Question 56706
f(n)=14*10^0.70n. 
What is the frictional force when the number of turns is: 0.5, 1, 3?
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f(0.5)=14*10^(35)=1.4*10^36 lbs. of frictional force
f(1)=14*10^0.7=1.4*10^1.7 = 1.4(10^1.7)=70.166... lbs. of frictional force
f(3)=14*10^(2.1)= 14*125.8925...=1762.4955... lbs. of frictional forch.
Cheers,
Stan H.