• Support PF! Buy your school textbooks, materials and every day products Here!

Effective spring constant

  • #1
174
0

Homework Statement



What is the effective spring constant for the system of the two springs, perfect pulley, and string shown on the left for it to be modeled by just one spring (constant keff) as shown on the right?

Homework Equations





The Attempt at a Solution



With k2 getting stretched by D, the pulley goes down by 2D. Therefore,
x1 = F/k1, x2 = F/k2
x1 + 2(x2) = F/k1 + 2F/k2

Solving this, I get k effective as

(2*k1*k2)/(k2+2*k1)

But this doesn't match the given answer. Where am I going wrong?
 

Attachments

Answers and Replies

  • #2
PeterO
Homework Helper
2,425
46

Homework Statement



What is the effective spring constant for the system of the two springs, perfect pulley, and string shown on the left for it to be modeled by just one spring (constant keff) as shown on the right?

Homework Equations





The Attempt at a Solution



With k2 getting stretched by D, the pulley goes down by 2D. Therefore,
x1 = F/k1, x2 = F/k2
x1 + 2(x2) = F/k1 + 2F/k2

Solving this, I get k effective as

(2*k1*k2)/(k2+2*k1)

But this doesn't match the given answer. Where am I going wrong?
If your maths is OK, your "assumptions" must be faulty.

The only assumption I see is "With k2 getting stretched by D, the pulley goes down by 2D."

Can you justify that statement?

Peter
 
  • #3
PeterO
Homework Helper
2,425
46

Homework Statement



What is the effective spring constant for the system of the two springs, perfect pulley, and string shown on the left for it to be modeled by just one spring (constant keff) as shown on the right?

Homework Equations





The Attempt at a Solution



With k2 getting stretched by D, the pulley goes down by 2D. Therefore,
x1 = F/k1, x2 = F/k2
x1 + 2(x2) = F/k1 + 2F/k2

Solving this, I get k effective as

(2*k1*k2)/(k2+2*k1)

But this doesn't match the given answer. Where am I going wrong?
Investigate with extremes to see some possible results:

If k1 >>> k2, Then applying an extra force ΔF will hardly move the pulley at all, instead k2 will be stretched an amount D, and the end of the string will go down an equal distance.

So if k1 >>> k2 Keff ≈ k2

If k1 <<< k2, then applying an extra force ΔF will hardy move the spring k2, and the mechanical advantage of the pulley will give a spring constant of ≈ 0.5 k1

Does your answer - or the supposedly correct one - correctly predict those two conditions.
 
  • #4
174
0
If your maths is OK, your "assumptions" must be faulty.

The only assumption I see is "With k2 getting stretched by D, the pulley goes down by 2D."

Can you justify that statement?

Peter
When I said the pulley goes down by 2D, I mean the length on both sides of pulley changes by that much. x1 = 2x2. I'm sure that's correct.
 
Last edited:
  • #5
PeterO
Homework Helper
2,425
46
When I said the pulley goes down by 2D, I mean the length on both sides of pulley changes, x1 = 2x2. I'm sure that's correct.
This is an entirely different statement to your original!

When the pulley goes down by 2D, that means k1 has stretched by 2D.

For this to happen there are several possibilities; here a three examples.

1: k2 hardly changed length, so the end of the string moved down 4D

2: k2 changes its length by D, so the end of the string moved down 5D

3: k2 changes its length by 2D, so the end of the string moved down 6D

plus any other of the million possibilities.

You need to consider how much k1 and k2 stretch when you apply a force F, or perhaps and extra force ΔF, to the string - and then work out how far the end of that string moves.
 
  • #6
PeterO
Homework Helper
2,425
46
When I said the pulley goes down by 2D, I mean the length on both sides of pulley changes by that much. x1 = 2x2. I'm sure that's correct.
Oh, and I did say IF your maths is OK.

You didn't show how you got from x1 + 2(x2) = F/k1 + 2F/k2 to (2*k1*k2)/(k2+2*k1) so I can't comment on whether you made an arithmetic error of not.
 
  • #7
PeterO
Homework Helper
2,425
46
When I said the pulley goes down by 2D, I mean the length on both sides of pulley changes by that much. x1 = 2x2. I'm sure that's correct.
Note that for the pulley to go down 2D, the only think you can be certain of is that k1 has stretched by 2D.
There will be other movements: the top end of spring k2 will move up some distance X, and the "free" end of the string will thus move down 4D+X.
 
  • #8
174
0
Note that for the pulley to go down 2D, the only think you can be certain of is that k1 has stretched by 2D.
There will be other movements: the top end of spring k2 will move up some distance X, and the "free" end of the string will thus move down 4D+X.
Okay, so what's your approach to this problem. Because, I'm totally clueless about this problem.
 
  • #9
PeterO
Homework Helper
2,425
46
Okay, so what's your approach to this problem. Because, I'm totally clueless about this problem.
When you apply a force to either spring, F = K1x1 and F = K2x2 will tell you how far each spring would extend because of that force.

We also know that F = KeffXtotal will give you the effective spring constant.

If Force, F, is applied to the string, then tension in the string will tell you that Force, F, is applied to Spring K2, while the pulley means a total force of 2F will act on spring K1

The first 2 formulae mentioned above will tell you the extension of each spring - call them E1 and E2.

The setup means the pulley will descend E1 (it is attached to the bottom of spring K1), while the upper end of spring K2 will extend up by E2.

So under the influence of an applied Force F, the string will move 2E1 + E2 (this is the total extension Xtotal mentioned above.

