Solving Force on a Hanging Sign

  • Thread starter Thread starter fball558
  • Start date Start date
  • Tags Tags
    Forces Sign
Click For Summary
SUMMARY

The discussion centers on calculating the tension in a wire supporting a 50 kg sign, with the conclusion that the question is flawed due to the assumption of rigidity. The correct answer, 173 N (option E), is derived by assuming the sign can pivot freely at the lower left corner. Participants emphasize the importance of considering torque and moments in such problems, noting that the pivot can exert a force without generating torque.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of torque and moments in physics
  • Ability to calculate weight using the formula mg
  • Familiarity with the concept of pivot points in mechanics
NEXT STEPS
  • Study the principles of torque and how it affects static equilibrium
  • Learn about free-body diagrams and their application in solving mechanics problems
  • Investigate the concept of rigid body dynamics in physics
  • Explore advanced problems involving tension in cables and supports
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of problem-solving in static equilibrium scenarios.

fball558
Messages
143
Reaction score
0
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)
 

Attachments

  • sign.jpg
    sign.jpg
    13.3 KB · Views: 621
Physics news on Phys.org


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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
8
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 28 ·
Replies
28
Views
4K
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
4
Views
3K