Terminal Velocity of a bobsled

[muddyjch]

Homework Statement

A bobsled has a frontal area of 0.86 m^2 and a coefficient of drag of 0.267. The air density is 1.043 kg/m^3. Total mass of the sled and the bobsled driver and pushers is 827 kg. Friction in the runners is negligible. The track slopes down at an angle of 0.078 radians. What will be the terminal speed (m/s)?
m=827kg g=9.81m/s^2 p=1.043kg/m^3 Cd=0.267 A=0.86m^2

Homework Equations

Fg=mg
Fd=.5pCdAv^2

The Attempt at a Solution

Fg=Fd but i do not get the right answer. Am I missing something is the angle just to tell you that it is going downhill?

mg=.5pCdAv^2
827*9.81=.5*1.043*0.267*0.86*v^2
8112.87=.11974683v^2
v^2=67750.1859548
v=260.289
He wants us to leave all decimals until the final answer which he wants rounded to 3 places.

 

[willem2]

For something falling straight down you would have Fg = Fd at terminal velocity.

For a sled on a slope Fg and Fd do not have the same direction. Only the component of Fg parallel to the slope is equal to the drag.

 

[muddyjch]

So if i understand correctly Fg=mg sin 0.078rad. Thanks for your help!