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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 µ such that the sign will remain in place? ...?

Respuesta :

syed514
the reason I use the moments is for a problem to be in static equilibrium the moments about any point is zero the reason I take the moments about where the rod touches the wall is the unknown horizontal force and force of friction are produce zero moments at that point the only contributing forces are the tension in the guy wire and the weight acting down the weight can be considered a point acting at a distance 3 meters from the wall along the rod since the sigh is attached to the rod why i used y' is to reduce variables; the moment is the same y' * T = y T(vertical); just wanted to find total T in one step sum M about O = 0 = W*3m - T(y') solve for T then use Net Force X = 0 therefore R(horizontal) = T(horizontal) use Net Force Y = R(vertical) + T(vertical) - W = 0 the reaction horizontal is where the rod meets the wall and is the normal force or from a materials point of veiw where the rod is being compressed into the wall the reaction vertical is the force of friction along the end of the rod on the wall you need on final formula force of friction = µ * normal force( which is R(horizontal))

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