What is the Correct Formula for Static Friction on an Inclined Box?

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Homework Help Overview

The discussion revolves around determining the correct expression for the frictional force acting on a box with mass "m" when a force "F" is applied at an angle "theta" above the horizontal. The box has a static friction coefficient "u," and participants are exploring the relationship between the applied force, normal force, and frictional force in the context of static equilibrium.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss two potential expressions for the frictional force, considering the role of the normal force and the balance of forces when the box remains stationary. There is a focus on understanding how the applied force's components interact with the frictional force.

Discussion Status

The conversation is ongoing, with participants actively questioning the assumptions regarding the normal force and its relationship to the applied force. Some guidance has been offered regarding the need to consider the vertical and horizontal components of the applied force, but no consensus has been reached on the correct formula.

Contextual Notes

Participants are grappling with the implications of the applied force's angle and its effect on the normal force, which complicates the determination of the frictional force. There is an acknowledgment that the normal force is not simply equal to the weight of the box due to the presence of the vertical component of the applied force.

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Homework Statement



On a box (of mass "m"), force "F" is applied at "theta" degree angle above the horizontal. The box has static friction coefficient of "u." When this force is applied, and the box remains stationary, what expression describes the frictional force?

Homework Equations



F=uN??
(N is normal force)

The Attempt at a Solution



I came down to 2 choices:
F=umg
or
F=uFcos(theta)

I initially thought the former, because I keep remember being drilled the equation F=uN, and about how it's all about the normal force when it comes to friction.

But then when I thought about it, forces must balance out... so if there's no net acceleration, applied force (i.e. Fcos(theta)) MUST balance out frictional force, no?

And now I'm just confusing myself.
 
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Right, yes the forces must balance out.

When you split the force into vertical and horizontal components what do you get?
 
right, so if the box isn't moving (yet), then for vertical, we just have normal force = gravity force (i think?), which is just mg=mg... then for horizontal, we have applied force (Fcos(theta))=frictional opposing force (u*F*cos(theta))...

i think i get the confusion. so the normal force equation is only relevant for *maximum* static frictional force equation, where that's the force at which the box moves b/c it overcomes static friction. yes?
 
yes but you see, your normal reaction is not just N=mg since the vertical component of the force Fy is present and depending how the force is applied, the normal reaction will change.
 

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