Now use F = KeffXtotal to solve for Keff.
 
  • #10
174
0
If Force, F, is applied to the string, then tension in the string will tell you that Force, F, is applied to Spring K2, while the pulley means a total force of 2F will act on spring K1

.
Thank you for your great explanation!
I still haven't understood why force F applied on k2 translates to 2F on k1

Also, solving it your way gives me
[itex] (2k1k2)/(2k2 + k1) [/itex]

Is that right?
 
  • #11
PeterO
Homework Helper
2,425
46
Thank you for your great explanation!
I still haven't understood why force F applied on k2 translates to 2F on k1
Once Force F is applied, the tension in the string means both sides of the pulley are pulled down with force F - giving a total of 2F



Also, solving it your way gives me
[itex] (2k1k2)/(2k2 + k1) [/itex]

Is that right?
I can't be certain that is right - I have never actually found the answer - that was always up to you, but it has a chance if it passes the "extremes" testing. [that was an earlier post of mine].

If K1 is very large (the spring is very stiff) it may be considered to be just a piece of string.
When you apply a force, the pulley stays put, and the system is effectively just a single spring, K2

In your answer, the denominator 2k2 + k1 is effectively just k1 as this value renders 2k2 insignificant.
The expression is thus
[itex] (2k1k2)/(k1) [/itex]
which simplifies to 2k2
Pity about that leading 2

If instead K2 is very large (that spring is very stiff), then K2 can be replaced with a piece of string - and the mechanical advantage of the pulley means the effective spring constant is K1/2.

when K2 >> K1, 2k2 + k1 is effectively just 2k2
he expression is thus
[itex] (2k1k2)/(2k2) [/itex]
which simplifies to k1

Where has the half gone??

It would be good if there was not a 2 in the numerator of your answer??

WIthout seeing your step by step derivation, I cannot see where (if at all) you have gone wrong.
 
  • #12
174
0
Once Force F is applied, the tension in the string means both sides of the pulley are pulled down with force F - giving a total of 2F





I can't be certain that is right - I have never actually found the answer - that was always up to you, but it has a chance if it passes the "extremes" testing. [that was an earlier post of mine].

If K1 is very large (the spring is very stiff) it may be considered to be just a piece of string.
When you apply a force, the pulley stays put, and the system is effectively just a single spring, K2

In your answer, the denominator 2k2 + k1 is effectively just k1 as this value renders 2k2 insignificant.
The expression is thus
[itex] (2k1k2)/(k1) [/itex]
which simplifies to 2k2
Pity about that leading 2

If instead K2 is very large (that spring is very stiff), then K2 can be replaced with a piece of string - and the mechanical advantage of the pulley means the effective spring constant is K1/2.

when K2 >> K1, 2k2 + k1 is effectively just 2k2
he expression is thus
[itex] (2k1k2)/(2k2) [/itex]
which simplifies to k1

Where has the half gone??

It would be good if there was not a 2 in the numerator of your answer??

WIthout seeing your step by step derivation, I cannot see where (if at all) you have gone wrong.
But they haven't mentioned anything about one spring being stiffer than the other.
How do i solve it in those conditions where they haven't mention anything about that.

Also, I still haven't understood why Force F translates to 2F on reaching the pulley. Anyways, appreciate your effort. Guess it's me. I'm slow.
 
  • #13
ehild
Homework Helper
15,396
1,803
First you have the pulley with relaxed springs, as the pulley is considered weightless. (In reality, it has weight but it is small, and only very slightly stretches the springs. )

Apply some force - you can hang a weight on the hook at the left piece of string (with mass much greater than that of the pulley). Both springs will stretch till new equilibrium is established. Spring 1 stretches by x1, spring 2 stretches by x2. The pulley goes down by x1, that makes both pieces of string 2 shorten, and the hook descends by 2x1. Spring 2 stretches by x2, that makes the right piece of string shorten and the left piece getting longer: The hook final displacement is X= 2x1+x2.

There is equilibrium. So the tension in the string 2 is T2=F. The tension in the upper string is T1. The net force on the pulley is T1-2T2=0, as the weight of the pulley can be ignored.

T1 tension stretches the string 1 by x1=T1/k1, T2 stretches spring 2 by x2=T2/k2. Can you proceed from here?

ehild
 

Attachments

  • #14
174
0
First you have the pulley with relaxed springs, as the pulley is considered weightless. (In reality, it has weight but it is small, and only very slightly stretches the springs. )

Apply some force - you can hang a weight on the hook at the left piece of string (with mass much greater than that of the pulley). Both springs will stretch till new equilibrium is established. Spring 1 stretches by x1, spring 2 stretches by x2. The pulley goes down by x1, that makes both pieces of string 2 shorten, and the hook descends by 2x1. Spring 2 stretches by x2, that makes the right piece of string shorten and the left piece getting longer: The hook final displacement is X= 2x1+x2.

There is equilibrium. So the tension in the string 2 is T2=F. The tension in the upper string is T1. The net force on the pulley is T1-2T2=0, as the weight of the pulley can be ignored.

T1 tension stretches the string 1 by x1=T1/k1, T2 stretches spring 2 by x2=T2/k2. Can you proceed from here?

ehild
Yes. THANKS A LOT!

Exactly the answer I was looking for!
 

Related Threads for: Effective spring constant

  • Last Post
Replies
0
Views
10K
  • Last Post
Replies
2
Views
914
Replies
0
Views
2K
Replies
7
Views
5K
Replies
1
Views
8K
Top