ES201 - Examination 2 Winter Adams and Richards NAME BOX NUMBER

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ES201 - Examinatin 2 Winter 2003-2004 Adams and Richards NAME BOX NUMBER Please Circle One : Richards (Perid 4) ES201-01 Adams (Perid 4) ES201-02 Adams (Perid 6) ES201-03 Prblem 1 ( 12 ) Prblem 2 ( 24 ) Prblem 3 ( 24 ) Prblem 4 ( 40 ) Ttal (100) INSTRUCTIONS Clsed bk/ntes exam. (Unit cnversin page prvided) Help sheet allwed. ( 8-1/2 x 11" sheet f paper, ne side ) Laptps may be used; hwever, n pre-prepared wrksheets r files may be used. 1) Shw all wrk fr cmplete credit. Start all prblems at the ANALYSIS stage, but clearly label any infrmatin yu use fr yur slutin. Prblems invlving cnservatin principles MUST clearly identify the system and shw a clear, lgical prgressin frm the basic principle(s). Fr FULL CREDIT, yu must clearly shw and label all mmentum transfers n a free-bdy diagram fr yur system. Dn't expect us t read yur mind as t hw r why yu did smething in the slutin. Clearly indicate hw yu arrived at yur answer and hw yu used the given infrmatin in the prcess. Always crunch numbers last n an exam. The final numerical answer is wrth the least amunt f pints. (Especially if all we wuld have t d is plug in the numbers int a well-dcumented slutin.) 2) Useful Rule f Thumb (Heuristic): (100 pint exam)/(90 min) 1 pint/minute. That means a 10 pint prblem is nt wrth mre than 10 minutes f yur time (at least the first time arund). 3) Please remain seated until the end f class r everyne finishes. (Raise yur hand and I ll pick up yur exam if yu have ther wrk yu need r want t d.) USEFUL INFORMATION SI USCS Mlar Mass [kg/kml; lbm/lbml] Ideal Gas Cnstant: R u = 8.314 kj/(kml-k) = 1545 (ft-lbf)/(lbml- R) Air 28.97 = 1.986 Btu/(lbml- R) O 2 32.00 Standard Acceleratin f Gravity: g = 9.810 m/s 2 = 32.174 ft/s 2 N 2 28.01 Density f liquid water = 1000 kg/m 3 = 62.4 lbm/ft 3 H 2 2.016 = 1.94 slug/ft 3 CO 2 44.01

Length 1 ft = 12 in = 0.3048 m = 1/3 yd Mass 1 m = 100 cm = 1000 mm = 39.37 in = 3.2808 ft 1 mile = 5280 ft = 1609.3 m 1 kg = 1000 g = 2.2046 lbm 1 lbm = 16 z = 0.45359 kg 1 slug = 32.174 lbm Temperature Values (T/K) = (T/ R) / 1.8 (T/K) = (T/ C) + 273.15 (T/ C) = [ (T/ F) 32 ]/1.8 (T/ R) = 1.8(T/K) (T/ R) = (T/ F) + 459.67 (T/ F) = 1.8(T/ C) + 32 Temperature Differences ( T/ R) = 1.8( T / K) ( T/ R) = ( T/ F) ( T/K) = ( T/ C) Vlume 1 m 3 = 1000 L = 10 6 cm 3 = 10 6 ml = 35.315 ft 3 = 264.17 gal 1 ft 3 = 1728 in 3 = 7.4805 gal = 0.028317 m 3 1 gal = 0.13368 ft 3 = 0.0037854 m 3 Vlumetric Flw Rate 1 m 3 /s = 35.315 ft 3 /s = 264.17 gal/s 1 ft 3 /s = 1.6990 m 3 /min = 7.4805 gal/s = 448.83 gal/min Frce 1 N = 1 kg m/s 2 = 0.22481 lbf 1 lbf = 1 slug ft/s 2 = 32.174 lbm ft/s 2 = 4.4482 N Pressure 1 atm = 101.325 kpa = 1.01325 bar = 14.696 lbf/in 2 1 bar = 100 kpa = 10 5 Pa 1 Pa = 1 N/m 2 = 10-3 kpa 1 lbf/in 2 = 6.8947 kpa = 6894.7 N/m 2 [lbf/in 2 ften abbreviated as psi ] Energy 1 J = 1 N m 1 kj = 1000 J = 737.56 ft lbf = 0.94782 Btu 1 Btu = 1.0551 kj = 778.17 ft lbf 1 ft lbf = 1.3558 J Energy Transfer Rate 1 kw = 1 kj/s = 737.56 ft lbf/s = 1.3410 hp = 0.94782 Btu/s 1 Btu/s = 1.0551 kw = 1.4149 hp = 778.17 ft lbf/s 1 hp = 550 ft lbf/s = 0.74571 kw = 0.70679 Btu/s Specific Energy 1 kj/kg = 1000 m 2 /s 2 1 Btu/lbm = 25037 ft 2 /s 2 1 ft lbf /lbm = 32.174 ft 2 /s 2

