Loading presentation...

Present Remotely

Send the link below via email or IM


Present to your audience

Start remote presentation

  • Invited audience members will follow you as you navigate and present
  • People invited to a presentation do not need a Prezi account
  • This link expires 10 minutes after you close the presentation
  • A maximum of 30 users can follow your presentation
  • Learn more about this feature in our knowledge base article

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.


Make your likes visible on Facebook?

Connect your Facebook account to Prezi and let your likes appear on your timeline.
You can change this under Settings & Account at any time.

No, thanks

The Physics of Skiing

No description

luke andrews

on 6 October 2013

Comments (0)

Please log in to add your comment.

Report abuse

Transcript of The Physics of Skiing

Going down a hill on an angle The Physics of Skiing Transformation of Energy h Gravitational Potential Energy (PEgrav)(J) = mgh

(mass*gravitational field strength (9.8 N/kg)*height A skier's motion is the result of the transformation of the gravitational potential energy they a have as a result of their elevated position on the top of a ski hill into kinetic energy. Once a skier begins their descent they posses both kinetic and potential energy

Kinetic energy in joules can be calculated with the formula KE=1/2mv^2 Linear Downhill Motion There are four forces to consider when a skier is moving down a hill in a straight line:
-The force of gravity
-The normal force
-The force of kinetic friction
-The drag force The Force of Gravity The force of gravity can be broken down into two components, one in the x direction acting parallel to the slope and one in the y direction acting perpendicular to the slope The component acting parallel to the slope provides the downward acceleration of the skier The Normal Force The normal force is the contact force of the ground pushing upwards on the skies

The normal force is equal in magnitude but opposite in direction to the component of the gravitational force acting perpendicular to the slope

Fg(y)=Fg cosθ Fn=Fg cosθ The force of gravity can be found by multiplying the skier's mass by the acceleration due to gravity (9.8m/s^2 for near earth objects) The force of Kinetic Friction The force of kinetic friction acts in a direction opposite to the movement of the skier, it is the result of the resistance caused by the contact between the skies and the snow

The magnitude of the force of kinetic friction can be found through the formula Ffk=μkFn where μ is the coefficient of kinetic friction and Fn is the normal force.

The coefficient of kinetic friction between waxed skies and snow is typically around 0.05 Drag Force Drag force is the result of air resistance opposing the movement of the skier, thus it acts in the opposite direction to the skier's motion

Can be calculated by applying the formula FD=1/2 Cd ρAv^2
In this formula:

Cd is the coefficient of drag for the skier. It is not a constant but varies in response to changes in speed, and object size. For a skier it is typically in the range of 0.4 to 0.1

ρ is the mass density of the air, again this can vary depending on the altitude of the ski run, the humidity, and the temperature

A is the frontal area of the skier perpendicular to the skier’s velocity

v is the velocity of the skier Net Force Together, these equations allow the net force acting on a skier in linear downhill motion to be found: Understanding the forces In understanding how the forces act on a skier we first must change our perspective of the directions our XY axis.
The Y axis is perpendicular to the hill, in other words it is sticking out of the hill.
The X axis is perpendicular to the velocity of the skier. The equipotential line is parallel to the ground and imporatant to calculate the angels acting on the skier B is the angle from the skis to the equipotential line

V is the velocity of the turning skier

R is the radius of the turn

a is the slope of the hill

and g is acceleration due to gravity Ski Jumping - the sport skiing off an incline to try to get as far down the ski-hill as possible

- Involves two steps; the first is increasing the speed on the decline before the jump, and the second is creating lift during the jump to increase the distance traveled while airborne

- The first step is achieved by using an aforementioned method, crouching to reduce frontal projected area, so that the drag caused by air flow is reduced.

- the second step is for the skier to lean forward, reducing the frontal projected area of the body, while pointing the ski tips upwards and in a V shape, increasing their frontal projected area.

- the effect of the skier making a small angle with the direction of the air flow causes the drag force to split into components- one going opposite of the velocity of the skier and one going vertical- lift
- lift increases the amount of time that the skier can remain airborne, and thus the distance traveled down the ski hill Ski Turns - When making a turn a skier try to minimize the amount of snow resistance the acts upon the skis.

- They usually turn in one of two ways, the first being a long skid around the turn, and the second being a clean carve on the edge of the ski.

- When a turn is skidded the skier angles the skis so that the bottom of the skis hit the snow head on, an increase in snow resistance that results in a decreased speed

- carving is when the skier turns on an edge, and it requires a more delicate technique than skidding. The next step is the most important...
drawing the FBD Because of our XY axis setup, no movement occurs in the Y axis so the normal force and the Y component of gravity cancel each other out F lift Example if a = 20 degrees, B = 60 degrees, L= 11 m, s = 0.4m,
v = 5m/s, the sidecut radius = 14.8m, F1 = N1 = 0
the angel of tilt = 30 degrees

the skiers mass doesn"t need to be known because it cancels out in the equation.

If we sub these numbers in we can solve for the skiers angel of lean, which is 88.3 degrees Formulating the equation To solve this problem we can use Newton's second law, Fnet = Ma
From that we can derive the equation F1 + F2 - Mg sin a cos B = m a

We sub in V^2/R for a2 to get the equation F1 + F2 - Mg sin a cos b = m v^2/r

sin a cos b = sin o - L Cos O- L

since we are in rotational equilibrium the equation must equal to 0

We now create our final equation: (F1 + F2)sin 0 l - N1 cos 0 l - N2(Lcos O + s)= 0

Full transcript