Holding a Cone Up: Minimum Force & θ Explained

  • Thread starter Thread starter physicsguy10
  • Start date Start date
  • Tags Tags
    Cone
Click For Summary
SUMMARY

The discussion focuses on calculating the minimum normal force required to hold an upside-down cone using two fingers, with the mass of the cone denoted as m and the static coefficient of friction as u. The angle at the tip of the cone is represented as 2θ. The user derived the relationship between the friction force and the normal force, concluding that the applied force F must exceed (mg/2)(cosθ/u - sinθ) to maintain equilibrium. This analysis is crucial for understanding the mechanics of static friction in inclined scenarios.

PREREQUISITES
  • Understanding of static friction and its coefficient (u)
  • Basic knowledge of forces, including gravity (mg)
  • Familiarity with trigonometric functions (sinθ, cosθ)
  • Ability to manipulate algebraic equations for force analysis
NEXT STEPS
  • Study the principles of static equilibrium in physics
  • Explore advanced applications of friction in inclined planes
  • Learn about the derivation of forces in two-dimensional motion
  • Investigate the role of normal force in various mechanical systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as educators seeking to enhance their understanding of frictional forces in practical applications.

physicsguy10
Messages
4
Reaction score
0

Homework Statement



With two fingers, you hold an cone motionless, upside down. The mass of the cone is m, and the static coefficient u. The angle of the tip, when viewed from the side, is 2θ. What is the minimum normal force required to hold the cone up (with each finger)? And, in terms of u, what is the minimum value of θ that allows you to hold up the cone?

Homework Equations



Friction force = uN
Gravity = mg
etc...

The Attempt at a Solution



I included a terrible paint drawing of my progress so far:

cone.jpg


The small "f" denotes the friction force, and i have split my mg force into mgsinθ and mgcosθ. Considering just one side, for the cone to remain motionless, I would assume that the friction force f must = (mgcosθ)/2. and f itself = uN. Also, N = F+mgsinθ/2, where F is the applied force from the one finger.

So now i have u(F+mgsinθ/2) = (mgcosθ)/2. Does this make sense? And if so, how do I find the minimum normal force that I need to apply with each finger?
 
Last edited:
Physics news on Phys.org
I expanded, with uN > f, that the finger force F must be at a minimum:

F > (mg/2)(cosθ/u - sinθ)

I would absolutely be indebted to anyone who would take the time to verify this.
 
Last edited:
Shameless bump :(
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 36 ·
2
Replies
36
Views
6K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
1
Views
1K