Calculate the contact force between the skis and the track

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SUMMARY

The discussion focuses on calculating the contact force between skis and a parabolic track for a 75-kg ski jumper, using the equation of the parabola, y = (12/225)x^2. The jumper's gravitational potential energy at point A is expressed as Vg_A = mg(12), while at point O, the velocity is determined using T_O = 1/2mv_O^2. Participants emphasize the need for additional equations to derive the velocity at point O and subsequently calculate the contact force accurately.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy equations
  • Familiarity with parabolic equations and their applications in physics
  • Knowledge of Newton's laws of motion
  • Basic skills in algebra and solving equations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion on curved tracks, particularly in the context of sports physics and engineering applications.

Alexanddros81
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Homework Statement


14.37 A 75-kg ski jumper starts his run in the position shown on the parabolic
track. The equation of the parabola is ##y = (12/225)x^2##, where x and y are in meters.
Neglecting friction, calculate the contact force between the skis and the track at O.

Fig P14_37.jpg


Homework Equations

The Attempt at a Solution


[/B]
At pont A (start of run):

##Vg_A=mg(12)##
##T_A=0##

At pont O:

##Vg_O = 0##
##T_O=1/2mv_O^2##

Something is missing?
Any hints how to continue?
 

Attachments

  • Fig P14_37.jpg
    Fig P14_37.jpg
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Alexanddros81 said:
Something is missing?
Indeed ! Some relevant equations featuring the variables and given information !
 
Alexanddros81 said:
how to continue?
So can you find the velocity at O?
What else do you need to know to figure out the force there?
 

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