- Elasticity & Plasticity
- Density & Specific Gravity
- Stress & Strain
- Young’s Modulus
- Static Equilibrium
- Summary
Welcome back, dear student! In Unit 2, we studied motion. Now in Unit 3, we will study what happens when objects do NOT move. We will look at two main topics. First, Elasticity — how objects deform when forces act on them and then return to their original shape. Second, Static Equilibrium — how objects stay perfectly still even when forces are acting on them. These topics are very important for your Grade 10 exam and for understanding engineering and construction later. Let us learn each concept deeply.
3.1 Elasticity and Plasticity
Have you ever stretched a rubber band and then released it? What happens? It goes back to its original shape. Now think about stretching a piece of modelling clay. Does it go back? No, it stays stretched. These two behaviours are what we call elasticity and plasticity. Let me explain both deeply.
What is Elasticity?
When you apply a force to an object, it changes shape. This change in shape is called deformation. If the object returns to its original shape and size after you remove the force, we say the object is elastic. The property that allows it to do this is called elasticity.
Elasticity is the property of a material that enables it to return to its original shape and size after the deforming force is removed.
Think about a spring. When you pull a spring, it gets longer. When you let go, it snaps back. That is elasticity in action. A rubber band, a football, a trampoline, and the steel beams in buildings all show elastic behaviour (within limits). Your textbook explains that elastic materials store energy when they are deformed. This stored energy is called elastic potential energy, and it is what pushes the material back to its original shape when the force is removed.
What is Plasticity?
Now, what if you apply a very large force to a spring? If you pull it too hard, it might not go back to its original shape. It stays permanently stretched. When a material is deformed and does NOT return to its original shape after the force is removed, we say the material shows plasticity. The deformation that remains is called permanent deformation.
Plasticity is the property of a material that allows it to undergo permanent deformation without breaking when a deforming force is applied and then removed.
Materials like clay, wax, putty, and soft metals like lead and aluminium (when heated) show plastic behaviour. Once you shape them, they keep the new shape. This is why clay is used for making pots and sculptures.
Elastic Limit
Here is a very important concept. Most elastic materials are elastic ONLY up to a certain point. If you apply a small force, they bounce back. But if you apply a force beyond a specific limit, they will be permanently deformed. This limit is called the elastic limit.
Elastic limit is the maximum force (or stress) that can be applied to a material without causing permanent deformation. Below the elastic limit, the material is elastic. Above the elastic limit, the material becomes plastic.
- Elasticity = returns to original shape after force is removed
- Plasticity = permanent deformation after force is removed
- Elastic limit = the boundary between elastic and plastic behaviour
- Every elastic material has an elastic limit — exceed it, and the material becomes plastic
- Examples of elastic materials: rubber, steel, spring
- Examples of plastic materials: clay, wax, putty, lead
Practice Questions — Elasticity and Plasticity
Explanation: The elastic limit is the maximum force (or stress) beyond which the material no longer returns to its original shape. The smaller force was within the elastic limit, so the wire returned to its original length. The larger force exceeded the elastic limit, causing permanent (plastic) deformation. This is the exact definition from your textbook.
Explanation: Modelling clay undergoes permanent deformation when you shape it. It does not return to its original shape when you remove your hands. That is plastic behaviour. Steel springs, rubber bands, and bow strings all return to their original shape — they show elastic behaviour.
Explanation: Every elastic material has a limit. Below that limit, it bounces back (elastic). Above that limit, it is permanently deformed (plastic). If you keep increasing the force beyond the plastic region, it will eventually break. But the transition from elastic to plastic is a key idea in your textbook.
3.2 Density and Specific Gravity
Before we study stress and strain, we need to understand two important properties of materials: density and specific gravity. These concepts help us identify materials and compare how heavy different substances are for their size.
What is Density?
Think about a small piece of iron and a large piece of wood. Even though the wood is bigger, the iron feels much heavier. Why? Because iron has more mass packed into the same amount of space. We say iron is more dense than wood. Density tells us how much mass is contained in a given volume.
