T: What is the Minimum Distance for No Slipping in Static Equilibrium?

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The discussion focuses on determining the minimum distance from the wall for a rod supported by a cable at a 37-degree angle to avoid slipping, given a coefficient of static friction of 0.5. The rod is 4 meters long and has an object of equal weight positioned at a distance xm from the wall. Key equations for static equilibrium are provided, including the sum of forces in the x and y directions, as well as the torque equation. Participants express uncertainty about how to incorporate the wall force and the distribution of weight on the rod. The conversation emphasizes the importance of understanding the normal force exerted by the rod against the wall to calculate static friction accurately.
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a rod of length 4m and weight Fg is supported at one end by a cable attached to a wall that makes an angle of 37 degrees with the rod. an object of equal weight is a distance of xm from the wall. the coefficient of static friction between the wall and the rod is .5 determine the minimum distance x from the wall for no slipping.

taking R as the wall force and T ans the tension in the cable

sum Fx = Rx-Tcos37=0
sum Fy = Ry+Tsin37-Fs-2Fg=0
sum torque = 4Tsin37-2Fg-xFg=0

not sure how to deal with the wall force
 
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nameVoid said:
a rod of length 4m and weight Fg is supported at one end by a cable attached to a wall that makes an angle of 37 degrees with the rod. an object of equal weight is a distance of xm from the wall. the coefficient of static friction between the wall and the rod is .5 determine the minimum distance x from the wall for no slipping.

taking R as the wall force and T ans the tension in the cable

sum Fx = Rx-Tcos37=0
sum Fy = Ry+Tsin37-Fs-2Fg=0
sum torque = 4Tsin37-2Fg-xFg=0

not sure how to deal with the wall force
I'm not sure where the weight is. Is it hanging from the rod?

Anyhow, the force of static friction depends on the normal force on the wall ie. into the wall. What is the force toward the wall exerted by the rod?

AM
 

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