Solving Force on a Hanging Sign

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Homework Help Overview

The problem involves analyzing the forces acting on a hanging sign with a specified mass. Participants are tasked with determining the tension in the wire supporting the sign, which is described as being rigidly attached to a wall.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the implications of the sign being rigidly attached and question the assumptions regarding the pivot point and the forces involved. Some discuss calculating the weight of the sign and using trigonometric functions, while others challenge the validity of the problem setup.

Discussion Status

The discussion is active, with participants raising concerns about the phrasing of the problem and the assumptions made regarding the sign's attachment. Some guidance has been offered regarding taking moments about the pivot point, but no consensus has been reached on the correct interpretation of the problem.

Contextual Notes

There are conflicting interpretations of the sign's attachment and its ability to pivot, which affects the calculations and assumptions about the forces acting on it. The original poster's calculations and the proposed answers are also under scrutiny.

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hanging sign forces??

Homework Statement



PIC IS ATTACHED

A businessman is hanging a sign outside his store in the
fashion shown in the figure below. The sign has a mass of
50 kg. For a sign of this mass, what is the tension in the
wire? The sign is rigidly attached to the wall at the lower
left corner.

A. 346 N
B. 260 N
C. 490 N
D. 87 N
E. 173 N



The Attempt at a Solution



i am stuck. only thing i though of was find mg then find the sin of the angle
so i did mg sin (45)
490 sin (45) got 370.
which is not even close to letter E (which is the right answer)
 

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If it's rigidly attached as it says in the question - then the tension in the wire is zero.
ie the sign would stay there without the wire

Normally in these question you assume the weight acts at the centre of the sign and take moments about the point that it joins the wall (assumed to pivot) to find the upward force at the outside end
 


The question is flawed.
Answer E is obtained if you assume the sign
can pivot freely about the bottom left corner.of the sign.
(hardly "attached rigidly"!)

Remembor the pivot can exert a force on the sign, but no torque.

David
 


davieddy said:
The question is flawed.
Answer E is obtained if you assume the sign
can pivot freely about the bottom left corner.of the sign.
(hardly "attached rigidly"!)

Remembor the pivot can exert a force on the sign, but no torque.

David


so if we assume the sign can pivot, how would you go about getting answer E??
 


Take moments about the pivot
 

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