Hw24 - 5 en the di 1 ccefficient cf Saluticn 3 = 111 e E = man FE 32_2 l 3 I” 2F = mat Ft 32.2 I 1“ 2F = ma N 3 = n | E.E.’ where the

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Unformatted text preview: 5 en the di 1: ccefficient cf Saluticn 3 = 111 e+ E}; = man: FE 32_2 l 3 I”; + 2F: = mat: Ft _ 32.2 I +1“ 2F = ma : N - 3 = n | E: .E.’ where the z—cmnponent cf eccelereticn :- ie :erc since the hlcck hee nc I“) vertical meticn, 2 u = 0.5 redje = ccnetent H = 0.5t redfe Therefore N'= 3 1b F; = U.U31D6 1h rt = {3.1315533 11) 5 - Dynamics ear shewn maintains a At salutian ngfree*bfidy diagram ef the drivEr includes his weight W and the nermal fmflejvhetween the driver and the Using eeerdinates aleng and the read, the equatiens ef n the read. then a. When the ear gees ever the tep ef a hum? i E 2 I n p p where V = 103 km/h = 27_7773 mils and p = 9:} m Therefere i 732 Ans E N'= an 9.91 — .21;33_.._] = 99.9 n . . . . . . . . . . . . . . . . . . .. l 90 1 ' ht' 5 whufl‘ifi Only aheut 13 percent of the drlver's welg ' d then a a d1 1n the rea , i I: when the ear gees ever the hettem of P 3 _JL__ 9 . New 27.9173 m/s and P = 9“ m ' fins. __2_._-—-—'-'-'——- 2 9.9193 ] s 1471 H 90 f the car in the horizon-"5' coordinates along and normal to the path o plane of motionr the equations of motion are 2 F'coe 9 + N sin 9 =.m[95.3333 jgflfl) {-+ BF = ma n n +TEF=ma= Ncoefi-FeinB-mg=fl z a where the normal component of E . acceleration is an e v {9 acting toward the center of curvature {horizontally to the left}, V = 65 mi/h = 95.3333 ft/e, and the z—component of acceleration is - - - - -- zero since the car has no vertical motion. a. If F = D, then H ein 9 = lo‘ogaam -N CUE 9 = 32.2m -1 6 = tan {10.0933/32.2) = 17.41u it b' If 3 = D . then -F = 10.0933m N = 32.2“: mflbile re I ‘ determine the minimum .21.”; of frietien required the tires and the road 5.3. automobile will not skid, +sz=ma: NeeeB—Fsin3_mg=fl where the hermal eempenent of E II acceleratien ie an = v {p aeting toward the center ef curvature is zero since the ear has he vertical met ien . IfF=then v=65 kmfih= 13.0556 tn/e, then N sin 6 = 1.63eem N :95 3 = 9.81m 0.986.an + D_15391N = 3 35802111 -D.16391F + 0.93643N = 9-Blm F = 2 19346m N = loiaflng 213 . . . . . . . . . . “H fine. u = FIN - 2.19345f10 309? = 0- where r = 0.5 ft/s. r = 2n rad/5. and 3 = 1 H= 3L2 [a + 2(0.5)[2n}] IiII-IlI-I 5 force p exerted by the hag-P an aflgle « . Sphere when the haflp 1.5 rotating about a the tical diameter at a constant angular velocity directed tuwa d by the sphere r = 0_15 sin a]! Circular path travele [in a horiznntal plane: and the z—cnmponent 0 zero since the Sphere haE matinn . Therefore . ...
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This note was uploaded on 04/27/2009 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

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Hw24 - 5 en the di 1 ccefficient cf Saluticn 3 = 111 e E = man FE 32_2 l 3 I” 2F = mat Ft 32.2 I 1“ 2F = ma N 3 = n | E.E.’ where the

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