Static Equilibrium, Elasticity, and Gravitation

Big Picture

An object is in static equilibrium when it neither translates nor rotates — all forces and torques cancel. Elasticity describes how materials deform under stress and return to shape (or break). Gravitation is the universal attractive force between all masses, following an inverse-square law. Together these topics connect everyday structural safety to the motion of the cosmos.

ELI-10: Explain It Like I'm 10

A book sitting on a table is in equilibrium — gravity pulls down, the table pushes up equally, and nothing twists. That is the "no motion, no rotation" rule. Elasticity is why a rubber band snaps back but a paperclip stays bent. Gravity is why you stay on the ground, why the Moon orbits Earth, and why Earth orbits the Sun — everything pulls on everything else.

6.1 Conditions for Static Equilibrium

Core Idea

For an object to be in static equilibrium:

1. Translational equilibrium: \(\sum \mathbf{F} = 0\) (net force is zero).
2. Rotational equilibrium: \(\sum \tau = 0\) (net torque about any axis is zero).

Both conditions must hold. You can choose the pivot point for torque calculations freely — a smart choice simplifies the problem (e.g., place the pivot at an unknown force to eliminate it from the torque equation).

Worked Example 1: Ladder Against a Wall

Problem: A 5.0 m uniform ladder of mass 12 kg leans against a frictionless wall at 60° to the ground. Find the friction force at the base needed to prevent slipping.

Solution: Choose the base as pivot. Forces: weight \(mg\) at center (2.5 m along ladder), normal from wall \(N_w\) at top, friction \(f\) at base, normal from ground \(N_g\) at base. Torques about base: \(mg(2.5)\cos 60^\circ - N_w(5.0)\sin 60^\circ = 0\). Solve: \(N_w = \frac{mg(2.5)\cos 60^\circ}{5.0\sin 60^\circ} \approx 34\ \text{N}\). Force balance: \(f = N_w = 34\ \text{N}\).

Worked Example 2: Beam Supported at Two Points

Problem: A uniform beam of mass 20 kg and length 4.0 m rests on two supports, one at each end. A 15 kg box sits 1.0 m from the left end. Find the upward normal force at each support.

Step 1 — Draw the free-body diagram. Forces acting on the beam:

• Weight of beam: \(W_b = m_b g = 20 \times 9.8 = 196\ \text{N}\), acting at the center (2.0 m from left).
• Weight of box: \(W_{\text{box}} = 15 \times 9.8 = 147\ \text{N}\), acting 1.0 m from left.
• Normal force at left support: \(N_L\) (unknown, upward).
• Normal force at right support: \(N_R\) (unknown, upward).

Step 2 — Choose a strategic pivot. Place the pivot at the left support. This eliminates \(N_L\) from the torque equation because its lever arm is zero there.

Step 3 — Apply rotational equilibrium \(\sum \tau = 0\). Taking counterclockwise as positive:

\[ -196(2.0) - 147(1.0) + N_R(4.0) = 0 \]

\[ N_R(4.0) = 196(2.0) + 147(1.0) = 392 + 147 = 539 \]

\[ N_R = \frac{539}{4.0} = 134.75 \approx 135\ \text{N} \]

Step 4 — Apply translational equilibrium \(\sum F_y = 0\).

\[ N_L + N_R - 196 - 147 = 0 \]

\[ N_L = 196 + 147 - 134.75 = 208.25 \approx 208\ \text{N} \]

Check: The left support bears more weight because the box is closer to it — physically sensible.

Why this method is powerful: The choice of pivot at the left support made the algebra far simpler. If we had chosen the center of the beam, both \(N_L\) and \(N_R\) would appear in the torque equation, requiring a system of two equations. The "smart pivot" trick — placing the pivot at an unknown force — is the single most useful strategy in statics problems.

ELI-10: Explain It Like I'm 10

A ladder does not fall because all pushes and twists cancel out. Gravity tries to rotate the ladder flat; friction at the base and the wall's push resist that. If friction is too weak, the twist from gravity wins, and the ladder slides out. Choosing a smart pivot — like the point where the ladder touches the ground — makes the math easier because forces acting right at the pivot create zero twist there.

