What Is the Coefficient of Kinetic Friction in Brenda's Wallpaper Project?

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SUMMARY

The coefficient of kinetic friction for Brenda's wallpaper project is calculated using the forces acting on a sanding block weighing 2.0N, which she pushes with a force of 3.0N at a 30-degree angle to the vertical. The normal force (Fn) acting on the block is 2.0N, and the y-component of the applied force is calculated as 2.6N. The resulting coefficient of kinetic friction (Uk) is derived from the equation Uk = F/N, yielding a value of 1.30. This value is incorrect as it exceeds the maximum possible coefficient of friction, indicating a miscalculation in the interpretation of forces.

PREREQUISITES
  • Understanding of free body diagrams (FBD)
  • Knowledge of forces and components in physics
  • Familiarity with the equation for kinetic friction (F = UkN)
  • Basic trigonometry, specifically cosine functions
NEXT STEPS
  • Review the concept of normal force in vertical surfaces
  • Learn about the differences between static and kinetic friction coefficients
  • Practice drawing and analyzing free body diagrams for various scenarios
  • Explore the implications of force angles on friction calculations
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Students studying physics, particularly those focusing on mechanics and friction, as well as educators looking for practical examples of force analysis in real-world applications.

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Homework Statement


Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on the surface. The sanding block weighs 2.0N and Brenda pushes on it with a force of 3.0N at an angle of 30.0 degrees with respect to the vertical, and angled toward the wall. Draw a free body diagram for the sanding block as it moves straight up the wall at a constant speed. What is the coefficient of kinetic friction between the wall and the block?

Fn = 2.0N
Theta = 30.0
Fhand = 3.0N


Homework Equations


F = UkN
y-component = ycos(theta)


The Attempt at a Solution


I drew an FBD of a block on a vertical wall with Fn pointing in the downwards direction. I extended the top part of the block a little with a dotted line to show the continuation of the y-axis, and drew a line that represented the magnitude of the hand pushing at 30degrees.

My first attempt I tried 3.0cos(30.0) (because I reasoned that if you find the y-component to the force of the hand you would find the F needed for the equation F = UkN)
I got 2.6N from this, and reworked F = UkN so it would be Uk = F/N
I plugged in Uk = 2.6N/2.0N (Normal force of the block), and got 1.30 as an answer.

This answer seems a little large to me though, considering it's the coefficient of Static Friction. So I'm just wondering where I might have gone wrong?
 
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The wall is vertical so the normal force is horizontal and not equal to the weight of the block, which points downward.

A good drawing always helps. Can you show yours?

ehild
 

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