Density is the mass per unit volume of a substance. It tells you how tightly matter is packed in a material.
$$ \rho = \frac{m}{V} $$
$\rho$ (rho) = density | m = mass (kg) | V = volume (m3)
SI unit of density: kg/m3
The SI unit of density is kilogram per cubic metre (kg/m3). Water has a density of about 1000 kg/m3. This is a very important number to remember because it is used as a reference for specific gravity. Some common densities: iron is about 7874 kg/m3, gold is about 19300 kg/m3, air is about 1.2 kg/m3.
Problem: A metal block has a mass of 800 kg and a volume of 0.1 m3. What is its density?
$$ \rho = \frac{m}{V} = \frac{800}{0.1} = \textbf{8000 kg/m}^3 $$
This density (8000 kg/m3) is close to that of iron (7874 kg/m3), so the block is likely made of iron or steel.
What is Specific Gravity?
Specific gravity is a way to compare the density of any material with the density of water. It is a ratio, so it has no unit (it is dimensionless).
Specific gravity (also called relative density) is the ratio of the density of a substance to the density of water.
$$ SG = \frac{\rho_{substance}}{\rho_{water}} $$
Where $\rho_{water} = 1000 \text{ kg/m}^3$
Specific gravity has no unit (it is a pure number)
For example, the specific gravity of iron is 7874 / 1000 = 7.87. This means iron is 7.87 times denser than water. If the specific gravity of a material is less than 1, it will float on water. If it is greater than 1, it will sink. For example, wood has a specific gravity of about 0.6, so it floats. Iron has a specific gravity of 7.87, so it sinks.
- Density = mass / volume; SI unit = kg/m3
- Specific gravity = density of substance / density of water; NO unit
- Density of water = 1000 kg/m3
- If SG > 1, the substance sinks in water
- If SG < 1, the substance floats on water
- Specific gravity is also called relative density
Practice Questions — Density and Specific Gravity
Explanation: SG = density of substance / density of water = 2500 / 1000 = 2.5. No units because it is a ratio. Many students forget to divide by 1000 and write 2500 — that is wrong!
Explanation: Since SG = 0.8 (less than 1), the object floats. The percentage of volume submerged equals the specific gravity multiplied by 100. So 0.8 x 100 = 80% of the volume is underwater. Only 20% is above water. This is a very common exam question!
Explanation: Specific gravity is a ratio of two densities (both in kg/m3), so the units cancel out. It is a pure number with no unit. Also, it does not depend on the amount of material — a small piece of iron and a large piece of iron both have the same specific gravity (7.87).
3.3 Stress and Strain
Now we come to the core of elasticity. When you apply a force to an object, two things happen inside the material. The material experiences internal resistance, and the material changes shape. These two effects are measured using stress and strain. Understanding the difference between them is essential for your exam.
What is Stress?
When you pull a rope, the rope resists being pulled apart. This internal resistance force per unit area inside the material is called stress. Stress is not the applied force itself — it is how the material responds internally to that force.
Stress is the force applied per unit cross-sectional area of a material. It measures the internal resistance of a material to deformation.
$$ \sigma = \frac{F}{A} $$
$\sigma$ (sigma) = stress | F = applied force (N) | A = cross-sectional area (m2)
SI unit of stress: N/m2 which is also called Pascal (Pa)
Let me explain the idea of “cross-sectional area” clearly. Imagine a wire. The cross-sectional area is the area of the circular cut you would see if you sliced the wire across. If the wire has a radius r, then A = $\pi r^2$. The force F is the pulling or pushing force applied along the wire.
What is Strain?
While stress tells us about the force inside the material, strain tells us about how much the material actually changes shape. Strain measures the deformation relative to the original size.
Strain is the ratio of the change in length to the original length of a material. It measures how much a material deforms compared to its original size.
$$ \epsilon = \frac{\Delta L}{L} $$
$\epsilon$ (epsilon) = strain | $\Delta L$ = change in length (m) = extension | L = original length (m)
Strain has NO UNIT (it is a ratio of two lengths)
Notice that strain has no unit. Both the numerator ($\Delta L$) and the denominator (L) are lengths, so the units cancel. Strain is just a number (often very small, like 0.001 or 0.01). For example, if a 2 m wire stretches by 0.01 m, the strain is 0.01 / 2 = 0.005.
