coefficient of friction between the belt and the flwheel is 0.5, ermine the magnitude of the tension in each side of the cable and the maimum couple that can be stopped if the flwheel is rotating: a) lockwise 5) b) ounter clockwise 5) Note: You need to draw all required FBD s for this problem a) FBD of Band T T 80 mm 0 mm P = 00 N 80 mm M 0 000 T 80 T80 0 T T 00 N FBD of Drum: pt for correct T and T and pt for correct E 50 mm E E T T.5 / ).5 / ) T T e T 80) T e T. N and T 05.8 N Torque T T ).r 05.8 -.).5.75 N m b) If the rotation is counter clockwise, the onl difference is that T and T will be reversed but since both are 80 mm awa from the values of T and T, and will not change from part a. 5
Problem 0 points) Four forces act on a small airplane in flight as shown: its weight W, the thrust provided b the engine FT, the lift provided b the wings FL, and the drag resulting from moving through air FD. ) Epress all forces in artesian -) vector form 8) ) Find the vector -) form of the resultant of the four forces FR 6) ) The angle of the resultant FR from positive -ais ) ) The angle of the resultant FR from positive ais of the plane ) 5) What is the status of the plane select ) at this moment? Eplain ) a) Static equilibrium; b) Deaccelerating; c) ccelerating; d) onstant speed. Note: You need to show all work to receive full credit. Solution ) F T = 0-cos0 i + sin0 j) kn = -.85i +.7j kn p) F L = sin0 i + cos0 j) kn =.7i +.6j kn p) W = - 5j kn p) F D = i kn p) ) F R = F T + F L + W + F D = -.85i +.7j +.7i +.6j - 5j + i p) = -.68i + 0.j kn p) ) The angle of F R with respect to -ais is p) = 80 tan.. = 80 7. = 7.77 ) The angle of F R with respect to -ais is: p) = 7.77 + 0 = 8.77 5) ns: c), because there is a net force in almost the same direction as the motion of the plane. p)
Problem 5 0 points) In the sstem shown: a) Epress all the forces in the FBD in artesian Vector Form ) b) Epress all the moments in the sstem in artesian Vector Form ) c) Determine the resultant force of the sstem ) d) Replace the force sstem with a force and a couple at point 8) e) an the sstem be reduced to one force? Eplain ) oordinates of points : 0,.,.06); B.6,.,.06);.6,., 0); D 0,., 0) E 0, 0,.06); F.6, 0,.06); G.6, 0, 0); H.,.06, -.) 70 70 N force : F..5 pt for each force I.,.,0 a) 50N force : F 50 kˆ N.0,.06, -.) 0 î 0 0 b)0 N m N m. N m..5 pt for each moment c) R F F 0î 0ĵ 0kˆ N pts for components and pt for unit d) R is the same as before 0î 0ĵ -0kˆ) N.6,.,0 8î 6ĵ ĵ - 60 kˆ).0,.,.06 î ĵ 6kˆ N m rg F î 8ĵ 6kˆ) 8ĵ 7kˆ i.6ĵ.kˆ N m î 8ĵ 6kˆ). 0î pts for resultant, pt for unit, pts for cross product, and pts for result 0 00 i 6 N j - k - 6 6 e) R 0 -.6 0 -. 0 0-88 - 0 since the force and the moment are orthogonal. pts. - pts if dot product is not used. es the sstem can be reduced to a single force
Problem 6 0 points) bar is loaded and supported as shown. End of the bar is supported with a ball-and-socket and end B with a cable and a link. ) Draw complete and separate FBD of the bar in the bo given 7) ) Epress the cable tension T in artesian vector form ) ) Write equations-of-equilibrium for the bar 7) ) Solve for reaction components at supports and B if the result is along the negative ais directions, use - ), as well as the magnitude of the cable tension T ) Draw FBD here. Solution: ) The FBD is shown each force component is p) ) = =,, ),, ) ) Equations of equilibrium for the bar are: = + 0.8 750 = 0 [] p) = 0.88 = 0 [] p) = + 0. + 500 = 0 [] p) = 0.8 0.88 + 0. ) lb p) Because all forces not passing are concurrent at B, we can use resultant force components to represent these forces: R = 0.8T 750)i + -0.88T)j + 0.T 500 + Bz)k lb Because the -component passes through, onl - and z-components contribute moment about : = + 0. 500) 8 0.8 750) 8 = p to this or an intermediate steps), which is a vector equation and equivalent to the following scalar equations: + 0. 500) 8 = 0 [] p) 0.8 750) 8 =0 [5] p) ) Solving equations [] through [5] for the 5 unknowns: From [5], T = 58 lb; p) 0.5 p for each of the following solutions) from [], B z = -500 lb From [], 0.0 lb; From [], = lb; From [], z 0.0 lb.