What Is the Terminal Speed of a Bobsled on a Slope?

AI Thread Summary
To calculate the terminal speed of a bobsled on a slope, the forces acting on the sled must be considered, specifically the gravitational force and drag force. The gravitational force acting down the slope is given by Fg = mg sin(θ), where θ is the slope angle. The drag force is calculated using Fd = 0.5 * p * Cd * A * v^2. The correct approach involves setting the component of gravitational force parallel to the slope equal to the drag force, leading to the equation mg sin(θ) = 0.5 * p * Cd * A * v^2. Understanding that the angle indicates the slope's effect on gravitational force is crucial for solving the problem accurately.
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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.
 
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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.
 
So if i understand correctly Fg=mg sin 0.078rad. Thanks for your help!
 
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