Engineering Analysis EGN 3321
University of Central Florida
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This 5 page Class Notes was uploaded by Lora Metz on Thursday October 22, 2015. The Class Notes belongs to EGN 3321 at University of Central Florida taught by Seetha Raghavan in Fall. Since its upload, it has received 80 views. For similar materials see /class/227556/egn-3321-university-of-central-florida in General Engineering at University of Central Florida.
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Date Created: 10/22/15
1 mm MgtugbantT shawanxgu exsusedtapu 39heS MgbugeE wahampe R mm bug xs mme Assume m mp dues mafthxi gclz l i 59 31 fncumal effects mm mm w 1 w n Itvlfuttv itlglvlm A mm m F H th nmmmmu mm mm gtyrm umle and huge mm Hm impumvc mm L pdhemmulhcmglm umllhmngw mrmullnthgt mm mm vul39 m umch WWW m mcwm uwmmwddummmcv m mwmg m nl cmamc wowdurc 0r dmwmg hc mmm mpulw m unomumum mm w mm m n 15er Imlrrlwliwv u thmmmr Nmmg um Mr mm Wv ham 7 w v 7 mm 35m m m m u nuv xgwl 7 100 ISO1 1 7 Hamx Am ns mqu wprqcmi In mybom s 0qu w km m mm mvubc m um mun and Show mm m ovuqu Wm n m m x 2 A boy having a mass of walks to the front B and bottom of the tob oggan Qiip F m A l as it r11 ul rum piman 40 Kg standing on the back ofa 15 Kg toboggan which is originally at rest Ifhe 39 L quot 39 39 between the 4 stops and the groundice Solution ll Free Body Diagram The unknown frictional force of the boy s shoes on the bottom of the tohoggan can be excluded from the analysis if the toboggan and boy on it are considered as a single system In this way the frictional force F becomes internal and the conservation of momentum applies Fig 15 121 Conservation of Momentum Since both the initial and nal momenta of the system are zero ecause the initial and nal velocities are zero the system s momentum must also be zero when the boy is at some intermediate point between A and B Thus 1 Here the two unknowns 1 and 17 represent the velocities of the boy moving to Lhe left and the tohoggan moving to he right Both are measured from a Lu11 inertial reference on the ground At any instant the position of pomt A on the toboggan and the postion ot39 the boy must be determined by integration Since 11 lsall then mbzlsh mm 0Assuming the initial position ofpoint A to be at Lhe origin Fig lS lZc then at the final position we have imbab my 2 0 Since Aquot s 2 m or n 2 Z r then 9 m1h mm 0 2 7mb2 7 5 m U 0 Ana Zmb mb m 3 Knowing that the coef cient of friction between the tires and the road is 080 for the automobile mass is 2000 Kg shown determine the acceleration on a level road assuming frontwheel drive and the rearwheel are free to rotate 20 points F rant wheel drive 151 1 2MA 2MA6H 25 mNB 15 mW 05 mma NB 06W 7 02ma Thus FB ukNB 08006W 027 048mg 016mc7 LZFX 2Fxeff FB ma 048mg 016m5 mg 048g 1165 a w981ms2 40593 ms2 116 or a 406 ms2 4 Ifbar AB has an angular velocity 03 4rads determine the velocity of the slider block C at the instant shown 15 points r5 0150 06 mfs rmc IL sin 1le sin rL HC 31641 m r5 IL sin EII 51113 rs Ll m H w 3 rad 03 rt LB16413 104 my39s gt 5 Just after the fan is turned on the motor gives the blade an angular acceleration 0L 20 en 6 rad52 where t is in seconds Determine the speed ofthe tip P ofone ofthe blades when t 3 s Howmany revolutionshas the hladeturned in 3 5 When t 0 the blade is at rest