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Feb7-13, 01:40 PM
i couducted an experiment trying to calculate the drag of a toy car going down a slope, i measured this at different distances 2.0m to 0.5m at 0.1m intervals along a 15° slope. i measured the time using light gates. i put one light gate at the end of the slope and one where the car started from e.g. 2.0m or 0.5m
how do i work out the drag?
results for 2m are 1.24s
i tried working out the actual average velocity and then the theoretical average velocity
i tried working out the actual veloity by using vectors and suvat
2cos15 = 1.93
2sin15 = 0.52
i worked out the vertical acceleration and also the horizontal acceleration
how do i work out the actual velocity and the theoretical velocity down the slope?
Feb8-13, 04:23 PM
There is a force pulling the car vertically downwards due to the gravitational attraction of earth.
This force can be split up into two components, a component pulling the car into the slope (perpendicularly to the slope), and a force pulling the car down the slope (parallel to the slope)
The total vertical force = mass of the car * 9.81 = F
so from this you can work out the components.
The maximum theoretical velocity assuming no friction or air resistance can be worked out using the force pulling the car parallel to the slope.
parallel acceleration = parallel F / M
then use SUVAT equations to find the velocity.
However the actual force pulling the car down parallel to the slope will be slightly less, because there will be a force in the opposite direction due to friction and air resistance.
so parallel acceleration = (parallel F - (air resistance force + friction force) )/ M
the friction force will equal a constant of friction between the tires and the slope multiplied by the perpendicular force pulling the car into the slope.
F(friction) = u * perpendicular force
it will be difficult to find u, as you now have two unknowns, the friction constant, and the force due to air ressistance, however you can work out what the net retardant force equals
net retardant force = (air resistance force + friction force)
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