# Statics/rotational dynamics:hanging sign

• LoN
In summary, we can determine the coefficient of friction between the rod and the wall by setting the torques from the weight of the sign and the friction force equal to each other and solving for m. The correct answer is 0.2696.
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?

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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.

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.

## What is statics?

Statics is a branch of mechanics that deals with objects at rest or in a state of constant motion. It involves the analysis of forces and moments acting on objects to determine their equilibrium.

## What is rotational dynamics?

Rotational dynamics is a branch of mechanics that deals with the movement and behavior of objects that rotate around an axis. It involves the analysis of forces and torques to determine the motion of rotating objects.

## What is a hanging sign?

A hanging sign is a type of sign that is attached to a pole or structure by a hanging mechanism, such as chains or ropes. It is commonly used for advertising or directional purposes and is subject to forces and moments due to its weight and wind resistance.

## How do you calculate the forces on a hanging sign?

To calculate the forces on a hanging sign, you must first determine the weight of the sign. Then, you can use the principles of statics to analyze the forces and moments acting on the sign, such as tension in the hanging mechanism and wind forces. These forces can be calculated using equations and diagrams based on the sign's geometry and the principles of equilibrium.

## How does the shape of a hanging sign affect its stability?

The shape of a hanging sign can greatly affect its stability. A sign with a wider base and a lower center of gravity will be more stable than a tall and narrow sign. Additionally, a symmetrical sign will experience more equal forces and moments, making it more stable than a sign with an irregular shape. The weight distribution of the sign also plays a significant role in its stability.

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