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Unformatted text preview: 104. REASONING AND SOLUTION The figure to the left below shows the forces that act on the sports car as it accelerates up the hill. The figure to the right below shows these forces resolved into components parallel to and perpendicular to the line of motion. Forces pointing up the hill will be taken as positive.
N P mg sin θ mg F N P mg cos θ a. The acceleration will be a maximum when P = fsmax . From the forces fsmax − mg sin θ = ma along the line of motion: The force fsmax is equal to µsFN. The normal force can be found from the forces perpendicular to the line of motion: FN = mg cos θ Then µs (mg cos θ ) − mg sin θ = ma
a = g ( µs cos θ – sin θ ) = 9.80 m/s 2 ( 0.88cos18° – sin18° ) = 5.2 m/s 2 ( ) b. When the car is being driven downhill, P (= fsmax ) now points down the hill in the same direction as (mg sin θ ). Taking the direction of motion as positive, we have fsmax + mg sin θ = ma Following the same steps as above we obtain µs (mg cos θ ) + mg sin θ = ma
a = g ( µs cos θ + sin θ ) = 9.80 m/s 2 ( 0.88cos18° + sin18° ) = 11 m/s 2
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- Spring '08
- Euclidean vector, mg sin, mg cos