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? μ =

Answer 1

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      (1)

ΣFy = 0    (↑)  ⇒   Ff + Ty - FL - FR = f + Ty - Wsign = 0     (2)

Στ = 0    ⇒     - Ff*(5 m) + FL*(4 m) = - Ff*(5 m) + (Wsign/2)*(4 m) = 0      (3)

[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°)      (4)

To get Ff use equation (3)

Ff = (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 = Ff / N = (2/3) tan(23°)  = 0.2829

The answer (in this particular case) does not depend on the weight of the sign.