Finding the Coefficient of Friction using Torque, and Forces

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SUMMARY

The discussion focuses on calculating the coefficient of friction required to prevent a 4.0 m long bar from slipping when supported by a string. Key equations used include τ = r*F*sinθ for torque and the principles of static equilibrium, where the sum of forces and torques equals zero. Participants analyze the forces acting on the bar, including the gravitational force and tension components, to determine the necessary frictional force at the pivot point. The discussion highlights the importance of identifying the correct pivot point and the components of forces involved in the calculation.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with torque calculations
  • Knowledge of force components in physics
  • Ability to analyze free-body diagrams
NEXT STEPS
  • Study the concept of torque in-depth, focusing on τ = r*F*sinθ
  • Learn about static equilibrium and its applications in physics problems
  • Explore free-body diagram techniques for analyzing forces
  • Research methods for calculating coefficients of friction in various scenarios
USEFUL FOR

Physics students, mechanical engineers, and anyone involved in analyzing forces and torques in static systems will benefit from this discussion.

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1.

A 4.0 m long bar is suppotred by a string as shown in Figure 10-61. What must be the coefficient of friction between the bar and the floor if the bar is on the verge of slipping

Image:
3ebcd17e8b532329dde12bc2534a4a32.png



2.

τ = r*F*sinθ or τ = r*F, where the force is perpendicular to the radius. Summation of forces; assuming that it, and the summation of torque is equal to 0 (static equilibrium).


3.

I attempted the problem by figuring out that the component of the Force of Gravity perpendicular to the bar, was needed for torque, as well as the component of Tension in the string, again, perpendicular to the bar. I am having trouble finding that component of Tension.

I also used summation of forces and figured out that in the y-direction, we had a normal force at the pivot point, the force of gravity acting on the centre of the bar, and the y-component of the tension force.

Essentially, I found the x-component of tension, and that it would be the only other force, including friction at the pivot point, in the x-direction.

The chosen pivot point was the point where the bar made contact with the ground.
 
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