Springs connected to a mass in series?

  • #36
erobz said:
I think you messed up the algebra or the application of the equations ( I can't tell because you don't shot the steps you took)
k1*k2/(k1 + k2)
 
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  • #37
flinnbella said:
k1*k2/(k1 + k2)
Good. Now divide by numerator and denominator by ##k_1 \cdot k_2## and you will see the start of a more general form that should look familiar if you've taken circuits.
 
  • #38
flinnbella said:
Ok I get that the total displacement of the mass is delta_x + delta_y, and this can be represented by a single spring instead of two. However, I don't see a relationship between the spring constants.
Note how much you can stretch two rubber bands (in parallel) at once.
Now, tie two ends of those bands (in series) and stretch both as one.

At the point that you are applying similar force as before, there will be much more stretching (x of balance) for the second case.

Parallel series springs.jpg
 
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  • #39
Lnewqban said:
Note how much you can stretch two rubber bands (in parallel) at once.
Now, tie two ends of those bands (in series) and stretch both as one.

At the point that you are applying similar force as before, there will be much more stretching (x of balance) for the second case.

View attachment 332054
Oh this makes some more sense, not 100% understanding but it helps. How did you make the graphic?
 
  • #40
flinnbella said:
Oh this makes some more sense, not 100% understanding but it helps. How did you make the graphic?
What is still unclear?

Left graphic:
Balance load equals mg.
Both, the strong and the weak spring, must deflect the same in order to produce a combined reactive force equaling mg.
Reactive individual force of strong spring is greater than the reactive individual force of weak spring, but summation of both is the balance load mg.

Could you try describing the right graphic?
 
  • #41
flinnbella said:
k1*k2/(k1 + k2)
So, if this is k_eq , what do you get for 1/k_eq ?

To restate that:

If ##\displaystyle \quad k_{eq}=\dfrac{k_1 \cdot k_2}{k_1 + k_2} ## ,

then what is ##\displaystyle \quad \dfrac 1 {k_{eq} } \ ? ##
 
  • #42
SammyS said:
So, if this is k_eq , what do you get for 1/k_eq ?

To restate that:

If ##\displaystyle \quad k_{eq}=\dfrac{k_1 \cdot k_2}{k_1 + k_2} ## ,

then what is ##\displaystyle \quad \dfrac 1 {k_{eq} } \ ? ##
1/keq = (k1 + k2)/(k1*k2)
 
  • #43
Lnewqban said:
What is still unclear?

Left graphic:
Balance load equals mg.
Both, the strong and the weak spring, must deflect the same in order to produce a combined reactive force equaling mg.
Reactive individual force of strong spring is greater than the reactive individual force of weak spring, but summation of both is the balance load mg.

Could you try describing the right graphic?
I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.
 
  • #44
flinnbella said:
1/keq = (k1 + k2)/(k1*k2)
Yeah, but now you can turn that into something familiar (resistors in parrallel) by dividing each term in the numerator by the denominator.
flinnbella said:
I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.
They don't both exert a force on the mass. The FBD of the mass only includes its weight and the force from sping 2. That force is transmitted through spring 2 and is applied to spring 1. Then that force is transmitted through spring one to the fixed structure.
 
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  • #45
flinnbella said:
I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.
What prevents spring 2 to fall down to the ground, together with the attached mass, is a force of magnitude mg pulling them up.
The net force on that spring 2 is then zero, and it remains in balance.
Exactly the same thing happens to spring 1 (which is attached to ground and to spring2-mass).

Parallel parallel springs.jpg


Series series springs.jpg
 
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  • #46
flinnbella said:
1/keq = (k1 + k2)/(k1*k2)
Simplify that.

(k1 + k2)/(k1*k2) = k1/(k1*k2) + k2/(k1*k2) = ?
 
  • #47
flinnbella said:
I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.
The mass exerts a force on spring 2 and spring 2 must exert the same force on spring 1.

The fascinating thing is that since both spring displacements are added up such that the total displacement is always larger than any of the springs alone, the equivalent spring that would replace both springs is also always less stiff than the softest of the springs.
 
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  • #48
jack action said:
The mass exerts a force on spring 2 and spring 2 must exert the same force on spring 1.

The fascinating thing is that since both spring displacements are added up such that the total displacement is always larger than any of the springs alone, the equivalent spring that would replace both springs is also always less stiff than the softest of the springs.
Why does spring 2 exert same force on spring 1? I don't think those are force action-reaction pairs.
 
  • #49
flinnbella said:
Why does spring 2 exert same force on spring 1? I don't think those are force action-reaction pairs.
1695036279985.png


The hand pulls on the spring with force ##F##.

What force does the wall apply to the spring? What force does the spring apply to the wall? Remember, we are talking about "massless" springs. Free the spring from external bodies and sum external forces i.e. ##\sum F_{ext}= \cancel{m}^0 a = 0 ##
 
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  • #50
flinnbella said:
Why does spring 2 exert same force on spring 1? I don't think those are force action-reaction pairs.
Yes, there are.

Doing the free body diagrams (FBD) for each component (the ceiling ##c##, the spring ##k_1##, the spring ##k_2## and the mass ##m##):

fbd.png
By definition, from one FBD to the next:
$$F_c = F_{k1}$$
$$F_{k1} = F_{k2}$$
$$F_{k2} = F_m$$
mass FBD:
$$F_m = mg$$
Therefore:
$$F_c = F_{k1} =F_{k2} = F_m = mg$$
The force ##mg## goes through every component.
 
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  • #51
jack action said:
Yes, there are.

Doing the free body diagrams (FBD) for each component (the ceiling ##c##, the spring ##k_1##, the spring ##k_2## and the mass ##m##):

By definition, from one FBD to the next:
$$F_c = F_{k1}$$
$$F_{k1} = F_{k2}$$
$$F_{k2} = F_m$$
mass FBD:
$$F_m = mg$$
Therefore:
$$F_c = F_{k1} =F_{k2} = F_m = mg$$
The force ##mg## goes through every component.
No, that can't be right. F K1 would be directed upward, not downwards, because the spring will be stretched downwards, and so FK1 will apply a restoring force upwards.
 
  • #52
flinnbella said:
No, that can't be right. F K1 would be directed upward, not downwards, because the spring will be stretched downwards, and so FK1 will apply a restoring force upwards.
On the thing that is attached to below ( i.e. spring 2 ) it is applying a restoring force ##F_{k2}##, which is equal and opposite the force that is applied to spring 1. (##F_{k1} = - F_{k2}##)
 
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  • #53
jack action said:
The force mg goes through every component.
This is true but that means they all have magnitude mg but they are not all mutually action-reaction pairs. There are three such pairs but the others are known equal and opposite because each spring is not experiencing acceleration or deformation.
 
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  • #54
hutchphd said:
This is true but that means they all have magnitude mg but they are not all mutually action-reaction pairs. There are three such pairs but the others are known equal and opposite because each spring is not experiencing acceleration or deformation.
And in the case of the oscillator( which is @flinnbella original problem) the forces on opposite ends of each spring are equal and opposite because the springs are “massless”.
 
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  • #55
Yes I had forgotten the Original Post!
 
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  • #56
hutchphd said:
Yes I had forgotten the Original Post!
Yeah… we have been walking around a little bit in hopes to get a flag in the ground…somewhere.
 
  • #57
flinnbella said:
No, that can't be right. F K1 would be directed upward, not downwards, because the spring will be stretched downwards, and so FK1 will apply a restoring force upwards.
Those are only the magnitudes. You can see the direction of the vectors on the drawing where I already assumed that the forces acting on each component are equal and opposite.
 
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