Thermodynamics question 1 (question about mass)

AI Thread Summary
The discussion focuses on calculating the power needed to overcome drag and rolling resistance forces using specific formulas. The user is confused about the units of force (lbf) and mass (lbm) and how to properly convert them for calculations. Key equations include the rolling resistance force (F_r) and drag force (F_d), with the total power (P) being the sum of these forces multiplied by velocity (V). The user seeks clarification on handling the "lbf" unit and confirms they are beginning to understand the problem. The conversation highlights the challenges of working with imperial units in thermodynamics calculations.
Saladsamurai
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Homework Statement



Alrighty then :smile:

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.

F_r=f\cdot W=Rolling resistance force
F_d=C_dA\frac{1}{2}\rho v^2=Drag force
Power=P=F\cdot V
Area=A=23.3 sq ft
C_d=.34
f=.02
\rho=.08\frac{lb}{ft^3}
v=55\frac{mi}{h}
weight=w=3550lbf <------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.

P=(\sum F)\cdot V

\Rightarrow P=(f*w+C_dA\frac{1}{2}\rho v^2)v

\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})I cannot tell what to do with that "lbf" ? I know that tyhis is really simple, but I am not used to these units :redface:

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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Don't worry guys. I think I got it :smile:
 
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