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?
i got .227 but its the wrong answer?
IT is not clear to me where the sign is, so here are general directions.
Locate the center of gravity for the sign.
Take the wire attaching, and at the attachment point, break tension into a vertical component and a horizontal component.
Assign at the wall two forces, vertical (which is friction), and horizontal (which is the same as the horizontal component of wire tension at the other end.
Now sum vertical forces:
Wire vertical force+rod wall vertical force=sign weight
Now sum moments, about any point. I will choose the point at which the sign is attached;
wirevertical force*L = wallverticalforce* (5-L)
you will determine L from the diagram, which I don't understand.
You now have the equations needed to solve the question.
how would i find that? divide the mass by the weight of the wire?
do i need to do the x,y for the rod?
To determine the smallest value of the coefficient of friction (µ) required to hold the sign in place, we need to consider the forces acting on the sign.
1. Weight of the sign:
The weight of the sign can be determined using the formula: Weight = mass x acceleration due to gravity. Given that the mass of the sign is 1050 kg and acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the sign.
Weight = 1050 kg x 9.8 m/s^2
2. Tension in the guy wire:
The guy wire is holding the sign at an angle of 23° to the horizontal. We can calculate the component of tension acting in the horizontal direction.
Tension (horizontal) = Tension x cos(angle)
3. Normal force from the wall:
Since the wall is rough and the sign is in contact with it, there will be a normal force exerted by the wall on the sign, perpendicular to the wall. This normal force will counteract the weight of the sign to keep it in place.
4. Frictional force:
The frictional force acting between the sign and the wall will be equal to the product of the coefficient of friction (µ) and the normal force.
To find the smallest value of µ that will keep the sign in place, we need to find the maximum value of the frictional force. This occurs when the sign is on the verge of sliding down the wall.
Now, let's summarize the forces acting on the sign:
Vertical forces:
Weight of the sign (acting downward)
Normal force (acting upward)
Horizontal forces:
Tension in the guy wire (acting to the right)
Frictional force (acting to the left)
To find the smallest value of µ, we need to set up an equation based on the forces in the horizontal direction. The equation is as follows:
Tension (horizontal) - Frictional force = 0
Tension (horizontal) = Frictional force
Tension (horizontal) = µ x Normal force
Tension x cos(angle) = µ x weight of the sign
Now substitute the values into the equation:
Tension = µ x weight of the sign / cos(angle)
Finally, plug in the known values:
Tension = µ x (1050 kg x 9.8 m/s^2) / cos(23°)
This equation relates the tension in the guy wire to the coefficient of friction. By solving the equation for µ, you can find the smallest value required for the sign to remain in place.