The beam example shows the same idea. By putting the pivot right under one support, that support's force drops out of the twist equation entirely. Then you only need one equation to find the other support's force. It is like solving a puzzle by standing in exactly the right spot.

6.2 Elasticity

Core Idea

Materials deform under stress. The relationship between stress (force per area) and strain (fractional deformation) is described by elastic moduli.

Physics and Mathematics

• Stress = \(\frac{\text{force}}{\text{area}}\). SI unit: pascal (Pa = N/m²).
• Strain = fractional change in dimension (dimensionless).

Young's modulus \(Y\) (tensile/compressive): \[ \frac{F}{A} = Y\frac{\Delta L}{L_0} \]

Shear modulus \(S\): \(\frac{F}{A} = S\frac{\Delta x}{L_0}\)

Bulk modulus \(B\): \(\Delta P = -B\frac{\Delta V}{V_0}\)

Higher modulus → stiffer material. Steel has a much higher Young's modulus than rubber.

Worked Example: Stretching a Steel Wire

Problem: A steel wire of diameter 2.0 mm and original length 3.0 m is used to hang a 50 kg mass. How much does the wire stretch? (Young's modulus for steel: \(Y = 2.0 \times 10^{11}\ \text{Pa}\))

Step 1 — Identify the force. The tension in the wire equals the weight of the hanging mass: \[ F = mg = 50 \times 9.8 = 490\ \text{N} \]

Step 2 — Calculate the cross-sectional area. The wire is cylindrical with radius \(r = 1.0\ \text{mm} = 1.0 \times 10^{-3}\ \text{m}\): \[ A = \pi r^2 = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6}\ \text{m}^2 \]

Step 3 — Compute the stress.

\[ \text{Stress} = \frac{F}{A} = \frac{490}{3.14 \times 10^{-6}} = 1.56 \times 10^8\ \text{Pa} \]

Step 4 — Apply Young's modulus to find strain.

\[ \frac{\Delta L}{L_0} = \frac{\text{Stress}}{Y} = \frac{1.56 \times 10^8}{2.0 \times 10^{11}} = 7.8 \times 10^{-4} \]

Step 5 — Compute the absolute stretch.

\[ \Delta L = \left(7.8 \times 10^{-4}\right) \times 3.0 = 2.34 \times 10^{-3}\ \text{m} = 2.3\ \text{mm} \]

Interpretation: A 3-meter steel wire supporting a 50 kg mass stretches by only about 2 millimeters — about the thickness of a coin. This tiny stretch, invisible to the naked eye, is why we need precise instruments to measure elastic deformation in stiff materials. The same force on a rubber band of equal dimensions would produce a far larger stretch because rubber's Young's modulus is about \(10^7\) Pa — roughly 20,000 times smaller.

Proportional Reasoning with Elasticity

You can often skip full calculations by reasoning proportionally. From \(\Delta L = \frac{FL_0}{AY}\):

• Double the force → double the stretch (for the same wire).
• Double the length → double the stretch (same force and area).
• Double the cross-sectional area → half the stretch (thicker wire stretches less).
• Double the diameter → area quadruples (area ∝ \(d^2\)), so stretch is one-quarter.

This kind of proportional reasoning is a powerful problem-solving shortcut, used throughout physics.

ELI-10: Explain It Like I'm 10

Pull on a rubber band — it stretches a lot. Pull on a steel wire with the same force — it barely stretches. Young's modulus is the "stiffness number." Rubber has a tiny modulus; steel has a huge one. Everything stretches a little when pulled; you just cannot see it with stiff materials. Stress is how hard the material is being pulled per unit area; strain is how much it stretches as a fraction of its original length.

6.3 Newton's Law of Universal Gravitation

Core Idea

Every mass attracts every other mass with a force proportional to the product of the masses and inversely proportional to the square of the distance between their centers.

