Effort-Saving Angle for Pulling Sled on Slope: Challenging Physics Homework

  • Thread starter Thread starter jgens
  • Start date Start date
  • Tags Tags
    Physics
Click For Summary
SUMMARY

The discussion centers on a physics homework problem involving a boy pulling a sled up a slope at an angle θ with a coefficient of friction of 0.10. The goal is to determine the optimal angle φ from the slope that minimizes the effort exerted by the boy. The solution involves setting up equations for the forces acting on the sled, leading to the derived formula φ = tan^(-1)[(μcot(θ) - 1)/2μ]. This formula provides a definitive method for calculating the angle that results in the least effort when pulling the sled.

PREREQUISITES
  • Understanding of basic physics concepts such as forces and friction
  • Familiarity with trigonometric functions and their applications in physics
  • Knowledge of algebraic manipulation for solving equations
  • Experience with vector resolution in two dimensions
NEXT STEPS
  • Study the principles of static and kinetic friction in physics
  • Learn about vector decomposition and its applications in force analysis
  • Explore advanced trigonometric identities and their relevance in physics problems
  • Investigate real-world applications of pulling forces on inclined planes
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the mechanics of forces acting on objects on inclined planes, particularly in relation to minimizing effort in physical tasks.

jgens
Gold Member
Messages
1,575
Reaction score
50

Homework Statement


A boy pulls his sled up a snowy slope of angle θ from the horizontal. If the coefficient of friction between the slope and sled is 0.10, at what angle, φ, from the slope should the boy pull the sled so that he exerts the least effort?


Homework Equations



N/A

The Attempt at a Solution



Well, I don’t think this is the correct approach to the problem but here it goes:

∑Fy = Fn + Fpsin(φ) - mgcos(θ) = 0 and ∑Fx = Fpcos(φ) - mgsin(θ) - μFn = 0 which, through some algebra, Fp = (mgcos(θ) - Fn)/sin(φ) = (mgsin(θ) + μFn)/cos(φ) and dividing the two equations results with 1 = tan(φ)(mgsin(θ) + μFn)/(mgcos(θ) - Fn) ⇒ cot(φ) = (mg sin(θ) + μFn)/(mgcos(θ) - Fn). Also, since the net force acting one the object is zero, the net work is zero as well; therefore, mgh = μFn*s or similarly mgsin(θ) = μFn. Applying this relationship I find cot(φ) = (2μFn)/(μFncot(θ) - Fn) = 2μ/(μcot(θ) - 1) which ultimately yields φ = cot^(-1)[2μ/(μcot(θ) - 1)] = tan^(-1)[(μcot(θ) - 1)/2μ].

Thanks.
 
Physics news on Phys.org
Homework Helpers: If I were to clarify my work, would that be beneficial perhaps?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
6
Views
4K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K