Hooke’s Law.

Presentation on theme: "Hooke’s Law."— Presentation transcript:

1 Hooke’s Law

2 Hooke’s Law In the 1600s, a scientist called Robert Hooke discovered a law for elastic materials. Hooke's achievements were extraordinary - he made the first powerful microscope and wrote the first scientific best-seller, Micrographia. He even coined the word ‘Cell’.

3 Hooke's Law, elastic and plastic behaviour
If a material returns to its original size and shape when you remove the forces stretching or deforming it (reversible deformation), we say that the material is demonstrating elastic behaviour. example is a spring A plastic (or inelastic) material is one that stays deformed after you have taken the force away. If deformation remains (irreversible deformation) after the forces are removed then it is a sign of plastic behaviour. example, crumpled paper If you apply too big a force a material will lose its elasticity. example, stretch a spring to far, break a hockey stick. Hooke discovered that the amount a spring stretches is proportional to the amount of force applied to it. This means if you double the force its extension will double if you triple the force the extension will triple and so on.

4 If you measure how a spring stretches (extends its length) as you apply increasing force and plot extension (x) against force (F); the graph will be a straight line.

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8 The elastic limit can be seen on the graph.
This is where it stops obeying Hookes law. Elastic limit can be seen on the graph. Anything before the limit and the spring will behave elastically. This is where the graph stops being a straight line. If you stretch the spring beyond this point it will not return to its original size or shape.

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10 Hooke’s Law Hooke's law:  is a principle of physics that states that the force (F) needed to extend or compress a spring by some distance. F = k ∆ x Where: F is the applied force (in newtons, N), x is the extension (in metres, m) and k is the spring constant (in N/m). Recall ∆ is final change minus initial change.

11 Stretching or compressing a spring
Stretching a Spring Compressing a Spring x = 0 x = 0 x x Fs Fs The more force applied to the spring the more you can stretch it or compress it.

12 Hold on a minute, K? Spring Constant?!
The spring constant measures how stiff the spring is. The larger the spring constant the stiffer the spring. The spring constant k is measured in Nm-1 because it is the force per unit extension. The value of k does not change unless you change the shape of the spring or the material that the spring is made of. k is measured in units of newtons per metre (Nm -1).

13 stiffer spring  _________ slope  _________ k
Example Comparing two springs that stretch different amounts. F spring B spring A Applying the same force F to both springs x xB xA Which spring stretches more? Which is stiffer? A Which spring has a greater slope? Which spring has a larger k value? B B B stiffer spring  _________ slope  _________ k greater larger

14 Known: 18cm = 0.18m, and weight is 56N F=kx 56=k(0.18) k= 311 N/m
Example #1 A sack of potatoes is hanging from a spring. The potatoes weigh 56N and their weight causes the spring to stretch 18cm. What is the spring constant? Known: 18cm = 0.18m, and weight is 56N F=kx 56=k(0.18) k= 311 N/m

15 Example #2 How much force is needed to stretch a spring 0.25 m when it’s constant is 95 N/m? F= 95 x 0.25 = N (negative depending on the direction) What is the spring constant when a 15kg ham is hung on a spring that goes 20cm? Note: We had mass not weight!!! So 15 x 9.8 = 147 N 147=k(0.2) so divide each side by 0.2 and we get 735 N/m

16 Example # 3 A weight of 8.7 N is attached to a spring that has a spring constant of 190 N/m. How much will the spring stretch? Equation: w/o weight w/ weight Fs = kx 8.7 N = (190 N/m) x x = 4.6 x 10-2 m x 8.7 N

17 Example #4 A spring is 0.38m long. When it is pulled by a force of 2.0 N, it stretches to 0.42 m. What is the spring constant? Assume the spring behaves elastically. Extension, ∆x = Stretched length – Original length = 0.42m – 0.38m = 0.04 m F = k ∆ x 2.0N = k x 0.04m So, k = 2.0 N 0.04 m = 50 N m-1

18 Example #5 Ex: A student stretches an elastic band with a spring constant of 50.0 N/m by 15 cm. How much force are they applying?

19 Example #6 Ex: Al McInnis uses a wooden stick with a spring constant of 850 N/m. What is the distortion on the stick if he exerts 525 N while taking a slapshot?

20 Example #7 A 65 kg girl sits in a human sling shot that has a spring constant of 10.5 N/m. If the sling is stretched by 45 m, what is her initial acceleration when released?

21 This can be seen very plainly by comparing the effect of kicking a football, which squashes as you kick it giving a big collision time

22 Andy O Brian with both the Ball and his nose showing elastic behaviour

23 followed by kicking a brick
followed by kicking a brick. The brick doesn't squash, giving a very quick collision time and a very painful foot. This is why airbags and crumple zones can reduce injuries (these are both parts of a car designed to squash rather than be rigid). So to reduce injuries in a collision, always slow down in as long a time as possible. This is why you bend your legs when landing after a jump and why parachutists roll when they hit the ground.

24 Key Definitions Hooke’s Law = The amount a spring stretches is proportional to the amount of force applied to it. The spring constant measures how stiff the spring is. The larger the spring constant the stiffer the spring. A Diagram to show Hooke’s Law F = k ∆ x