How can I find the critical mu of friction?

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Homework Help Overview

The discussion revolves around determining the critical coefficient of static friction (μ) for a box being pushed at an angle θ with a force F. Participants analyze the forces acting on the box, including friction and gravitational forces, while trying to reconcile their findings with a textbook answer.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore equations of motion and forces in both the x and y directions, questioning the relationship between the applied force, friction, and the angle of application. Some participants express confusion over the textbook's answer and seek clarification on the assumptions made in the problem setup.

Discussion Status

The discussion is ongoing, with participants sharing their equations and reasoning. Some have provided insights into the implications of the problem's conditions, such as the effect of increasing force on the coefficient of friction. There is no explicit consensus on the correct approach or answer yet.

Contextual Notes

Participants note the importance of understanding the problem's setup, including the direction of the applied force and the conditions under which the box remains stationary. The original problem statement from a textbook is referenced to clarify the context of the discussion.

  • #31
student34 said:
I see; cosθ ≥ μ*sinθ. But that is so beyond any kind of critical thinking that the textbook has ever expected in any other questions. This is very humbling.

You just have to read a problem text carefully.

The problem said
If μ(static) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of μ(static)

Assume that μ is given. The woman can just set the cart moving if her force is

F = (μ*m*g)/(cosθ - μ*sinθ).

If μ=1/tan(θ) the required force becomes infinite, the women can not exert infinite force no matter how hard she pushes. If μ is greater than that the equation can not be fulfilled by a positive F.

ehild
 
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  • #32
The answer is as follows: as the woman pushes harder and harder, the effect of gravity becomes less important overall, in proportion to her pushing, in deciding what happens to the box. As we approach this extreme point, or the limit, since mg <<< Fsinø, You can essentially take mg+Fsinø = Fsinø. Now reducing becomes simple algebra.

In other words, in the case that she is pushing infinitely hard, or somewhere near there, if mu_s is cotø, the box will not budge.
 
  • #33
PhysicsLord said:
mg+Fsinø = Fsinø
I assume that is not what you meant.
Anyway, the student seems to have arrived at the answer in post #27, three years ago.
 

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