Confused which formula to use

  • #1
anthonyk2013
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I have a question I'm having trouble with.

A wire is stretched 1.2 mm by a force of 240n. Calculate the force needed to stretch the wire 2.0mm assuming the elastic limit is not exceeded.

I'm new to physics and I have only used formulas for finding stress, strain and youngs modulas. Should I be able to find the value using one of those formulas?
 

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  • #2
arildno
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Use the fundamental tenet of Hooke's law (for which the limit of elasticity is not exceeded), that the force must be proportional to the stretching of the wire.

The simplest use of this is that the ratio between the two forces must equal the ratio of the two stretchings.
 
  • #3
anthonyk2013
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I use f=-KX?
 
  • #4
arildno
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That is indeed Hooke's law!

Since you are not bothering about the directions of accelerations here, you can disregard the minus sign, and say that the magnitude of the force "f" equals "k" times the magnitude of the stretching X.
That is:
F=kX.

Now, you know both the force in the first instance, and the stretching in that instance.
Can you solve for "k", so that you can use that in order to find the magnitude of the force in the second instance?
 
  • #5
anthonyk2013
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240=K.1.2
K=240x1.2=288N ?
 
  • #6
arildno
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If you have the equation:
240=K*1.2

How do you solve for K?

Remember that we can switch about the order of factors in a product, and also what we term the LHS and RHS of an equation.

Thus, the above equation can also be written as:

1.2*K=240

Also, do NOT use as multiplication sign "." when you also use "." as decimal point sign!
 
  • #7
anthonyk2013
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If you have the equation:
240=K*1.2

How do you solve for K?

Remember that we can switch about the order of factors in a product, and also what we term the LHS and RHS of an equation.

Thus, the above equation can also be written as:

1.2*K=240

Also, do NOT use as multiplication sign "." when you also use "." as decimal point sign!

Ok won't do that again.

So K=240*1.2=288N
 
  • #8
arildno
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Really?

How do you solve the equation:

2*x=4?
 
  • #9
anthonyk2013
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Sorry my bad, need to get my head in the game here.
K=240/1.2=200.
 
  • #10
arildno
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Now we are in agreement! :smile:

Now that you know the K-value is 200 N/mm, what must the applied force be in the second instance, when you know the stretching is equal to 2.0 mm?
 
  • #11
anthonyk2013
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F=k*x
F=200*2=400N/mm squared ?
 
  • #12
arildno
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How does (N/mm)*mm become?
 
  • #13
anthonyk2013
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How does (N/mm)*mm become?

Need to concentrate no mm.

400N
 
  • #14
arildno
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Yup! :smile:
 
  • #15
arildno
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Now, here's how you can do it without bothering to solve for "K" first:

We have, for the first instance, F_1=k*x_1
and for the second instance, F_2=k*x_2

Dividing the second equation by the first, we have:

F_2/F_1=(k*x_2)(k*x_1)=x_2/x_1

That is:
F_2/F_1=x_2/x_1, or, multiplying with F_1

F_2=(x_2/x_1)*F_1
 
  • #16
anthonyk2013
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Now, here's how you can do it without bothering to solve for "K" first:

We have, for the first instance, F_1=k*x_1
and for the second instance, F_2=k*x_2

Dividing the second equation by the first, we have:

F_2/F_1=(k*x_2)(k*x_1)=x_2/x_1

That is:
F_2/F_1=x_2/x_1, or, multiplying with F_1

F_2=(x_2/x_1)*F_1

Thanks very much for your help. I won't feel too bad in class Thursday evening.
Cheers
 
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