183_Physics ProblemsTechnical Physics

183_Physics ProblemsTechnical Physics - Chapter 6 P6.67 (a)...

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Unformatted text preview: Chapter 6 P6.67 (a) If the car is about to slip down the incline, f is directed up the incline. ∑ Fy = n cos θ + f sin θ − mg = 0 where n= b g b g θ t 2 v min yields R b Rg tan θ − µ s v min = n f mg µ s mg and f = . cos θ 1 + µ s tan θ cos θ 1 + µ s tan θ ∑ Fx = n sin θ − f cos θ = m Then, f = µ sn gives t θ mg n cos θ g f sin θ . 1 + µ s tan θ f cos θ n sin θ When the car is about to slip up the incline, f is directed down the incline. Then, ∑ Fy = n cos θ − f sin θ − mg = 0 with f = µ sn yields mg µ s mg mg n= and f = . cos θ 1 − µ s tan θ cos θ 1 − µ s tan θ b In this case, g ∑ Fx = n sin θ + f cos θ = m If v min = (c) v min = b Rg tan θ − µ s b 1 + µ s tan θ 1 − µ s tan θ g = 0 , then g t θ n 2 v max , which gives R Rg tan θ + µ s v max = (b) b g f . θ t mg µ s = tan θ . a100 mfe9.80 m s jatan 10.0°−0.100f = 1 + a0.100 f tan 10.0° a100 mfe9.80 m s jatan 10.0°+0.100f = 1 − a0.100f tan 10.0° n cos θ 2 n sin θ 8.57 m s f cos θ 2 v max = 16.6 m s f sin θ mg FIG. P6.67 185 ...
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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