Coefficient of Friction of a wooden box

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Homework Help Overview

The discussion revolves around deriving equations related to the coefficients of friction (static and kinetic) for a wooden box being pulled up an inclined plane at a constant speed. Participants are exploring the relationships between normal force, frictional force, and the forces involved in the scenario.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive equations for the normal force (N) and frictional force (Ff) based on the forces acting on the box. There are questions regarding the correctness of these derivations and the relationships between the forces and the coefficients of friction.

Discussion Status

Some participants have provided insights into the relationships between the forces and the coefficients of friction, suggesting that the reading on the spring scale can be related to both static and kinetic friction. However, there is no explicit consensus on the correctness of the derived equations, and multiple interpretations of the relationships are being explored.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for deriving the equations. There is also a discussion about the angle of the force relative to the incline, which affects the calculations.

goj2
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1. If a wooden box was pulled up the inclined plane at a constant speed using a spring scale. How would an equation be derived for uk, us, Ff, and N.


Homework Equations


So to get N. The equation would be N=mg-F (sin (theta)).
with F being the reading on the spring scale.
Ff would be the opposite of the pulling force F, from what is read in the spring scale.
So when pulling upward (tan (thetha)) is not used to find uk or us.
So I need help because I don't know wat the equation would be for Uk or Us. And if I am deriving N or Ff the right way.

The Attempt at a Solution

 
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It appears that the only thing you need now is the relation between the force of friction Ff and uk (or us).
 
Is this right?

If F=ukN
with F being the reading on the spring scale when the box is being pulled at constant speed

F=usN
with F being the reading on the spring scale when the box starts to move.
 
goj2 said:
Is this right?

If F=ukN
with F being the reading on the spring scale when the box is being pulled at constant speed

F=usN
with F being the reading on the spring scale when the box starts to move.

Since you stated that the force makes an angle other than zero with the incline, then the force of friction (static/kinetic) equals F*cos(theta), where theta is the angle between the force and the incline, and F is the reading on the scale. Plug thta into the equations above, combine with the expression you got for N, and you should get your relation.
 

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