# A Recap on Hooke's Law When going over Power, Work Done and Energy, we briefly went over Hooke's law and the spring force as a means of speeding things up. We don't want you hunched over on your desk, tears pouring like a faulty fountain over physics revision after all! So, we'll be (hopefully) extending your knowledge on Hooke's law in this document. As a reminder, this is Hooke's Law, in terms of the spring force: $F = -kx$ Where $F$ is the normal spring force acting on itself, $x$ is the extension of the spring and $-k$ is the *spring constant*, an intrinsic property of springs. Each spring has a **restoring force** that acts on it - aiming to pull the object back to it. This is known as **equilibrium**, and can be described with two rules - where the net force and net torque on an object must be 0: $F_{net} = 0 \,, \tau_{net} = 0$ >[!Example]- Problem Taster # Indeterminate Structures - Intro to Elasticity Ever wondered why some school chairs are wobblier than others? For the most part - they're missing parts, and with education budget cuts creates the perfect environment for students to lean back on their chairs. It's time to think of this in the way a physicist would - past the grave head injuries sustained from going just a little too far on the incline. The question we pose is: *How do we model an unbalanced object?* The forces the chair exerts on the ground are different on each leg, meaning that we have a sizeable number of constants to have to deal with if we so choose. So, what's the solution to this question? Note how in every other model we've made, we've assumed that the bodies are *rigid* - that they do not deform regardless of the forces they sustain. While this might seem obvious to the layman, to model such a non-rigid object will and is very difficult to this day. With wobbly chairs, if we put a block of steel on it, the weight of the steel would be large enough to deform the legs so that all legs (including the shorter one) touch the floor - meaning the table wouldn't wobble and this would become a normal question. However, how do we find the forces on *each* individual leg? # Elasticity When we discussed [[Types of Bonding (Chemistry)|intermolecular bonds]], we talked about how atoms stuck together to form large solids - through intermolecular forces. For a solid, this force can be interpreted as a spring between two molecules - which is our measure of elasticity. Strictly put, elasticity refers to the level of stretching the intermolecular "spring" can do before dislocating. As a result, all bodies are to some extent elastic, though we usually use this term to refer to the most elastic objects on the spectrum. Elastic and plastic deformations of objects can be generalised into their components through the descriptions of the three types of stresses - shown below. The equation for stress is: $s = \frac{F}{A}$ Where $F$ is the deforming force applied perpendicularly over a unit area $A$. Strain is the unit deformation in any arbitrary unit of length, $\Delta x$. We need to introduce a constant to help us unify the two ideas together, given by the term *modulus*: $stress = modulus \times strain$ No, mental strain is not a form of stress, unless your brain is actively deforming! By the way, the force at which the object begins plastic deformation is known as the **yield strength**, $S_{y}$, and the force at which it ruptures (completely breaks) is known as the **ultimate strength** $S_u$ - quite the collection of badass names if I'd say. ## Strain & Solid Compression - Tensile Stress >Tensile stress occurs when an object experiences a force that leads to linear stretching and deformation. ![[Pasted image 20231008181113.png]] Let's go back to the chair leg example from before, treating it as a rod, when a compressive is applied every part of the rod actually experiences a strain, or linear deformation. Diagram above if you need it! Therefore, strain can be interpreted as dimensionless. Therefore, the modulus of the stress on the object will have the same unit as the stress, giving us the equation: $\frac{F}{A} = E \frac{\Delta L}{L}$ Where: - $E$ is the Young's Modulus - $\frac{\Delta L}{L}$ is the ratio of the change in the rod's length with the original length - $F$ is the compressive/stretching force - $A$ is the *effective area* - or perpendicular area the force acts on. *Note that this change in length is given as a percentage*. Note that this basically means the stress on the object is the pressure we exert on it! To measure this, a **strain gage** is used, which has varying electrical properties dependant on strain. ## Shear - Shearing A **shearing force** dictates a force that acts parallel to the plane of the object, not perpendicular. This leads to a deformation in the object that increases its *width*, with a net change of $\Delta