Dynamics question.... where did this formula come from?

In summary: In this case, because the wire is horizontal, the only force on the bit of wire is the tension in the wire. So T = FN - mg.
  • #1
calivianya
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Hi there,

I found a thread from 2004 that perfectly answered my homework question here, but I don't understand how the person who responded came up with the answer. I would have just replied to the thread and resurrected it, but it says the thread is no longer open for replies. I'll paraphrase it here:

Original question: Arlene is to walk across a high wire strung horizontally between two buildings 10 m apart. The sag in the rope when she is at the mid-point should not exceed 10 degrees. If her mass is 50 kg what must be the tension in the rope?

Original answer: The sum of forces in the vertical direction would look like this: 0 = 2Tsin(10) - 490.5, T = 1412.3384 N

I actually had the exact same problem with the exact same numbers, but I have no idea why 1412.33 N is the right answer.

I know the equation where ΣFy = FN - mg

I also get that the ΣFy is 0 if there is no movement in the y direction.

0 = FN - mg.
mg = 50kg X 9.8m/s2 = 490 N
0 = FN - 490 N...

How do you know that FN = 2Tsin(θ)? I can't find that equation anywhere in the dynamics chapter in my book, my notes, or my teacher's powerpoint. This is supposed to be homework from the dynamics chapter, and I would have been totally unable to solve this problem if I hadn't found this website. I also tried googling the formula and didn't find anything helpful.

Do I just need to take this for granted or is there somewhere I can see how the 2Tsin(θ) was derived?
 
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  • #2
Hi calivianya,

Welcome to Physics Forums!

calivianya said:
How do you know that FN = 2Tsin(θ)? I can't find that equation anywhere in the dynamics chapter in my book, my notes, or my teacher's powerpoint.
You won't find every equation for every problem already "canned and ready to eat" in your text. What you'll find is techniques for finding those equations for situations that you come across. In this case you'll want to start with the first step recommended for *every* problem: Draw the Free Body Diagram and label the forces acting. Then it's just a matter of geometry/trigonometry to apply your force summation rules.
 
  • #3
gneill said:
Hi calivianya,

Welcome to Physics Forums!You won't find every equation for every problem already "canned and ready to eat" in your text. What you'll find is techniques for finding those equations for situations that you come across. In this case you'll want to start with the first step recommended for *every* problem: Draw the Free Body Diagram and label the forces acting. Then it's just a matter of geometry/trigonometry to apply your force summation rules.
Thanks so much. I guess the geometry/trigonometry is where I struggle the most - I took AP Calculus in 2005 and that's the last time I had math. I took geometry in 2003 so that's really been a long ways back. It's difficult to figure out what I need to do without remembering the math.
 
  • #4
calivianya said:
0 = FN - mg.
That is the equation you get from applying ΣF=ma to Arlene. To get T into the picture you have to apply the same principle to the bit of wire she is standing on.
Because the wire is exerting an upward force FN on Arlene, she is exerting a downward force of the same magnitude on that bit of wire.
The other forces on that bit of wire come from the tension in the wire. Draw the freebody diagram for that.
 

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The formula for potential energy, PE=mgh, was first introduced by Galileo Galilei in the 16th century. He believed that the energy of an object at rest has the potential to do work when it is moved to a higher position. This concept was later expanded upon by others, including Sir Isaac Newton and Robert Hooke.

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