171_PartUniversity Physics Solution

171_PartUniversity Physics Solution - Applying Newtons Laws...

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Applying Newton&s Laws 5-43 5.95. IDENTIFY: Apply m = F a ! ! to the automobile. SET UP: The "correct" banking angle is for zero friction and is given by 2 0 tan v gR β = , as derived in Example 5.23. Use coordinates that are vertical and horizontal, since the acceleration is horizontal. EXECUTE: For speeds larger than 0 v , a frictional force is needed to keep the car from skidding. In this case, the inward force will consist of a part due to the normal force n and the friction force rad ; sin cos . f nf m a ββ += The normal and friction forces both have vertical components; since there is no vertical acceleration, cos sin . nfm g −= Using s f n μ = and 2 2 0 rad (1.5 ) 2.25 tan , v v ag R R == = these two relations become s sin cos 2.25 tan nn m g βμ and s cos sin m g . Dividing to cancel n gives s s sin cos 2.25 tan . cos sin βμ β + = Solving for s and simplifying yields s 2 1.25 sin cos 11 . 2 5 s i n = + . Using 2 2 (20 m s) arctan 18.79 (9.80 m s )(120 m) ⎛⎞ ° ⎜⎟ ⎝⎠ gives s 0.34. = EVALUATE: If s is insufficient, the car skids away from the center of curvature of the roadway, so the friction in inward. 5.96. IDENTIFY: Apply m = F a ! ! to the car. The car moves in the arc of a horizontal circle, so rad, = aa !! directed toward the center of curvature of the roadway. The target variable is the speed of the car. rad a will be calculated from the forces and then v will be calculated from 2 rad /. av R = (a) To keep the car from sliding up the banking the static friction force is directed down the incline. At maximum speed the static friction force has its maximum value ss . f n = SET UP: The free-body diagram for the car is sketched in Figure 5.96a. EXECUTE: y y Fm a = s cos sin 0 g But , f n = so s cos sin 0 m g s cos sin mg n = Figure 5.96a x x a = sr a d sin cos m a a d (sin cos ) nm a Use the y F equation to replace n : a d s (sin cos ) cos sin mg ma 22 s rad s sin cos sin 25 (0.30)cos25 (9.80 m/s ) 8.73 m/s cos sin cos25 (0.30)sin25 + ° = −° ° rad / R 2 = implies 2 rad (8.73 m/s )(50 m) 21 m/s. va R = (b) IDENTIFY: To keep the car from sliding down the banking the static friction force is directed up the incline. At the minimum speed the static friction force has its maximum value . f n =
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5-44 Chapter 5 SET UP: The free-body diagram for the car is sketched in Figure 5.96b. The free-body diagram is identical to that in part (a) except that now the components of s f have opposite directions. The force equations are all the same except for the opposite sign for terms containing s . μ Figure 5.96b EXECUTE: 22 s rad s sin cos sin25 (0.30)cos25 (9.80 m/s ) 1.43 m/s cos sin cos25 (0.30)sin25 ag βμ β βμ β ⎛⎞ −° ° == = ⎜⎟ + ° ⎝⎠ 2 rad (1.43 m/s )(50 m) 8.5 m/s. va R = EVALUATE: For v between these maximum and minimum values, the car is held on the road at a constant height by a static friction force that is less than s . n When s 0, rad tan . = Our analysis agrees with the result of Example 5.23 in this special case.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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171_PartUniversity Physics Solution - Applying Newtons Laws...

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