x$. Now, instead of a ratio with the changes in length, the change in width is used as a numerator instead, giving us: $\frac{F}{A} = G \frac{\Delta x}{L}$ Ankle twists are actually a prime example of this - when you slide on grass, eventually the force on your tendons is so great that they can't handle it anymore and they "rupture" - or deform. Thank god for a self-repairing body! Just wish collagen creation was a natural process :(. %%diagram! this one isn't obvious%% *Note that this change in width is also given as a percentage*. ### Fluid Compression - Compressive Stress >[!Tip]- Mixing Young's Modulus and the Bulk Modulus up >As the title says, remember that the bulk modulus is only effective for objects which are being compressed by a [[Fluid Dynamics Basics (Mechanics)|fluid]]! Pretty easy to mix this up, given the names. ![[Pasted image 20230911145633.png]] Imagine a diver, flippers on, respirator attached, the weight of three atmospheres on their back ... it's a marvel that we've been able to transcend much of our biological engineering by combining our minds together, though perhaps we shouldn't continue to test nature - nitrogen narcosis is a bit of a problem after all! [^3]Note that three atmospheres of pressure act on their body - corresponding to a depth of 20 metres. We'll be using a large submersible [^4] - this time unmanned, with its hull almost completely filled with devices and experimental tools. This time, the pressure actually takes the form of stress, being a compressive force on the edges of the sub. Meanwhile, the submersible experience a strain given by a resultant change in the **volume** of the object, giving us a strain ratio of $\frac{\Delta V}{V}$. The equation for bulk modulus is as shown: $p = B \frac{\Delta V}{V}$ Where: - $p$ refers to the pressure of the fluid medium on the object - $B$ is the bulk modulus - given with the same units as pressure - $V$ is the original volume of the object *Note that this change in volume is also given as a percentage*. [^3]: Did I mention that I dive? Absolutely love it - though it can get a little scary with the currents sometime. [^4]: Not a *Titan* reference. Trust me. --> make sure that if there is a *beam with a cross section* the initial length is taken at the middle of the beam, while the extension is taken further outwards! # Example Problems >[!Example]- Module 12-2 Problem 35 >A cubical box is filled with sand and weighs 890 N. We wish to “roll” the box by pushing horizontally on one of the upper edges. > >(a) What minimum force is required? >(b) What minimum coefficient of static friction between box and floor is required? >(c) If there is a more efficient way to roll the box, find the smallest possible force that would have to be applied directly to the box to roll it. (Hint: At the onset of tipping, where is the normal force located?) > >**Solution a)** > >![[Pasted image 20230910112714.png]] > >When you push something at an angle, the normal force is not going to be acting across the object as the object probably won't be entirely in contact with the surface. To make the object roll, we're going to need to have a clockwise moment higher than the moment caused by the normal force (in the diagram) - where we let the length of the object be equal to L: > >$\overrightarrow{F}\times \frac{L}{2} = 890 N \times \frac{L}{2}$ >$\overrightarrow{F} = 890N$ > >**Solution b)** > >c) i'm going to hold off on this one... >[!Example]- Module 12-3 Problem 43 >A horizontal aluminium rod 4.8 cm in diameter projects 5.3 cm from a wall. A 1200 kg object is suspended from the end of the rod. The shear modulus of aluminium is 3.0$\times 10^{10}$ N/m2. Neglecting the rod’s mass, find >a) the shear stress on the rod and >b) the vertical deflection of the end of the rod. > >**Solution a)** >Let's practice using the shear stress equation! Remember that it is: >$\frac{F}{A} = G \frac{\Delta x}{L}$ >We only want to know the leftmost term - so we can just break it down into its constituents where the force is the weight of the object $mg$ and the area is the open top $\pi r^2$: >$\frac{F}{A} = \frac{mg}{\pi r^2} = \frac{1200\times9.81}{\pi 0.048^2} = 6.5 \times 10^6 Nm^{-2}$ >**Solution b)** >We know the shear modulus, we can use the shear modulus equation to get our answer ($\Delta x$): >$\frac{F}{A} = G \frac{\Delta x}{L}$ >$6.5 \times 10^6 Nm^{-2} = 3.0 \times 10^{10} \left( \frac{\Delta x}{0.053} \right) \approx 1.15 \times 10^{-5}$ >[!Example]- Module 12-3 Problem 47 > # What's Next? ...and you're done with the fundamentals! Congratulations, this is almost 20000 words of content that you've absorbed! The next step is to head to the Gravitation chapter, go here to do so: [[Newton's Law of Gravitation - The Intro]] Or you could just navigate back to the main page by going here: [[The (Incomplete) Physics Almanac]]