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## Physics library

### Course: Physics library > Unit 5

Lesson 2: Springs and Hooke's law# What is Hooke's Law?

Learn about elasticity and how to determine the force exerted by a spring.

# What is a spring?

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.

There is nothing particularly magical about the shape of a coil spring that makes it behave like a spring. The 'springiness', or more correctly, the

*elasticity*is a fundamental property of the wire that the spring is made from. A long straight metal wire also has the ability to ‘spring back’ following a stretching or twisting action. Winding the wire into a spring just allows us to exploit the properties of a long piece of wire in a small space. This is much more convenient for building mechanical devices.# What happens when a material is deformed?

When a force is placed on a material, the material stretches or compresses in response to the force. We are all familiar with materials like rubber which stretch very easily.

In mechanics, the force applied per unit area is what is important, this is called the

*stress*(symbol sigma). The extent of the stretching/compression produced as the material responds to stress is called the*strain*(symbol \epsilon). Strain is measured by the ratio of the difference in length delta, L to original length L, start subscript, 0, end subscript along the direction of the stress,*i.e.*\epsilon, equals, delta, L, slash, L, start subscript, 0, end subscript.Every material responds differently to stress and the details of the response are important to engineers who must select materials for their structures and machines that behave predictably under expected stresses.

For most materials, the strain experienced when a small stress is applied depends on the tightness of the chemical bonds within the material. The stiffness of the material is directly related to the chemical structure of the material and the type of chemical bonds present. What happens when the stress is removed depends on how far the atoms have been moved. There are broadly two types of deformation:

. When the stress is removed the material returns to the dimension it had before the load was applied. The deformation is reversible, non-permanent.**Elastic deformation**. This occurs when a large stress is applied to a material. The stress is so large that when removed, the material does not spring back to its previous dimension. There is a permanent, irreversible deformation. The minimal value of the stress which produces plastic deformation is known as the**Plastic deformation***elastic limit*for the material.

Any spring should be designed and specified such that it only ever experiences elastic deformation when built into a machine under normal operation.

# Hooke's law

When studying springs and elasticity, the 17ᵗʰ century physicist Robert Hooke noticed that the stress vs strain curve for many materials has a linear region. Within certain limits, the force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring. This is known as Hooke's law and commonly written:

Where F is the force, x is the length of extension/compression and k is a constant of proportionality known as the

*spring constant*which is usually given in start text, N, slash, m, end text.Though we have not explicitly established the direction of the force here, the negative sign is customarily added. This is to signify that the

*restoring force*due to the spring is in the opposite direction to the force which caused the displacement. Pulling down on a spring will cause an extension of the spring**downward**, which will in turn result in an**upward**force due to the spring.It is always important to make sure that the direction of the restoring force is specified consistently when approaching mechanics problems involving elasticity. For simple problems we can often interpret the extension x as a 1-dimensional vector; in this case the resulting force will also be a 1-dimensional vector and the negative sign in Hooke’s law will give the correct direction of the force.

When calculating x, it is important to remember that the spring itself will also have some nominal length L, start subscript, 0, end subscript. The total length L of a spring under extension is equal to the nominal length plus the extension, L, equals, L, start subscript, 0, end subscript, plus, x. For a spring under compression, it would be L, equals, L, start subscript, 0, end subscript, minus, x.

**Exercise 1:**A 75 kg person stands on a compression spring with spring constant 5000, space, start text, N, slash, m, end text and nominal length 0, point, 25, space, start text, m, end text. What is the total length of the loaded spring?

**Exercise 2a:**You are designing a mount for moving a 1 kg camera smoothly over a vertical distance of 50 mm. The design calls for the camera to slide on a pair of rails, with a spring supporting the camera and pulling it up against the tip of an adjustment screw as shown in Figure 1. The nominal length of the spring is L, start subscript, 0, end subscript, equals, 50, space, start text, m, m, end text.

**What is the minimum spring constant required for this design?**

**Exercise 2b:**What is the minimum elastic limit required by your spring?

# Young's modulus and combining springs

Young's modulus (also known as the elastic modulus) is a number that measures the resistance of a material to being elastically deformed. It is named after the 17ᵗʰ century physicist Thomas Young. The stiffer a material, the higher its Young's modulus.

Young's modulus is usually given the symbol E, and is defined as:

Young's modulus can be defined at any strain, but where Hooke's law is obeyed it is a constant. We can directly obtain the spring constant k from the Young's modulus of the material, the area A over which the force is applied (since stress depends on the area) and nominal length of the material L.

This is a very useful relationship to understand when thinking about the properties of combinations of springs. Consider the case of two similar ideal springs with spring constant k which can be arranged to support a weight end-to-end (series) or jointly (parallel) as shown in Figure 2.

