A very difficult problem(hooke's law)

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In summary, the problem involves a mass being held by three extensible strings with different spring constants. The string in the middle has a length of l, and the goal is to find the tension of the three strings. However, there is not enough information given to solve the problem accurately. The only given information is the length of the middle string, and it is not possible to determine the tension without knowing the mass or the natural length of the strings.
  • #1
DarkNess_wtc
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a mass is held by 3 extensible strings with spring constant k1,k2 and k3 respectively.
Now, the string in the middle is having a length of l
that's all the info. given
find the tension of the 3 strings

please help!

http://imageshack.us/photo/my-images/821/helptd.jpg

http://imageshack.us/photo/my-images/821/helptd.jpg
 
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  • #2
Please post the full problem exactly as given.

Show what you've done so far and where you are stuck.
 
  • #3
that's all the information given.

I just don't know how the use the l
 
  • #4
To be honest I don't think it's possible, seeing as though it's asking for the tension... This would require a quantity for mass which is not given.

If anything the answer would be in terms of K2.
 
  • #5
i think the answer is in terms of l,k1,k2,k3 and theta
 
  • #6
think or know?
 
  • #7
know
 
  • #8
Well, if the string in the middle is having a length of l, there's high probability that all 3 strings are having the same lenght. So, when they extend, they all do it by x cm.
Now, the sum of the three tensions equals the weight of the object held by the three strings: T1+T2+T3=m*g, where m is the mass and g is the gravitational constant. But according to hooke's law, T1=k1*x, T2=k2*x and T3=k3*x.
Then, it becomes: k1*x+k2*x+k3*x=m*g<=>x*(k1+k2+k3)=m*g<=> x=(m*g)/(k1+k2+k3).
Once you found out x, you can find the three tensions.
Did I explain it well enough?
 
  • #9
how can you know they having the same length and all do it by x cm.
 
  • #10
I agree. length can't be the same...

I know the answer but will just give hints for now...

Use pythag for lenghts.
 
  • #11
The problem gives the length of the middle string, right?
Why didn't it give the length of any other? Because then there would be a chance that the mass didn't stay perfectly horizontal.
But you're right about the x cm. I can't be sure of that.
 
  • #12
however, the natural length is not given. they might not be the same
 
  • #13
Yes, you're right.
I can't figure it out, though. How can I use pythag, when all I've got is the length of the middle string?
 
  • #14
As far as I can see, there is not enough information to solve this problem. After all, you can disconnect the middle spring and still have the others support it; alternatively, you can have the middle spring support as much of the weight as you like.

Where did you get this problem?
 

Related to A very difficult problem(hooke's law)

1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to extend or compress a spring by some distance is directly proportional to that distance.

2. Why is Hooke's Law considered a difficult problem?

Hooke's Law can be difficult to apply in real-world situations because it assumes that the spring is perfectly elastic and has no mass. In reality, most springs have some mass and do not behave completely elastically.

3. What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring.

4. How is Hooke's Law used in everyday life?

Hooke's Law is used in many everyday objects, such as mattresses, trampolines, and door hinges. It is also used in various engineering and scientific applications, such as in the design of suspension systems for vehicles.

5. What happens to a spring when the force applied to it exceeds its elastic limit?

When the force applied to a spring exceeds its elastic limit, the spring will permanently deform and will no longer return to its original shape when the force is removed. This is known as plastic deformation and can result in the spring breaking or becoming unusable.

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