Answer:
μminimum = 0.2829
Explanation:
Draw a free body diagram of the rod.
Split all forces on the rod into their vertical and horizontal components.
Choose a pivot point so that the torque equation only has one unknown.
ΣFx = 0 (→) ⇒ N - Tx = 0 <em> (1)
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ΣFy = 0 (↑) ⇒ Ff + Ty - FL - FR = f + Ty - Wsign = 0 <em>(2)</em>
Στ = 0 ⇒ - Ff*(5 m) + FL*(4 m) = - Ff*(5 m) + (Wsign/2)*(4 m) = 0 <em>(3)</em>
[If you choose the axis of rotation at the end of the rod furthest away from the wall]
Trigonometry tells us that Ty/Tx = tan(23°) <em>(4) </em>
To get Ff use equation <em>(3)
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Ff = (2/5) Wsign
To get Ty use equation <em>(2)
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Ty = (3/5) Wsign
To get Tx use equation <em>(4)
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Tx = (3/5) Wsign / tan(23°)
To get N use equation <em>(1)</em>
N = (3/5) Wsign / tan(23°)
Since f ≤ μ N for static friction,
μminimum = Ff / N = (2/3) tan(23°) = 0.2829
The answer (in this particular case) does not depend on the weight of the sign.