Coefficient of static friction of uniform rod

AI Thread Summary
The discussion revolves around calculating the minimum coefficient of static friction for a uniform rod in static equilibrium, with one end against a rough wall and the other connected to a fixed point. The equations provided indicate that the tension in the cord and the normal force must balance the weight of the rod and any additional weight applied. The initial attempt at solving the problem led to an incorrect equation due to a sign error in the torque calculation. Correcting this error is crucial for finding the accurate minimum coefficient of static friction, which is expressed in terms of the rod's length, the distance from the pivot, and the angle with the cord. The final answers for the coefficient of static friction are μ≥tanθ and μ≥(2l-d)tanθ/(l+d).
maurice2705
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Homework Statement


A uniform rod AB has a length l and a weight W0. End A is in contact with a rough wall, but it is not fixed. A massless cord connects end B and is fixed to the wall at point C. The rod AB is now horizontal and the angle formed between the cord and rod is θ as shown in the figure. (a) In order to achieve static equilibrium, what is the minimum value of the coefficient of static friction between the rod and the wall? (b) If a block of weight W=W0/2 is hang on the rod with the hanging point a distance d from the end point A, what is then the minimum value of the coefficient of static friction between the rod and the wall? Express all your results in terms of l, d, W0, and θ. (ANS: μ≥tanθ ; μ≥(2l-d)tanθ/(l+d) )


Homework Equations


Which point should we regard as the pivot?


The Attempt at a Solution


Set A as the pivot, then
Tsinθ+μN=W0
Tcosθ=N
Tsinθ*l+W0*l/2=0
Then I get μ=-3tanθ. What's wrong?
 
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maurice2705 said:
... Tsinθ*l+W0*l/2=0
...What's wrong?

All correct except the + in the torque equation.
Try replacing the + by -.
 
Thank you! I'd try again
 
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