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

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Homework Help Overview

The discussion revolves around a physics problem involving a box sliding on inclined ramps with friction. Participants are tasked with determining the maximum angle of incline at which the box will not slide back down, given specific coefficients of static and kinetic friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various trigonometric functions and equations related to friction, attempting to apply them to the problem. There are discussions about the relevance of static and kinetic friction equations, and participants express confusion about the correct approach to take.

Discussion Status

The conversation includes attempts to clarify the equations needed for static equilibrium and the forces acting on the box. Some participants provide guidance on breaking down forces into components and summing them correctly, while others express uncertainty about their understanding of the concepts involved.

Contextual Notes

Participants mention feeling overwhelmed by the material and express a desire for clarity on the relevant equations and concepts. There is a recognition of the need to work through the problem step by step, with some participants questioning their previous attempts and assumptions.

  • #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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