Max Angle of Incline for Box Sliding w/ Friction on Ramps

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SUMMARY

The maximum angle of incline for a box sliding with friction on ramps is determined using static and kinetic friction coefficients. Given a coefficient of static friction of 0.7 and kinetic friction of 0.5, the correct approach involves applying Newton's second law and resolving forces into components. The final equation derived is tan(theta) = 0.7, leading to a maximum angle of approximately 34.99 degrees for static equilibrium. This analysis is essential for understanding the dynamics of objects on inclined planes.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to resolve forces into components on inclined planes
  • Familiarity with trigonometric functions (sine, cosine, tangent)
NEXT STEPS
  • Study the derivation of forces on inclined planes using free-body diagrams
  • Learn about the applications of static and kinetic friction in real-world scenarios
  • Explore advanced topics in dynamics, such as friction in non-inertial frames
  • Practice solving problems involving inclined planes and friction using various coefficients
USEFUL FOR

Students in physics courses, particularly those studying mechanics, as well as educators and tutors looking to enhance their understanding of friction and motion on inclined surfaces.

  • #31
the forces in the x direction are fgx and fst

9.8msin(theta) + (9.8m)cos(theta) * .7
 
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  • #32
fgx and fst are opposing forces, when you say "sum up the forces" you don't add the magnitudes. You have to set one direction to be positive and the other negative, so when they oppose forces one is negatively acting on the object and the other is positively acting on the object.
 
  • #33
this is sooo confusing! thank you soooooo much for helping me !

so would it be 9.8msin(theta) - (9.8m)cos(theta) * .7
 
  • #34
yes it would be, now that you know what the sum of forces is recall what they are equal to from post 19-20.
 
  • #35
9.8msin(theta) - (9.8m)cos(theta) * .7 = 0

9.8msin(theta) = (9.8m)cos(theta) * .7

9.8msin(theta) /(9.8m)cos(theta)= .7

sin(theta)/cos(theta) =.7

tan(theta) = .7
 
  • #36
yes sir, and now you have the answer. ;)
 
  • #37
thank you sooo much! you are the best ever!
 
  • #38
No problem.
 

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