What is the Normal Force in Terms of Mass on a Ski Jump Training Hill?

  • Thread starter Thread starter Uiik
  • Start date Start date
  • Tags Tags
    Force Normal
Click For Summary
SUMMARY

The discussion centers on calculating the normal force on a skier at various positions on a ski jump training hill, specifically in terms of mass. The problem involves applying Newton's equations, with key parameters including a height change (Δh) of -100m and a radius of curvature (R) of 40m. The skier's speed is derived using the formula v² = -2gΔh, resulting in a calculated speed of 44.27 m/s. The resolution clarifies that the normal force can be expressed in terms of mass, eliminating the need for a numerical value.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with circular motion equations
  • Knowledge of gravitational force calculations
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of normal force in circular motion scenarios
  • Learn about the relationship between mass and gravitational force
  • Explore angular motion equations and their applications
  • Investigate the effects of friction and air resistance on motion
USEFUL FOR

Students in physics, particularly those studying mechanics, as well as educators looking to clarify concepts related to forces and motion in practical scenarios.

Uiik
Messages
3
Reaction score
0

Homework Statement



A photo of this problem is available at: http://imgur.com/l5lnI56
(the visual is needed to understand the question)

Problem text: A ski jump training hill can be thought of as featuring a near vertical plunge followed by a gradually decreasing (in magnitude) slope until it completely levels as shown in Fig. 1. A circle has been drawn to indicate the radius of curvature of the curve part way down the hill. Take Δh = -100m and R = 40m. Neglecting air resistance and friction, please answer the following question:

Write down Newton's equations and find the normal force at each position. After falling through a height Δh the speed of the skier (squared) is v2=-2gΔh


Homework Equations



Fnet = ma
ω = v/r
α = at/r
ar = v2/r = ω2r
Fc = mv2/r


The Attempt at a Solution



ƩFx, i = 0
ƩFy, i = -Fg
--> since any contact between the object and the surface is negligible at i, there is no normal force.

ƩFx, ii = Fg(sinθ)
ƩFy, ii = N + Fc - Fgcosθ = maii

ƩFx, iii = 0
ƩFy, iii = N - Fg = 0
N = Fg

So at this point, there is a bit of a problem. When simplifying forces at points ii and iii, it becomes clear that mass is needed.
e.g.
N = Fg
N = mg -> what is m?

I can only assume that the answer comes from use of angular motion equations.

At this point, I have the following information:
radius (40m)
height (-100m) --> used to find velocity using given formula --> velocity
angle is 45 degrees.

I calculated v to be 44.27m/s (using given velocity formula and height)
I calculated ω to be 1.10677rad/s (using the given velocity formula and ω=v/r)
I calculated ar to be 49m/s2 (using v2/r)

I know that (defining x as the tangent to the circle, and y as toward the middle of the circle) in the y direction, the forces acting are (N + Fc) - 9.8mcos45. In the x direction, only 9.8msin45 is acting (force due to gravity down the incline at the point of contact).

Other pieces of important that seem important to the solution can't be calculated without m.
at = 9.8sin45
α = 9.8msin45/40 = 0.245msin45

At this point, I'm not sure what other steps to take to find mass, or if mass can be canceled out in some sort of a proportional reasoning setup. I would appreciate a suggestion to help me figure out the rest. I'm not looking for a step-by-step solution, or even an answer.

If any part of my post is unclear/inappropriately stated, let me know.
 
Physics news on Phys.org
Perhaps I'm wrong but I can't see a way to calculate a value for the normal force without knowing the mass either. I haven't checked your working.
 
This is resolved. Thanks to CWatters.

Professor forgot to mention that he wasn't expecting a numerical answer, but instead, he wanted us to express the normal force in terms of mass (no need to find the magnitude of the mass). That made this a 5-minute question instead of a many-hour question.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
7K
Replies
17
Views
2K
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
6
Views
2K