Prblem 1 (12 pts) a) Cnsider a unifrm pressure P acting n the left side f a hemisphere and a circular plate as shwn in the figure. The hemisphere and circular plate have identical diameters, D hemisphere = D plate. Hw des the net frce due t the pressure cmpare fr the tw cases? (Circle ne.) i) F a > F b ii) F a < F b iii) F a = F b iv) Nne f the abve. b) What are the dimensins f angular mmentum? What is a typical set f units fr angular mmentum? c) Tw frces, bth with magnitude F, are ppsitely directed and separated by a cnstant distance d as shwn in the figure. Hw des the net mment due t these tw frces abut pint P cmpare t that abut pint Q? (Nte: a > b, θ < φ) i) M P > M Q ii) M P < M Q iii) M P = M Q iv) Nne f the abve. F d F θ b Q a φ P

Prblem 2 (24 pts) A jet f water with density ρ hits a hinged flap with a mass m as shwn in the figure. The velcity f the water f bth the incming and utging jet is V jet. The incming water jet is circular with a diameter d. Knwn dimensins are given in the figure. a) Find the angle θ that the statinary flap makes with the hrizntal. Express yur answer in terms f the knwn quantities. b) Find the hrizntal and vertical reactin frces at the pin cnnectin A. Yu may assume that θ is knwn frm part (a). L 1 L 2 L 3 A d θ G V jet (Nte: The centerline f the exiting water jet is cllinear with A and G.) V jet

Prblem 3 (24 pts) Tw blcks rest n an inclined plane with θ = 35 as shwn in the figure. Blck A has mass m A = 13.5 kg and blck B has mass m B = 40 kg. The cefficients f static and kinetic frictin between all surfaces are µ S = 0.3 and µ k = 0.2, respectively. Initially the blcks are statinary and are supprted by a stp blck and fixed-length wire, as shwn n the figure. When the stp blck is remved the blck B immediately starts t mve because the angle θ is large enugh t prduce mtin. Find the acceleratin f blck B and the tensin in the wire immediately after the stp blck is remved. Wire A B Stp Blck θ

Prblem 4 (40 pts) M s Dairy has entered the annual Dairy Drag Race at the State Fair. His drag racer is a fully-laded milk truck shwn in the figure. When fully laded, the milk truck weighs 5000 lbf. The truck is a rearwheel drive vehicle, and the frnt tires prvide negligible frictinal drag when rlling. The maximum tractin between the tires and the rad ccurs when there is n slip between the tires and the rad, i.e. the frce between the rad and the tires is due t static frictin. The static cefficient f frictin between the rubber tires and the cncrete pavement is µ S = 0.80. (a)(16 pts) Determine the reactins between the tires and the rad at pints A and B, in lbf, when the truck is statinary. (b) (24 pts) Determine the maximum acceleratin pssible fr the fully-laded, rear-wheel drive milk truck, in ft/s 2 r in g s. Als determine the crrespnding reactins at pints A and B, in lbf. Is there any danger f the truck tipping under these cnditins? A M s Dairy G B d A = 4 ft d B = 6 ft h = 4 ft h d A d B