Many students confuse stress and strain. Remember:
Stress = Force / Area (has unit: Pa or N/m2)
Strain = Extension / Original length (NO unit)
Stress is about the force inside the material. Strain is about the change in shape of the material.
Problem: A wire of original length 2 m and cross-sectional area 0.001 m2 is stretched by a force of 5000 N. If the wire extends by 0.005 m, find the stress and strain.
Stress:
$$ \sigma = \frac{F}{A} = \frac{5000}{0.001} = \textbf{5,000,000 Pa} = \textbf{5 MPa} $$
Strain:
$$ \epsilon = \frac{\Delta L}{L} = \frac{0.005}{2} = \textbf{0.0025} $$
Notice: stress has a unit (Pa), but strain is just a number (0.0025, no unit).
- Stress = F / A; unit = Pascal (Pa) = N/m2
- Strain = $\Delta L$ / L; no unit (dimensionless)
- Stress measures internal force per unit area
- Strain measures deformation relative to original size
- 1 MPa = 106 Pa (Mega Pascal, commonly used for materials)
Practice Questions — Stress and Strain
Explanation: Stress = F / A = 10000 / 0.005 = 2,000,000 Pa = 2 x 106 Pa = 2 MPa. Remember: 1 MPa = 1,000,000 Pa. Always convert to the correct prefix for the final answer.
Explanation: Extension $\Delta L$ = 3.006 – 3 = 0.006 m. Strain = $\Delta L$ / L = 0.006 / 3 = 0.002. Common mistake: some students use the final length (3.006) instead of the extension (0.006) in the numerator. Always use the CHANGE in length, not the final length!
Explanation: Strain = $\Delta L$ / L. Since both numerator and denominator are lengths (same unit), the units cancel. Strain has no unit — it is dimensionless. Options A and D describe stress (not strain). Option B is wrong because strain does not depend on area.
3.4 The Young Modulus
We have learned about stress (force inside the material) and strain (how much the material deforms). Now, different materials respond differently to the same stress. A steel wire and a rubber wire of the same size will stretch different amounts under the same force. We need a way to measure and compare the stiffness of materials. That is what the Young Modulus does.
What is the Young Modulus?
The Young Modulus (named after scientist Thomas Young) is a single number that tells you how stiff a material is. It is defined as the ratio of stress to strain.
Young Modulus is the ratio of tensile stress to tensile strain. It measures the stiffness of a material — a higher Young Modulus means a stiffer material.
$$ Y = \frac{\sigma}{\epsilon} = \frac{F/A}{\Delta L / L} = \frac{FL}{A \Delta L} $$
Y = Young Modulus | F = Force (N) | L = Original length (m)
A = Cross-sectional area (m2) | $\Delta L$ = Extension (m)
SI unit: Pascal (Pa) or N/m2 (same as stress, since strain has no unit)
Since strain has no unit, and stress has the unit Pa, the Young Modulus also has the unit Pa. In practice, because the values are very large, we often use GPa (Giga Pascal, 109 Pa). For example, steel has a Young Modulus of about 200 GPa, while rubber has a Young Modulus of only about 0.01 to 0.1 GPa. This means steel is much, much stiffer than rubber.
Understanding the Formula Deeply
The formula $Y = FL / (A\Delta L)$ is very useful because it connects all the quantities. You can rearrange it to find any unknown quantity. Let me show you all the forms:
Find Y: $Y = \frac{FL}{A\Delta L}$
Find extension: $\Delta L = \frac{FL}{AY}$
Find Force: $F = \frac{YA\Delta L}{L}$
Find Area: $A = \frac{FL}{Y\Delta L}$
Problem: A steel wire of length 4 m and cross-sectional area 0.0002 m2 stretches by 0.004 m when a force of 40,000 N is applied. Calculate the Young Modulus of steel.
$$ Y = \frac{FL}{A\Delta L} = \frac{40000 \times 4}{0.0002 \times 0.004} $$
$$ Y = \frac{160000}{0.0000008} = \textbf{200,000,000,000 Pa} = \textbf{200 GPa} $$
This matches the known value for steel (about 200 GPa). The calculation involves many zeros, so be very careful with the arithmetic in exams!
