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dont get it #2......?
A car of unknown mass traveling at 19.4 m/s is rounding a curve with radius 80m that's "perfectly banked" for the car. (I think "perfectly banked" means no friction force is needed to keep the car at a constant height. A component of the normal force alone does that.)
If the car speeds up to a constant 25 m/s what coefficient of static friction will be needed to keep the car in its circular path and not skid?
1 Answer
- AlgolLv 71 decade agoFavorite Answer
When there is no friction acting to hold the car in place (as you suggest, the normal force is serving this purpose), we can use Newton’s 2nd law to find the banking angle. The vertical component of the normal force is ncosθ, and it is balanced by the weight (mg) of the car. The normal force is then:
ΣF(y) = 0 = ncosθ - mg
n = mg/cosθ
The horizontal equation has the centripetal force equal to the horizontal component of the normal force:
ΣF(x) = mv²/r = nsinθ
Plugging the expression for n into the centripetal force equation, we get:
mv²/r = (mg/cosθ)sinθ
θ = tan⁻¹(v²/rg)
= tan⁻¹[(19.4m/s)² / (80.0m)(9.80m/s²)]
θ = 25.6°
I’ll skip the derivation of the relevant equation for the case where friction is present, but if you want it, I’ll provide it on request. The equation is:
μ = (v² - rgtanθ) / (v²tanθ + rg)
= [(25.0m/s)² - (80.0m)(9.80m/s²)tan25.6°] / [(25.0m/s)²tan25.6° + (80.0m)(9.80m/s²)]
= 0.230
Hope this helps.