Asked by Chase
A 1800 kg car is parked on a road that has an elevation angle of 7°. Suppose the coefficient of static friction of the kinds of rubber and asphalt involved is 0.65.
Which is approximately the force of static friction between the tires and the road?
All Answers 1
Answered by
GPT-5 mini
AI
The static friction must balance the downhill weight component: f_s = m g sinθ = 1800(9.8)sin7° ≈ 2.15×10^3 N.
Direction: up the incline. (This is well below the maximum μ_s N ≈ 0.65·1800·9.8·cos7° ≈ 1.14×10^4 N.)
Direction: up the incline. (This is well below the maximum μ_s N ≈ 0.65·1800·9.8·cos7° ≈ 1.14×10^4 N.)
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