RAB*F (201-10)3-451 +1.735) (2011-133) + (-103.455) Question 1 (26 points) M = ||||||F|| cine SIN (40) Ogget You are using your left hand to control the steering wheel of your car. You use this hand to exert 34.6K 34.5K = 69.1k a 3 N force on the wheel is shown. The inner radius of the wheel is 10 cm, while the outer radius of the wheel is 20 cm. With the steering wheel in its current position, the crossbars on the wheel make 40° angles with the horizontal, as shown. Gr 3N 40° 20 cm 10 cm 3(1.151+0.585) 3.451+1.735 = +1.15 +0.585 2010 17,32 a) What is the magnitude and direction of the moment produced about A by the force +8 applied at B? (12 points) •√20-102 = √300 ||rl|= 201-105 FI= fcos 40+ sinto 2.294+ 3 = 17,32 M = TAB=FB ||F|| 201-10 x 2.291 +31 (201)(2,291) + (201) (35) + (-105) (2.297) (-1057135) 44.22 kNm 45.0+60-22.71k b) Use the picture of the steering wheel to draw how you could apply a 3 N force to the steering wheel to create the largest possible moment about point A. (6 points) c) Calculate the moment vector resulting from what you drew in part b). (8 points) m=F (SIN 90) 69..1ks 90) = 69.1 K =69.1 m +6 +2 3N 2 Question 2 (40 points) A cable BG is attached to pipe AC. The tension in the cable is 50 N. Calculate the moment that the force in the cable produces about point D. a) Present your answer in vector notation. (30 points) b) Present your answer as a drawing of the components of the moment acting about the x, y, and z axes at point D. You can draw these components on the figure provided below, but make sure they are clearly labeled. If your figure has gotten crowded you can make a new picture of ACDE with coordinate axes and draw your components there. (6 points) ACD connects to DE via a screw oriented along DE. c) What magnitude of the moment you calculated in part a) will act to turn this screw? (2 pts) d) Will the force act to tighten this screw connection or loosen it? (2 pts) EA 15 ema 15 cm 50N Mpz 45 cm •Ma G(5,0, 30) em 20 cm Mox produces & M=rxfsh (10) =rx50 -7501-2250= 50 -157-455 M=FBG Coa-(20-5 11+ (45-0)) +(30-30) = 151 +45 X FBG = -50 N -SON (1ST + 45J)?! @-750-2250 m turns © -151-451 ORO to tighten Scrap 3 F= fu 41 +1.145 = 1.271+.167 700+3505 Question 3 (34 points) 14.0 √132500 6159 400 = 66 F 1001+350 364 (MOT +350ĵ) F = 400 1007 +3505 a) A bracket is loaded by FA, M. Fc, and F. Calculate an equivalent force and moment acting ..27 + .96ĵ F = 41 +1.145 at O. (25 points) M FD -300 mm- Fc 500 mm F 200 mm- MB 350 mm FA - 20 N x . 1 |||M.||= ||fc|| -20 N =400 Nmm \F\\ -10 N 20 vectors! 20N+20N +ION = 50N Mg=400 Nmm M = c Fc -vatur! 20 = 6000+ 10000ĵ +12 (3001-500;) mc=60001+10000ĵ . MD=rop x FD (01+500) 10 = 5000ĵ m = sooo MA - VOAX FA • Fo= 50N + 3 force@Mx? M = MA+ Mo+Mc+ = (0 +350) 20 70 00 = Mp ·7000, +400 + 160001 +10000) + 5000ĵ M=22000+6400 b) Would your answer to a) be the same if My was acting at point C instead of at point B? Circle YES or NO and explain your answer. (3 pts) applied Game. +3 Jes because No matter where the moment is in an equivalent force, the answick will be the c) Would your answer to a) be the same if My was replaced by two equal and opposite forces acting some distance apart from one another? Circle YES or NO If you answered NO, explain why not. If you answered YES, draw what the forces would look like, providing possible values for the force magnitudes and the separation distance. (6 pts) If the forces were equal and apposite, 40 the moment would be equal to 0. 4