Question 1047148: The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the care and the square of the speed. It 247-lb of force keeps a 1200-lb car from skidding on a curve of radius 400 ft. at 35 mph, what force would keep the same car from skidding on a curve of radius 850 ft. at 60 mph?
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the care and the square of the speed.
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F = k*(1/r^2)*W*(1/v^2) --- k = a constant, r = radius, v = speed
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Step 1, solve for k
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F = k*(1/400^2)*1200*(1/35^2) = 247
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Solve for k, then use k for the calculations.
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It 247-lb of force keeps a 1200-lb car from skidding on a curve of radius 400 ft. at 35 mph, what force would keep the same car from skidding on a curve of radius 850 ft. at 60 mph?
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Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! F=k*(1/R)*M*V^2
247=k*(1/400)*1200*35^2; (k can be different if we use ft/sec, but it will still give the same answer at the end)
247/3675=0.06721
F=0.06721*(1/850)*1200*60^2
F=341.59 pounds or 342 pounds.
Check. The radius increases by about 2.1 approximately but the speed increases by 1.7^2 or about 2.9 , so this should increase the force by approximately 1.38. The ratio is 1.38, which confirms it.
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