How to handle string intersection in static friction calculation?

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Homework Help Overview

The problem involves a mass supported by strings in a static friction scenario, where participants are tasked with determining the minimum coefficient of static friction required for equilibrium. The context includes forces acting on both a larger mass and a smaller mass, with specific angles and tension considerations in play.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on both the larger and smaller blocks, including tension and friction. There are attempts to set up equations based on the forces in the x and y directions. Some participants question the role of tension and its components in the equations.

Discussion Status

There are various equations presented, and some participants confirm the correctness of each other's approaches. However, the discussion reflects ongoing exploration of the relationships between tension, friction, and the angles involved, without reaching a definitive conclusion.

Contextual Notes

Participants are navigating through the implications of string intersections and the angles involved in the tension forces, which may affect their calculations. There is a focus on ensuring equilibrium conditions are met, but some assumptions about the setup are still being questioned.

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Homework Statement



a mass m is supported as shown in the figure by ideal strings connected to a rigid wall and to tal mass 3m sitting on a horizontal surface. what is the minimum coefficient of static friction between the larger mass and the surface that permits the system to remain in equilibrium in the situation shown

Homework Equations





The Attempt at a Solution



Big block

Fx = u3mg - T = 0

Little block & strongs

Fx = T - mgsin60 = 0

T = us3mg - mgsin60


I don't think this is correct. How do u deal with the string intersection
 

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For big block Fx = μ*3mg - T*sin30 = 0
For smaller block Fy = ...?
 
Why is the tension have the sin 30 doesn't the tension come directly in the x plain

so is Fy = mg - Tcos30 = 0
 
Correct.
Eliminate T and find the value of μ.
 
u3mg -Tsin30 = 0
mg - t cos30 = 0 t = mg/cos30
u3mg - mgsin30(cos30) = 0
u = tan30/3
 
Correct.
 

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