How much lift must be produced to lift a 190 pound person?
Well, the answer is anything
greater than 190 pounds of force, since that would mean the
force up is greater than the force down. So then that begs the
question, how do we get that force? Well, to
do that we need the both the velocity of the sledder, and the
area of the wings, where the force of lift is being applied.
So then, what is the velocity of the average sledder? Simple physics, that's how.
We can visualize the sledder going to down the hill as a block sliding down a ramp:
So then, what is the velocity of the average sledder? Simple physics, that's how.
We can visualize the sledder going to down the hill as a block sliding down a ramp:
where the mass is 86 kilograms, and the
angle theta, which is the slope of the hill, is roughly 25
degrees.
Not included in this picture is the force of friction, which is moving opposite to the direction of travel,
in this case, opposing mgsin(theta), or Fgx.
Frictional force, Fk, is calculated by taking the normal force, in this case Fgy, and multiplying it by the coefficient of friction.
The coefficient of friction in this case is kinetic, since the block (or sledder) is moving.
I am using the coefficient of friction of ice on steel, since I...uh, somebody...would be using a sled with metal runners.
the coefficient for steel on ice is .05, and that brings our frictional force to 38.30 N.
the resultant force, or Fgx - Fk, is approximately equal to 318.93 N.
Plugging this into the famous
F=ma
we reach an acceleration of 3.7 m/s.
I'm going to assume the average period of acceleration on the sledding hill is 5 second, before the hill levels off and you start to decelerate.
That means that an average sledder would reach speeds of 18 meters per second.
Or around 40 mph.
Damn. That's pretty fast.
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Not included in this picture is the force of friction, which is moving opposite to the direction of travel,
in this case, opposing mgsin(theta), or Fgx.
Frictional force, Fk, is calculated by taking the normal force, in this case Fgy, and multiplying it by the coefficient of friction.
The coefficient of friction in this case is kinetic, since the block (or sledder) is moving.
I am using the coefficient of friction of ice on steel, since I...uh, somebody...would be using a sled with metal runners.
the coefficient for steel on ice is .05, and that brings our frictional force to 38.30 N.
the resultant force, or Fgx - Fk, is approximately equal to 318.93 N.
Plugging this into the famous
F=ma
we reach an acceleration of 3.7 m/s.
I'm going to assume the average period of acceleration on the sledding hill is 5 second, before the hill levels off and you start to decelerate.
That means that an average sledder would reach speeds of 18 meters per second.
Or around 40 mph.
Damn. That's pretty fast.
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