- #1
Saladsamurai
- 3,020
- 7
Homework Statement
Alrighty then
I still cannot get used to texts that use lbf (lb force) and lbm (lb mass) and just plain old lb.
Metric is bomb.
I am trying to find the power needed too overcome drag and rolling resistance forces from this given info.
[itex]F_r=f\cdot W[/itex]=Rolling resistance force
[itex]F_d=C_dA\frac{1}{2}\rho v^2[/itex]=Drag force
Power=[itex]P=F\cdot V[/itex]
Area=A=23.3 sq ft
[itex]C_d=.34[/itex]
[itex]f=.02[/itex]
[itex]\rho=.08\frac{lb}{ft^3}[/itex]
[itex]v=55\frac{mi}{h}[/itex]
weight=[itex]w=3550lbf[/itex] <------how do I convert this term properly? I think I NEED TO
So, if P=F*V Total power is the sum of this product.
[tex]P=(\sum F)\cdot V[/tex]
[tex]\Rightarrow P=(f*w+C_dA\frac{1}{2}\rho v^2)v[/tex]
[tex]\Rightarrow P=(.02*3550lbf+.34*23.3ft^2*\frac{1}{2}*.08\frac{lb}{ft^3}*55^2\frac{mi^2}{h^2}*5280^2\frac{ft^2}{mi^2}*\frac{h^2}{3600^2s^s})*(55\frac{mi}{h}*\frac{5280ft}{mi}*\frac{h}{3600s})[/tex]I cannot tell what to do with that "lbf" ? I know that tyhis is really simple, but I am not used to these units
Little help? Thanks!EDIT: Is a lbf just a lbm*ft/s^2 ? I am assuming that density is given in lbm even though it does not specify...
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