Statics: Crate Problem - Solving for N = 132N

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Discussion Overview

The discussion revolves around a statics problem involving a crate, focusing on the calculation of forces, specifically the applied force in the horizontal direction and the frictional force. Participants explore the relationships between static and kinetic friction, as well as the conditions under which these forces apply.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant calculates a force of 132 N based on a given equation involving weights.
  • Another participant questions what the applied force in the horizontal direction is, suggesting confusion about whether the calculated force is frictional or applied.
  • A participant clarifies that the frictional force is dependent on the applied perpendicular force until a maximum value is reached.
  • There is a discussion about the conditions for static versus kinetic friction, with references to Newton's first law and the implications of the box being at rest.
  • One participant concludes that if there is no applied force, the frictional force would be zero.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the definitions and calculations of applied and frictional forces, indicating that multiple views remain on the interpretation of the problem and the relationships between the forces.

Contextual Notes

Participants rely on assumptions about static and kinetic friction coefficients and the conditions of the crate's motion, which may not be fully defined in the discussion.

sandmanvgc
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Homework Statement
Answer key say answer is d. I took combined weight and vertical component of 200 force to calculate frictional force would be 132 N. Am I missing something?
Relevant Equations
##F_f = uF_n##
(4/5)200 + 500 = 660
660(0.2) = 132 N
 

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But what’s the applied force in the horizontal direction
 
PhanthomJay said:
But what’s the applied force in the horizontal direction
120. Is that what they were asking for? Is frictional force not what I calculated above?

EDIT: 120 is the answer but I thought that would've been the applied force and not the Frictional force.

Or is the frictional force just equal to whatever the applied perpendicular force is until max possible value is reached?
 
Yes, for static friction, the friction force is less than or equal to uN, from equilibrium equation Newton’s first law. If the box is moving , then kinetic friction applies and the friction force is equal to uN,
Suppose the box was just resting there and there was no applied force at all. Using u =0.2, what would be the friction force for that situation?
 
PhanthomJay said:
Yes, for static friction, the friction force is less than or equal to uN, from equilibrium equation Newton’s first law. If the box is moving , then kinetic friction applies and the friction force is equal to uN,
Suppose the box was just resting there and there was no applied force at all. Using u =0.2, what would be the friction force for that situation?
So it would be 0
 

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