Hanging Sign Equations: Finding Tension and Force Exerted by Beam

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In summary, the conversation involved a person seeking help with a problem involving the magnitude of tension and force exerted by a beam. After drawing a free body diagram and using the equations Fty = Ft(sinθ) and Ftx = Ft(cosθ), the person initially got an incorrect answer due to plugging in the wrong value for Ft. With the help of another person, they were able to identify and correct the mistake, leading to the correct answer.
  • #1
Olivia Carey
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upload_2015-11-18_3-27-35.png


Equations I used:
Fty = Ft(sinθ)
Ftx = Ft(cosθ)

My attempt:

I drew a free body diagram that looked like this (the red are just components of the tension, I know they wouldn't usually be included on a free body).
upload_2015-11-18_3-33-20.png


Finding Magnitude of Tension:
Fg = 516 (I gathered from the problem)
Fg = Fty
Fty = 516
Fty = Ft(sinθ)
516 = Ft (sin35)
Ft = 899.6

Finding Magnitude of Force Exerted By Beam:
Ftx = F
Ftx = Ft(cosθ)
F = Ft(cosθ)
F = 516 (cos35)
F = 422.7

I don't know where I went wrong, but I could really use some help figuring it out! Thanks!
 
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  • #2
Olivia Carey said:
View attachment 92005
Finding Magnitude of Force Exerted By Beam:
Ftx = F
Ftx = Ft(cosθ)
F = Ft(cosθ)
F = 516 (cos35)
F = 422.7

Are you aware that you can see, that your answer can't be correct as F < Fty? By just comparing the lengths of the vectors it is obvious, that there must be a mistake.

Regarding the mistake: Your formulas are correct, you just plugged in a wrong value for Ft.
 
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  • #3
stockzahn said:
Are you aware that you can see, that your answer can't be correct as F < Fty? By just comparing the lengths of the vectors it is obvious, that there must be a mistake.

Regarding the mistake: Your formulas are correct, you just plugged in a wrong value for Ft.

Thank you so much! I have no idea how I didn't catch that, but I guess that's why having a fresh pair of eyes always helps! I plugged in the correct value for Ft and got 736.9, which was correct.
 

Related to Hanging Sign Equations: Finding Tension and Force Exerted by Beam

1. What is a balanced force?

A balanced force is a situation in which two or more equal and opposite forces act on an object, causing it to remain at rest or to move at a constant velocity. In other words, the forces are balanced and cancel each other out, resulting in no overall change in the object's motion.

2. How is a hanging sign an example of balanced forces?

A hanging sign is an example of balanced forces because the sign is acted upon by two forces: the force of gravity pulling it downwards and the force of tension in the rope or string holding it up. These two forces are equal and opposite, resulting in a state of equilibrium where the sign remains stationary.

3. What is the significance of balanced forces for a hanging sign?

The significance of balanced forces for a hanging sign is that it allows the sign to remain in a stable position without falling or moving. This is important for the function and safety of the sign, as it ensures that it will not cause any damage or harm.

4. Can the forces acting on a hanging sign ever become unbalanced?

Yes, the forces acting on a hanging sign can become unbalanced if there is a change in the magnitude or direction of one of the forces. For example, if the wind starts blowing and creates a horizontal force on the sign, the forces will become unbalanced and the sign may start to swing or fall.

5. How can we calculate the forces acting on a hanging sign?

The forces acting on a hanging sign can be calculated using Newton's Second Law, which states that the net force on an object is equal to its mass multiplied by its acceleration. By measuring the mass of the sign and the acceleration due to gravity, we can determine the force of gravity acting on the sign. The force of tension can also be calculated using trigonometric functions if the angle and length of the rope or string are known.

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