# HW7 - PROBLEM 16.5 Knowing that the coefficient of static...

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PROBLEM 16.5 1.5m ~ 1 m Knowing that the coefficient of static friction between the tires and the road is 0.80 for the automobile shown, determine the maximum possible acceleration on a level road, assuming (a) four-wheel drive, (b) rear- wheel drive, (c) front-wheel drive. SOLUTION (a) Four-wheel drive: Thus: 0.80mg = ma a = 0.80g = 0.80(9.81 mIs2) = 7.848 mIs2 a = 7.85 mIs2 .... (b) Rear-wheel drive: Thus: NA = OAW + 0.2ma FA = f.JkNB = 0.80(OAW + 0.2ma) = 0.32mg + 0.16ma -:!:- 'LFx = 'L (Fx )etf : FA = ma 0.32mg + 0.16ma = ma 0.32g = 0.84a a = 0.32 (9.81 mIs2) = 3.7371 mIs2 0.84 or a = 3.74 mIs2 ....

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,.., ~ 1 PROBLEM 16.5 CONTINUED (c) Front-wheel drive: ~ ~~ I - - = ~ ''''oi t ~ . ~ L0 G --G}) O~5". ~ r=:s t)/ f }I~ .1~8 f.5'~ "'" +} 'LMA = 'L(MAtff: (2.5 m)NB - (1.5 m)W = -(0.5 m)m£1 Thus: N B = 0.6W - 0.2m£1 FB = J.ikNB = 0.80(0.6W - 0.2m£1) = 0.48mg - 0.16m£1 0.48mg - 0.16m£1 = m£1 0.48g = 1.16£1 a = 0.48(9.81m/s2) = 4.0593m/s2 1.16 or a = 4.06m/s2- ~ - --
PROBLEM 16.7 ~.51b I' h G .-:1 ,. 36 in. A 50-lb cabinet is mounted on casters that allow it to move freely (,u = 0) on the floor. If a 25-lb force is applied as shown, determine (a) the acceleration of the cabinet, (b) the range of values of h for which the cabinet will not tip.

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