At What Angle Will This Box Begin to Slide?

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SUMMARY

The discussion centers on determining the angle θ at which a box begins to slide down an inclined plane, specifically using the equation ΣFx = max. The participants clarify that the frictional force provided in the problem should be used directly, rather than substituting it with the product of the coefficient of friction (μ) and the normal force (F_N). The calculations lead to the conclusion that θ equals 30°, derived from the equation 981sinθ = 490.5. The conversation highlights the importance of accurately interpreting given forces in physics problems.

PREREQUISITES
  • Understanding of Newton's Second Law (ΣFx = max)
  • Knowledge of free body diagrams
  • Familiarity with trigonometric functions, specifically sine
  • Concept of frictional force in physics
NEXT STEPS
  • Study the relationship between normal force and friction in inclined planes
  • Learn about static and kinetic friction coefficients
  • Explore advanced applications of free body diagrams in complex systems
  • Investigate the effects of varying angles on frictional forces
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators looking for examples of problem-solving in inclined plane scenarios.

x2017
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Homework Statement



IMG_9759.jpg

Homework Equations


ΣFx=max

The Attempt at a Solution


IMG_9760.JPG


I drew a free body diagram for the box and set up the equation to solve for θ, but I'm not sure how to do it without μ
 
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I don't think you need to substitute
##F_{friction}## with ##μ F_N##
They want you to use the frictional force they themselves have provided in the question.
 
Suraj M said:
I don't think you need to substitute
##F_{friction}## with ##μ F_N##
They want you to use the frictional force they themselves have provided in the question.

Oh my goodness I can't believe I missed that... Thank you!

ΣFx=max
mgx-Ffx=max
mgsinθ-Ffx=max
(100)(9.81)sinθ-490.5=(100)(0)
981sinθ=490.5
θ=sin-1(490.5/981)
θ=30°

Wow, that was an embarrassing mistake!
 
x2017 said:
Oh my goodness I can't believe I missed that... Thank you!

ΣFx=max
mgx-Ffx=max
mgsinθ-Ffx=max
(100)(9.81)sinθ-490.5=(100)(0)
981sinθ=490.5
θ=sin-1(490.5/981)
θ=30°

Wow, that was an embarrassing mistake!
In a way, your mistake is one that is reasonable to make.

In an actual situation, the friction force does depend upon the normal force, which depends upon the angle the surface make with the horizontal.

Having the friction force be a fixed value, is not realistic.
 

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