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?
4 answers
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.
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?