A truck having a projected area of 6.5 square metres travelling at 70 km per hour has a total resistance of 2000N. Of this 20% is due to rolling friction and 10% is due to surface friction. The rest is due to form drag. Calculate the coefficient of drag. Take ρ=1.25kg/m3ρ=1.25kg/m3 of air.

 A truck having a projected area of 6.5 square metres travelling at 70 km per hour has a total resistance of 2000N. Of this 20% is due to rolling friction and 10% is due to surface friction. The rest is due to form drag. Calculate the coefficient of drag. Take 

ρ=1.25kg/m3$\rho =1.25kg/m^3$ of air.

Solution:- 

Given: -

Area of truck, A=6.5m2$A=6.5m^2$

Speed of truck, V=70km/hr=70×100060×60=19.44m/s$V=70km/hr={\displaystyle \frac{70\times 1000}{60\times 60}}=19.44m/s$

Total resistance, FT=200N$F_T=200N$

Rolling friction resistance 20% of total force, i.e.,

FC=20100×2000=400N$F_C={\displaystyle \frac{20}{100}}\times 2000=400N$

Surface friction resistance is 10% of total force, i.e.,

FS=10100×2000=200N$F_S={\displaystyle \frac{10}{100}}\times 2000=200N$

Therefore, form drag, FD=200−FC−FS=2000−400−200=1400N$F_D=200-F_C-F_S=2000-400-200=1400N$

Where if FD$F_D$ is the form drag,

Then CD$C_D$ is the coefficient of form drag

1400=CD×6.5×1.25×19.4422$1400=C_D\times 6.5\times 1.25\times {\displaystyle \frac{{19.44}^2}{2}}$

∴CD=1400×26.5×1.25×19.44×19.44$\therefore C_D={\displaystyle \frac{1400\times 2}{6.5\times 1.25\times 19.44\times 19.44}}$

CD=0.912