Question

Sign SIGN BASE Hanging Sign PHYSICS IS FUN !! A sign has a mass of 1050 kg, a height h = 1 m, and a width W 4 m. It is held by a light rod of length 5 m that is perpendicular to a rough wall. A guy wire at 23° to the horizontal holds the sign to the wall. Note that the distance from the left edge of the sign to the wall is 1 m. Suppose we rely upon friction between the wall and the rod to hold up the sign (there is no hinge attaching the rod to the wall). What is the smallest value of the coefficient of friction u such that the sign will remain in place? 0.92 Submit Help 0.92 X Your submissions: Computed value: 0.92 Submitted: Thursday, November 21 at 7:23 PM Feedback: ch

Answer 1

Tsin23 RX Tcos223 3m PHYSICS IS FUN !! M

Take moment about O

$\sum M_{O}=0$

$Tsin23*5m-W*3m=0$

$Tsin23*5=W*3$

$T=\frac{W*3}{sin23*5}$

$T=\frac{mg*3}{sin23*5}$

$T=\frac{1050kg*9.81m/s^{2}*3}{sin23*5}$

${\color{Red} T=15817.27062N}$

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Consider all vertical forces

$\sum F_{y}=0$

$R_{y}+Tsin23-W=0$

$R_{y}=W-Tsin23$

$R_{y}=mg-Tsin23$

$R_{y}=1050kg*9.81m/s^{2}-15817.27062N*sin23$

${\color{Red} R_{y}=4120.2N}$

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Consider all horizontal forces

$F_{x}=0$

$R_{x}-Tcos23=0$

$R_{x}=Tcos23$

$R_{x}=15817.27062N*cos23$

${\color{Red} R_{x}=14559.87437N}$

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Frictional force = Coefficient of friction * Normal force

$f=\mu N$

$R_{y}=\mu *R_{x}$

$\frac{R_{y}}{R_{x}}=\mu$

$\frac{4120.2N}{14559.87437N}=\mu$

ANSWER: ${\color{Red} \mu=0.283}$

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