Question

Consider a carwith front cross-sectional area of A=(1.77 ± 0.02)m^2 and a drag coefficient of C=(0.38±-0.02) moving with a speed of v=(10.0±0.5)m/s. Suppose that the density of air is p=1.20 kg/m^3 at the sea level. calculate the force F=(1/2)CApv^2 on the car due to air friction.

Answer 1

This is an expression derived from Bernoulli's equation, so use all the values and substitute in the given equation, making sure that the dimensions are compatible.

Check dimensions (C is dimensionless)
F=(1/2)CAρv^2
=m^2*kg/m^3*(m/s)^2
=m-kg/s^2
= same unit as Newton (force unit).

So force
F=(1/2)0.38(1.77)(1.2)10^2
= 40.356 N
=40.4N approx.

However, since tolerances are given, we need to calculate the limits.
Calculate once each with upper and lower limits of tolerances gives
F(max)=47.4 N
F(min)=34.1 N

So the expected force is 40.4N that could vary between 34.1 and 47.4N due to tolerances of the given data.