Hooke's Law

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Hooke’s Law: Introduction

– Hooke’s law states that the extension of a spring is proportional to the applied force, provided that the force is not large enough to deform the spring permanently.
– Mathematically expressed as Force α extension.

Spring Constant

– Since Force α extension then Force / Extension = constant (k).
– The constant of proportionality (k) is called the spring constant.
F / e = k or Force (N) = k e.
– The spring constant is a measure of the stiffness of a spring. The greater the constant the stiffer the spring.
– The spring constant varies with the following;

Material – identical springs mad of different materials will have different constants i.e. steel and copper.
Diameter – the stiffness decreases with the increase in diameter.
Thickness of the wire – a spring made of a thicker wire is stiffer than the one made of thin wire of the same material.
Length of spring – a short spring is stiffer than a longer one.
Number of turns per unit length – a spring with higher number of turns per unit length is less stiff than the one with fewer turns per unit length.

Practice Example 1

1. If the springs shown below are similar and the constant of proportionality (k) is 100 Nm^-1, determine total extension in each arrangement.

Solution

k = 100 Nm^-1
extension = force / k = 10 ÷ 100 = 0.1 m = 10 cm. Extension of the lower spring = 10 cm, extension of the two parallel springs = 5 cm.
Total extension = 10 + 5 = 15 cm. Extension of the two lower springs = 5 cm
Middle spring extend by = 10 cm
Upper two springs extend by = 5 cm
Total extension = 5 + 10 + 5 = 20 cm.

The spring balance

– It is made up of a spring mounted in a metal or plastic casing.
– The spring is fitted with a pointer which moves along a calibrated scale divided into ten equal parts.

Practice Example 1

1. A load of 4 N causes a certain copper wire to extend by 1.0 mm. Find the load that will cause a 3.2 mm extension on the same wire. (Assume Hooke’s law is obeyed).

Solution

F α e, also F1 / F2 = e1 / e2
= F2 = (4 × 3.2) / 1.0
= 12.8 N.

Practice Example 2

2. A body of 200 g was hung from the lower end of a spring which obeys Hooke’s law. Given that the spring extended by 100 mm, what is the spring constant for this spring?

Solution

F = α e, F = k e. F = 200 x 10^-3
kg x 10 N /kg = 2 N.
Extension = 100 x 10^-3 m = 0.1 m.
Spring constant (k) = 2 / 0.1 = 20 N/m.

Practice Example 3

3. Two identical springs, whose spring constant is 6.0 N/cm, are used to support a load of 60 N as shown below. Determine the extension of each spring.

Solution

Since the springs are parallel their spring constant equals 2k.
Therefore extension = Force / k = 2 F / k = 60 / 2 x 6 = 5 cm.
Each spring will extend by 5 cm.

Darasa Huru

Hooke's Law

Hooke’s Law: Introduction

Spring Constant

Practice Example 1

Solution

The spring balance

Practice Example 1

Solution

Practice Example 2

Solution

Practice Example 3

Solution

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Darasa Huru

Hooke's Law

Hooke’s Law: Introduction

Spring Constant

Practice Example 1

Solution

The spring balance

Practice Example 1

Solution

Practice Example 2

Solution

Practice Example 3

Solution

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Hooke's Law

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