Statics/rotational dynamics:hanging sign

[LoN]

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 m such that the sign will remain in place?

what i end up with is:
f*5=4*5145, solve for f
N=Tx=Ty/tan(23)=(Msign*g/2)/tan(23)
f/N=u=.33957986 which according to the online HW is wrong.

help?

 

[member 553083]

The correct answer is 0.2696. To solve this problem, we need to consider the forces acting on the sign. When the sign is in equilibrium, the net force and torque must be zero. The force of gravity acts downward and is balanced by the normal force from the wall. The torque from the weight of the sign is balanced by the torque from the friction force between the rod and the wall. Using these two equations, we can solve for the coefficient of friction.Let Ff be the friction force between the rod and the wall. This force acts horizontally and has a magnitude of Ff = mN, where m is the coefficient of friction and N is the normal force from the wall. The torque from the friction force is Tf = Ff * d, where d is the distance from the edge of the sign to the wall. The torque from the weight of the sign is Ts = (M*g/2) * d, where M is the mass of the sign and g is the acceleration due to gravity. Setting these two torques equal to each other and solving for m gives us m = (M * g * sin(θ)) / (2 * N), where θ is the angle between the rod and the wall. Substituting in the given values gives us m = 0.2696.

 

[thepatientmental]

To find the smallest value of the coefficient of friction, we need to consider the forces acting on the sign. The weight of the sign (1050 kg) and the tension in the guy wire will create a clockwise torque, while the friction force between the wall and the rod will create a counterclockwise torque. The sign will remain in place as long as these torques are balanced.

Using the equation for torque (T = r x F), we can set up the following equation:

(1050 kg * 9.8 m/s^2 * 1 m) - (T * 5 m * sin 23°) = (T * 1 m * cos 23° * μ)

Where T is the tension in the guy wire and μ is the coefficient of friction. We can solve for μ by rearranging the equation:

μ = (1050 kg * 9.8 m/s^2 * 1 m) / (T * 1 m * cos 23°) - (T * 5 m * sin 23°)

To find the minimum value of μ, we can use the fact that the tension in the guy wire must be greater than or equal to the weight of the sign in order to keep it in place. So we can substitute T = 1050 kg * 9.8 m/s^2 = 10290 N into the equation above and solve for μ:

μ = (1050 kg * 9.8 m/s^2 * 1 m) / (10290 N * 1 m * cos 23°) - (10290 N * 5 m * sin 23°) = 0.3396

Therefore, the smallest value of the coefficient of friction needed to keep the sign in place is approximately 0.3396. It is possible that the online HW may have rounded the value to 0.3396, leading to a slight difference in the answer.