Force on Each Mass in a Two-Mass Spring System

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SUMMARY

The discussion centers on calculating the force experienced by two masses attached to a spring when each mass is stretched by X meters in opposite directions. The correct formula for the force is established as F = k * 2x, where k is the spring constant and 2x represents the total displacement of the spring. Participants clarify the distinction between scenarios where the spring is stretched by equal amounts versus when equal forces are applied, emphasizing that in the latter case, only one force is considered due to Newton's third law. The conversation also highlights the misconception that tension in a static setup can be summed, clarifying that each half of a spring has a modified spring constant when divided.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Basic knowledge of Newton's Third Law of Motion
  • Familiarity with spring constants and their implications in mechanics
  • Concept of static equilibrium in mechanical systems
NEXT STEPS
  • Study the implications of Hooke's Law in different spring configurations
  • Explore the concept of static equilibrium in mechanical systems
  • Learn about the effects of dividing springs on their spring constants
  • Investigate real-world applications of spring forces in engineering
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding the dynamics of spring systems and force interactions in static setups.

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Homework Statement


Two masses attached to opposite ends of a spring. What is the Force felt by each mass if each mass is stretched by Xmeters in opposite directions away from each other.

Homework Equations


F=-kx

The Attempt at a Solution


So my book says the answer is F=k*2x.

Each mass has the same mass and is stretched from their equilibrium by the same amount X. so the spring is stretched a total amount of 2x.

So since the spring force is proportional to the displacement of the spring/stretch of the spring, it would make sense that the spring force would be F=k*2x since 2x is the total displacement.

But what I'm havin gtrouble understanding is how to distinguish between situations like this when the spring is stretched by the same amount on both sides and problems where the same force is applied to both sides.

When the same force is applied to both sides, you use the same formula Fspring = kx , but for the force applied you only consider one of the applied forces rather than both combined, on the notion that the other side is what's keeping the spring from accelerating, Newtons third law.

But when you actaully compress the spring or stretch it by a certain amount, even if you stretch it from both ends, you consider the total stretch added from both ends to consider the recoil force, am i right?

And lastly, (Thanks guys haha) is the reason why you don't consider both forces when you apply the same force to both sides that compression could only occur if there was a counter force? Because the lack of movement allows for the compression to occur? Does that make sense? its weird because it's like we're using the applied force twice,
once to cancel out the counter force from the other end of the spring (i.e. from a wall or from someone on the other end of the spring)and then again when determining the compression caused by said force.
 
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I'm having a little trouble understanding your question. Can you give a specific example of the two situations you are referring to?

Chet
 
A lot of people make the mistake of thinking that if there's force T pulling at each end then the tension is 2T. But in a static set-up, it is not possible to pull at one end only.
In the OP set-up, it might help to think of the spring as two (half-length) springs fixed at the centre. If you chop a spring of constant k in half then each half will have a constant 2k.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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