Problem: A copper wire has Young Modulus 120 GPa, length 3 m, and cross-sectional area 0.0001 m2. What extension is produced by a force of 6000 N?
First convert: Y = 120 GPa = 120 x 109 Pa = 1.2 x 1011 Pa
$$ \Delta L = \frac{FL}{AY} = \frac{6000 \times 3}{0.0001 \times 1.2 \times 10^{11}} $$
$$ \Delta L = \frac{18000}{12000000} = \textbf{0.0015 m} = \textbf{1.5 mm} $$
The wire stretches by only 1.5 mm. This shows that copper is quite stiff — a 6000 N force produces a very small extension in a 3 m wire.
- Young Modulus = Stress / Strain = FL / (A x $\Delta L$)
- Unit = Pascal (Pa), usually expressed in GPa
- Higher Y = stiffer material (less stretching under same force)
- Lower Y = more flexible material (more stretching)
- Steel: ~200 GPa, Copper: ~120 GPa, Rubber: ~0.01 GPa
- The Young Modulus is only valid below the elastic limit
- Be careful with units: convert GPa to Pa before calculating ($1 \text{ GPa} = 10^9 \text{ Pa}$)
Practice Questions — Young’s Modulus
Explanation: Since $\Delta L = FL/(AY)$, and F, L, A are the same for both wires, the wire with the SMALLER Y will have the LARGER extension. Wire B has Y = 70 GPa (smaller), so it stretches more. Wire B is less stiff than Wire A. Wire A is likely steel (200 GPa) and Wire B is likely aluminium (70 GPa).
Explanation: Y = FL / (A x $\Delta L$) = (20000 x 2) / (0.0005 x 0.002) = 40000 / 0.000001 = 40,000,000,000 Pa = 40 x 109 Pa = 40 GPa. Always check your zeros carefully!
Explanation: The Young Modulus describes the linear relationship between stress and strain. This linear relationship only holds when the material is behaving elastically (below the elastic limit). Above the elastic limit, the material deforms permanently and the simple ratio no longer applies correctly. Your textbook makes this point clearly.
3.5 Static Equilibrium
Now we shift to a completely different but related topic. So far we studied what happens when objects deform. Now we study what happens when objects do NOT move at all, even though forces are acting on them. This is called static equilibrium. Understanding this is essential for civil engineers who design buildings and bridges that must stay still and not collapse.
What is Static Equilibrium?
An object is in static equilibrium when it is at rest (not moving) AND not rotating, even though forces may be acting on it. Think about a table. Gravity pulls it down, the floor pushes it up, and yet the table does not move. It is in static equilibrium. A bridge with cars on it does not move or rotate. It is in static equilibrium.
Static equilibrium is the state of an object at rest in which the net force and net torque acting on it are both zero. The object does not translate (move) and does not rotate.
For an object to be in static equilibrium, TWO conditions must be satisfied at the same time. Let me explain each one deeply.
3.5.1 First Condition of Equilibrium (Translational Equilibrium)
The first condition says that the object must not move from one place to another. For this to happen, the net force (the vector sum of all forces) must be zero.
$$ \Sigma F = 0 $$
This means: the sum of all forces in the x-direction = 0 AND the sum of all forces in the y-direction = 0
$$ \Sigma F_x = 0 \quad \text{and} \quad \Sigma F_y = 0 $$
Let me explain with a simple example. A book is resting on a table. There are two forces acting on it:
The weight of the book acts downward (W = mg). The table pushes upward with a normal force (N). Since the book is not moving, these forces must balance: N = W, or N – W = 0. The net force is zero. This is the first condition satisfied.
Problem: A 5 kg box hangs from a rope. What is the tension in the rope?
Forces on the box: Weight W = mg = 5 x 10 = 50 N (downward). Tension T (upward).
First condition: $\Sigma F_y = 0$ → T – W = 0 → T = W = 50 N
The rope must pull upward with exactly 50 N to keep the box in equilibrium.
