Engineering Tension force in two strings supporting a sign

[HareJare]

Homework Statement

A sign have the mass 4kg. It will be hung up on a wire that can take a traction force of 35N
How long must the distance h be? (more info in picture)

Relevant Equations

F = mg

First i calculated the sign Force which was 4*9.81 = 39.24N
which meant that the force already exceeds the traction force.
What i don t understand in the question is how the distance h will avoid this?

 

[berkeman]

Homework Statement:: A sign have the mass 4kg. It will be hung up on a wire that can take a traction force of 35N
How long must the distance h be? (more info in picture)
Relevant Equations:: F = mg

First i calculated the sign Force which was 4*9.81 = 39.24N
which meant that the force already exceeds the traction force.
What i don t understand in the question is how the distance h will avoid this?

There are essentially two wires sharing that load, so you are not over the limit if the wires are very long so they are almost vertical.

As the wires are made shorter and shorter, the tension (what you translated as "traction") goes up. Can you write the equation for how much it goes up with the increasing horizontal angles?

 

[Lnewqban]

Have you studied addition of vectors?

If not, please, see:
https://roperescuetraining.com/physics_angles.php

 

[HareJare]

There are essentially two wires sharing that load, so you are not over the limit if the wires are very long so they are almost vertical.

As the wires are made shorter and shorter, the tension (what you translated as "traction") goes up. Can you write the equation for how much it goes up with the increasing horizontal angles?

Thank your for your reply but i don't really understand yet

Are you saying that when the force F1x is increased the wire will eventually break?
And F1 will always stay the same right? its just the F1y and F1x that will change?

 

[berkeman]

Are you saying that when the force F1x is increased the wire will eventually break?
And F1 will always stay the same right? its just the F1y and F1x that will change?

The vertical forces will stay the same, and they will each be half of the weight of the sign. The vertical forces and the angles of the wires tell you what the total forces (tensions) are. The total tension in the wires is what will cause them to break if the tension force along the length of each wire gets too big.

 

[HareJare]

The vertical forces will stay the same, and they will each be half of the weight of the sign. The vertical forces and the angles of the wires tell you what the total forces (tensions) are. The total tension in the wires is what will cause them to break if the tension force along the length of each wire gets too big.

Okay i think i got it now thank you!

 

[berkeman]

Great. Can you show us the equation for the tension force in each wire as a function of the angle now?

 

[HareJare]

sin(x) = F1y/F1

 

[berkeman]

sin(x) = F1y/F1

I'd recommend using θ for the angle, since x and y are already defined as the horizontal and vertical axes.

Also, check your equation to be sure you are using the correct trig function of that angle θ. It would help if you could label the angle θ on your diagram so that it's easier for us to understand what angle you are calling θ.

BTW, the math symbols are now under the 3-dot menu in the middle of the toolbar above the Edit window (click on the little Greek Parthenon symbol, 3rd from the right).

 

[HareJare]

I'd recommend using θ for the angle, since x and y are already defined as the horizontal and vertical axes.

Also, check your equation to be sure you are using the correct trig function of that angle θ. It would help if you could label the angle θ on your diagram so that it's easier for us to understand what angle you are calling θ.

BTW, the math symbols are now under the 3-dot menu in the middle of the toolbar above the Edit window.

View attachment 281374

Yes okay i will think about this next time! Again thank you for your help