**What is the effective spring constant for the combination in each case?**In the series configuration, we can see that the combined spring is equivalent to one spring with double the length. The spring constant in this case must therefore be half that of an individual spring, k, start subscript, start text, e, f, f, e, c, t, i, v, e, end text, end subscript, equals, k, slash, 2.

In the parallel configuration, the length remains the same but the force is distributed over twice the area of material. This doubles the effective spring constant of the combination, k, start subscript, start text, e, f, f, e, c, t, i, v, e, end text, end subscript, equals, 2, k.

# Springs with mass

Consider the setup shown in Figure 3. A spring supports a 1 kg mass horizontally via a pulley (which can be assumed to be frictionless) and an identical spring supports the same mass vertically. Suppose the spring has mass of 50 g, spring constant k=200 N/m.

**What is the extension of the spring in each case?**In both cases, the force on the spring due to the mass has the same magnitude, m, g. So we might first assume that the extension is the same in both cases. It turns out that for a real spring this is not true.

The complication here is that the spring itself has mass. In the vertical case, the force of gravity acts on the spring in the same direction as the force due to the mass. So the mass of the spring adds to that of the weight. The extended spring is supporting a total weight of 1.05 kg which causes an extension of

In the horizontal case, the pulley has changed the direction of the force. The force due to the 1 kg weight acting on the spring is now

*orthogonal*to the force of gravity acting on the spring. So the extension of the spring is supporting only 1 kg. It therefore extendsThis difference can be quite significant and if not taken into account it can lead to incorrect results in the laboratory. In physics teaching laboratories, we often use

*spring balances*to measure force. A spring balance (Figure 4) is simply a spring with a pointer attached and a scale from which the force can be read.Because the manufacturers of spring balances expect their product to be used vertically (for example, by a fisherman measuring the mass of his fish) the scale is calibrated to take into account the mass of the spring and hook. It will give an incorrect

*absolute*result if used to measure a horizontal force. However, Hooke's law tells us that there is a linear relationship between force and extension. Because of this we can still rely on the scale for*relative*measurements when used horizontally. Some spring balances have an adjustment screw which allows the zero point to be calibrated, eliminating this problem.## Want to join the conversation?

- What is nominal length?(14 votes)
- nominal length = original/initial length(23 votes)

- I didn't understood what that minus in formula was for?(16 votes)
- It just tells us that the exerted by the spring will be in the opposite direction to the force applied.(8 votes)

- The spring constant is the same for all springs or different? explain.(10 votes)
- is spring constant is same on another planet?(8 votes)
- gravity won't change the rigidity of the spring, so I think it is the same on other planets(2 votes)

- I have a question on example 2a. If the nominal length is 50mm, and the compression is of 50mm then shouldn't the length under compression(L = Li- x = 50mm - 50mm = 0) become zero?(5 votes)
- The nominal length is 50m. But the spring is being extended not compressed and therefore, the formula is L = Li + x= 50 mm +50 mm = 100 mm(7 votes)

- I don't understand what the length notations of springs refers to. For example, strain is defined as change in length over initial length. I understand what change in length is, but what is the initial length referring to ? Is it the entire length of the spring ?

I'm referring to the so called nominal length or "initial length".(5 votes)- The spring has some length when it is relaxed, right? That's the initial length.(6 votes)

- In exercise 1, how come the F (force) is equal to mg, not mgh? When the person steps on, there is still some height (from the spring's nominal height of 0.25 m). I plugged this in and got 0.21325. I am so confused.(3 votes)
- I think you're confused as to the difference between gravitational force and potential energy. Close enough to the Earth's surface, we always feel the force of gravity (equal to mg).

Potential energy on the other hand is mgh. This energy can be converted into kinetic energy to make the object move, but energy and force are two different physical quantities.(7 votes)

- Why is k=E.A/L? if we put K = -F.x and E=stress/strain = (F/A)/(x/L) = F.L/(A.x)

then we get -F.x = F/x

Please help me(5 votes)- I think we are only interested in the absolute value. You need not include the negative sign.(2 votes)

- in exercise 1, when we divided F over K, why the K isn't negative as hooke's law said?(3 votes)
- Hooke's law doesn't say that the constant k is negative. The negative sign is only saying that the force is in the opposite direction of the displacement. Springs pull back on you, they don't push you more in the direction you are already pulling them.(2 votes)

- I didn't understand Exercise 2a. The equation says:

F = -kx

Here F = mg = 9.8N

9.8N = -kx

=> k = 9.8/-x

To get the minimum value of k, x must be the maximum extension, right? According to me, the maximum extension is 150mm - 50mm = 100mm. But the solution shows that the maximum extension is 50mm. Could somebody please explain this question as I have not understood. Thank you.(2 votes)- 2a states that the
*minimum*extension is equal to 50mm, which is the value needed to solve the question.(1 vote)