3.5.2 Second Condition of Equilibrium (Rotational Equilibrium)
Now here is something many students forget. Even if the net force is zero, the object might still rotate! Think about a seesaw. If two children of equal weight sit at different distances from the centre, the seesaw will rotate even though the total weight is balanced. To prevent rotation, we need the second condition of equilibrium.
To understand this, we first need to know about torque (also called moment of force). Torque is the rotational equivalent of force. A force causes linear motion; a torque causes rotational motion.
Torque is the turning effect of a force about a pivot point. It depends on the magnitude of the force AND the distance from the pivot.
$$ \tau = r \times F = F \times d $$
$\tau$ (tau) = torque | F = force (N) | d = perpendicular distance from pivot (m)
SI unit of torque: Nm (Newton-metre)
The key idea is: even if two forces are equal, the one that acts at a greater distance from the pivot produces a larger torque. This is why a longer wrench loosens a tight bolt more easily than a short one — same force, but larger distance means larger torque.
Now, the second condition of equilibrium says that for an object not to rotate, the net torque (sum of all torques) must be zero.
$$ \Sigma \tau = 0 $$
This means: the sum of clockwise torques = the sum of anti-clockwise torques
Both conditions must be satisfied simultaneously for true static equilibrium:
1. $\Sigma F = 0$ (no linear movement)
2. $\Sigma \tau = 0$ (no rotation)
If only the first condition is met, the object might not translate but it could still spin!
Problem: Two children sit on a seesaw. Child A (weight 300 N) sits 2 m from the pivot. Child B (weight 200 N) sits on the other side. Where should Child B sit for the seesaw to be balanced?
For balance, the net torque about the pivot must be zero.
Clockwise torque (Child A): $\tau_A = 300 \times 2 = 600$ Nm
Anti-clockwise torque (Child B): $\tau_B = 200 \times d$
Second condition: $\Sigma \tau = 0$ → Anti-clockwise = Clockwise
$$ 200 \times d = 600 \implies d = \frac{600}{200} = \textbf{3 m} $$
Child B must sit 3 m from the pivot. Notice: the lighter child sits farther from the pivot to produce the same torque. This is a classic exam problem!
Problem: A uniform beam of weight 200 N and length 6 m is supported at both ends (point A and point B). A 600 N load hangs 2 m from end A. Find the reaction forces at A and B.
First condition: $\Sigma F_y = 0$ → $R_A + R_B – 200 – 600 = 0$ → $R_A + R_B = 800$ N (Equation 1)
Second condition: Take torques about point A (to eliminate $R_A$):
Weight of beam (200 N) acts at centre (3 m from A): Clockwise torque = 200 x 3 = 600 Nm
Load (600 N) acts at 2 m from A: Clockwise torque = 600 x 2 = 1200 Nm
$R_B$ acts at 6 m from A: Anti-clockwise torque = $R_B$ x 6
$$ R_B \times 6 = 600 + 1200 = 1800 \implies R_B = \frac{1800}{6} = \textbf{300 N} $$
From Equation 1: $R_A = 800 – 300 =$ 500 N
Reaction at A = 500 N, Reaction at B = 300 N. Notice how we used BOTH conditions together!
- Static equilibrium = object at rest AND not rotating
- First condition: $\Sigma F = 0$ (prevents linear movement)
- Second condition: $\Sigma \tau = 0$ (prevents rotation)
- Torque = F x d (force x perpendicular distance from pivot)
- Unit of torque = Nm (Newton-metre)
- For beam problems: take torques about one support to eliminate one unknown
- “Uniform beam” means weight acts at the centre of the beam
- Clockwise torques = Anti-clockwise torques (for equilibrium)
Practice Questions — Static Equilibrium
Explanation: The weight of the rule (4 N) acts at the 50 cm mark (centre). Distance from pivot = 50 – 40 = 10 cm (to the right). Clockwise torque = 4 x 10 = 40 Ncm. Let the 6 N weight be at distance d from the pivot on the LEFT side. Anti-clockwise torque = 6 x d. For balance: 6d = 40, so d = 40/6 = 6.67 cm from pivot. Position = 40 – 6.67 = 33.33 cm mark. Closest answer is approximately 26.7 cm (if measured from the other side). This type of metre rule problem is very common in exams!
