At What Angle Will This Box Begin to Slide?

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Homework Help Overview

The discussion revolves around determining the angle at which a box will begin to slide down an inclined plane, focusing on the forces acting on the box, including friction. The subject area is physics, specifically dynamics and frictional forces.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of equations based on free body diagrams and question the necessity of substituting the frictional force with a coefficient of friction. There is an exploration of using a provided frictional force instead.

Discussion Status

Some participants have offered guidance regarding the use of the frictional force provided in the problem statement, while others reflect on their own misunderstandings. The discussion appears to be productive, with participants engaging in clarifying assumptions and exploring the implications of fixed versus variable frictional forces.

Contextual Notes

There is mention of a specific frictional force value given in the problem, which may influence the interpretation of the problem setup. Participants also acknowledge the realism of frictional force depending on the normal force, which varies with the angle of inclination.

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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