Frictional Force of a Crate of Potatoes: 13 kg, 40° incline

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Homework Help Overview

The problem involves a crate of potatoes with a mass of 13.0 kg positioned on a 40° incline. The coefficients of static and kinetic friction are provided, and the task is to determine the frictional force acting on the crate while it is at rest.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of static friction and the conditions for equilibrium, questioning the assumption that static friction equals its maximum value. There is an exploration of the forces acting on the crate, including weight, friction, and normal force.

Discussion Status

Some participants have provided guidance on considering the equilibrium of forces and the need to analyze the components acting parallel and perpendicular to the incline. There is an ongoing exploration of the correct approach to determine the frictional force.

Contextual Notes

Participants are navigating the concepts of static friction and equilibrium, with some expressing confusion about the forces involved and how to apply the relevant equations.

lalahelp
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Homework Statement


A crate of potatoes of mass 13.0 kg is on a ramp with angle of incline 40° to the horizontal. The coefficients of friction are μs = 0.78 and μk = 0.43. Find the frictional force (magnitude) on the crate if the crate is at rest.



Homework Equations


F=ma
Ff=μmgcosα


The Attempt at a Solution



Ff=.78*13*9.81*cos40
=76.2 N


Why is my answer wrong?
 
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You assumed that the static friction would be equal to its maximum possible value (μN). Instead, solve for the value of static friction considering that the crate is in equilibrium.
 
What equation would I use to do that?
 
lalahelp said:
what equation would i use to do that?
The net force is zero.
 
I don't understand it still
 
lalahelp said:
I don't understand it still
What forces act on the crate?
 
the mass and friction
 
lalahelp said:
the mass and friction
The weight, the friction, and the normal force. What direction does each act?
 
the weights acts straight down, friction to the right, and normal force up
 
  • #10
lalahelp said:
the weights acts straight down, friction to the right, and normal force up
Consider force components parallel to the incline and apply ΣF = 0.
 
  • #11
Ok thank you I got it !
 

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