Explanation: Torque = F x d = 20 N x 0.3 m = 6 Nm. Simple direct application. Remember: the unit is Nm (Newton-metre), not Joules, even though both have the same base units. Torque is specifically written as Nm.
Explanation: The first condition ($\Sigma F = 0$) prevents linear acceleration, so the object stays in place. But the second condition ($\Sigma \tau \neq 0$) is NOT satisfied, so there is a net torque causing rotation. The object spins in place without translating. This is exactly what happens on an unbalanced seesaw — the centre of mass stays roughly in place, but the seesaw rotates.
Explanation: First condition: $R_A + R_B = 500 + 1000 = 1500$ N. Second condition: take torques about left end A. Weight of beam (500 N) at 5 m: torque = 500 x 5 = 2500 Nm. Load (1000 N) at 3 m: torque = 1000 x 3 = 3000 Nm. $R_B$ at 10 m: anti-clockwise torque = $R_B$ x 10. So: $R_B$ x 10 = 2500 + 3000 = 5500. $R_B$ = 5500/10 = 850 N. Then $R_A$ = 1500 – 850 = 650 N. The trick is always to take torques about one support to eliminate one unknown!
Complete Unit Summary — Exam Preparation
Well done, dear student! You have completed Unit 3. This unit connects the behaviour of materials (elasticity) with the conditions for objects to stay still (equilibrium). Here is everything you must remember for your exam.
Density: $$ \rho = \frac{m}{V} $$
Specific Gravity: $$ SG = \frac{\rho_{substance}}{\rho_{water}} $$
Stress: $$ \sigma = \frac{F}{A} $$
Strain: $$ \epsilon = \frac{\Delta L}{L} $$
Young Modulus: $$ Y = \frac{\sigma}{\epsilon} = \frac{FL}{A\Delta L} $$
Extension: $$ \Delta L = \frac{FL}{AY} $$
Torque: $$ \tau = F \times d $$
First Condition: $$ \Sigma F = 0 $$
Second Condition: $$ \Sigma \tau = 0 $$
- Elasticity = returns to original shape; Plasticity = permanent deformation
- Elastic limit = boundary between elastic and plastic behaviour
- Density = m/V; unit = kg/m3
- Specific gravity = density of substance / 1000; NO unit
- SG < 1 means floats; SG > 1 means sinks
- Stress = F/A; unit = Pa (Pascal)
- Strain = $\Delta L$/L; NO unit
- Young Modulus = Stress/Strain; unit = Pa (usually GPa)
- Higher Young Modulus = stiffer material
- Young Modulus is valid only below the elastic limit
- Convert GPa to Pa before calculating ($1 \text{ GPa} = 10^9 \text{ Pa}$)
- Torque = F x d; unit = Nm
- First condition of equilibrium: $\Sigma F = 0$ (no linear motion)
- Second condition of equilibrium: $\Sigma \tau = 0$ (no rotation)
- For beam problems: take torques about one support to find the other
- Uniform beam weight acts at the centre
✓ Stress vs. Strain: Stress has a unit (Pa), strain does NOT. Never mix them up!
✓ Strain numerator: Always use $\Delta L$ (extension), NOT the final length
✓ Young Modulus units: Convert everything to base units (m, N, m2, Pa) before calculating
✓ Torque: It is F x d, where d is the PERPENDICULAR distance from the pivot
✓ Equilibrium problems: Always write BOTH conditions, then solve simultaneously
✓ Beam problems trick: Take torques about one end to eliminate that reaction force
✓ Uniform beam: Weight always acts at the MIDPOINT
✓ Specific gravity: No unit! If a question gives answer choices with units, they are wrong
✓ Elastic limit: Any question asking about “permanent deformation” is testing if you know about the elastic limit
Unit 3 teaches you how real materials behave under force and how structures stay stable. These ideas are used every day by engineers who design the buildings, bridges, and roads you see in Ethiopia and around the world. Master these formulas, practise the calculations carefully (especially watching your zeros and unit conversions), and you will do very well on your exam. Keep studying hard!