Question

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?

Answer 1

where does the wire attach?

Answer 2

a image would help geatly but the wire is most likely attached to the farside of the rod, opposite of the wall
take the moments about the end of the rod at the wall to find the tension in the guy wire; easier for me to use y' instead of y but you can do anyway you please.
Tension, T, in the guy wire is 15801.147003237769864697986040163 N for a static problem.
to that µ =[ weight of sign - T(vertical) ] / T(horizontal)
µ = 0.28298321080640316134902137529501 ( no units)

Answer 3

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))

Answer 4

CONCEPTUAL ANALYSIS
Category: Static Equilibrium.
The rod is static; there is no (angular or linear) acceleration.
The sum of the forces on the rod in any direction must be zero.
The sum of the torques about any point on the rod must be zero.
At first, we don't have any information about the forces from the strings attaching the rod to the sign, but we can figure them out by realizing that the sign is also in static equilibrium. Then you can conclude after doing the math that FL, the force from the left string and FR, the force from the right string are both equal to half the weight of the sign: FL = FR = Wsign/2.

STRATEGIC ANALYSIS
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.
QUANTITATIVE ANALYSIS
ΣFx = 0 Implies (1) N - Tx= 0.
ΣFy= 0 Implies (2) f + Ty - FL - FR= f + Ty - Wsign= 0.
Στ = 0 Implies (3) - f*(5 m) + FL*(4 m) = - f*(5 m) + (Wsign/2)*(4 m) = 0
[If you choose the axis of rotation at the end of the rod furthest away from the wall]
Trigonometry tells us that (4) Ty/Tx= tan(23°)
To get f use equation (3)
f = (2/5) Wsign
To get Ty use equation (2)
Ty = (3/5) Wsign
To get Tx use equation (4)
Tx = (3/5) Wsign / tan(23°)
To get N use equation (1) N = (3/5) Wsign / tan(23°)
Since f ≤ μ N for static friction, μminimum = f / N = (2/3) tan(23°)
The answer (in this particular case) doesn't depend on the weight of the sign!