Physics and Mathematics

\[ F_g = G\frac{m_1 m_2}{r^2} \]

Where:

• \(F_g\) = gravitational force (N)
• \(G = 6.67 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2\) (universal gravitational constant)
• \(m_1, m_2\) = masses (kg)
• \(r\) = distance between centers (m)

Near Earth's surface, this reduces to \(F_g = mg\) with \(g = GM_E/R_E^2 \approx 9.8\ \text{m/s}^2\).

Deriving \(g = 9.8\) from Universal Gravitation

The connection between the local \(g\) and the universal \(G\) is one of physics' most elegant unifications. Here is the explicit derivation:

Start with Newton's law for an object of mass \(m\) at Earth's surface, distance \(R_E\) from Earth's center:

\[ F_g = G\frac{M_E m}{R_E^2} \]

But near Earth's surface, we also write the weight as \(F_g = mg\). Equating the two:

\[ mg = G\frac{M_E m}{R_E^2} \]

Cancel \(m\) (all objects fall at the same rate in a vacuum — Galileo's insight, now explained by Newton):

\[ g = \frac{GM_E}{R_E^2} \]

Plug in values: \(M_E = 5.97 \times 10^{24}\ \text{kg}\), \(R_E = 6.37 \times 10^6\ \text{m}\), \(G = 6.67 \times 10^{-11}\):

\[ g = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^6)^2} = \frac{3.98 \times 10^{14}}{4.06 \times 10^{13}} \approx 9.80\ \text{m/s}^2 \]

This is not a coincidence — it is a derived result. \(g = 9.8\ \text{m/s}^2\) is simply what Newton's universal law predicts for Earth's particular mass and radius. On the Moon, \(g\) is different (≈1.6 m/s²) because \(M\) and \(R\) are different — but the same \(G\) applies everywhere.

Proportional Reasoning with the Inverse-Square Law

The inverse-square relationship \(F \propto 1/r^2\) is one of the most important patterns in physics. Mastering proportional reasoning with it saves time and builds intuition:

Why \(1/r^2\) specifically? The force spreads out over the surface of a sphere. A sphere's surface area grows as \(4\pi r^2\), so the force per unit area — the intensity — falls as \(1/r^2\). The same geometry explains why light dims, sound fades, and electric forces weaken with distance.

Worked proportional reasoning: If the gravitational force between two masses is 40 N when they are 2.0 m apart, what is the force when they are moved to 6.0 m apart?

Distance triples (\(2.0 \to 6.0\) m), so force is divided by \(3^2 = 9\):

\[ F_{\text{new}} = \frac{40}{9} \approx 4.4\ \text{N} \]

No need to plug into the full formula — the ratio method is faster and reveals the physics more clearly.

ELI-10: Explain It Like I'm 10

Everything pulls on everything else. You pull on the Earth just as hard as the Earth pulls on you — but you are so much lighter that you move, while Earth barely budges. The pull gets weaker very quickly with distance: if you double the distance, the pull becomes one-quarter as strong. This inverse-square pattern shows up in gravity, electric forces, light brightness, and sound intensity.

6.4 Gravitational Potential Energy

Core Idea

The universal form of gravitational potential energy (valid at any distance, not just near Earth's surface):

\[ U = -G\frac{m_1 m_2}{r} \]

Why negative? The zero is chosen at infinite separation (\(r \to \infty, U \to 0\)). As objects fall together, \(U\) becomes more negative (energy decreases), and kinetic energy increases. The sign is a convention — only changes in \(U\) are physical.

ELI-10: Explain It Like I'm 10

Think of gravity as a deep hole. At the bottom of the hole (close together), the potential energy is very negative — you need to add a lot of energy to climb out. Far away (the "top"), it is nearly zero. The negative sign just means you measure from the top of the hole, not the bottom.

6.5 Orbital Mechanics

Core Idea

An orbit is a continuous free fall. A satellite "falls" toward Earth but has enough tangential speed that Earth curves away beneath it.

Circular orbit speed: \(v = \sqrt{\frac{GM}{r}}\)

Escape velocity: \(v_{\text{esc}} = \sqrt{\frac{2GM}{r}}\) — the minimum speed needed to escape a body's gravity entirely.

Kepler's Laws:

1. Orbits are ellipses with the central body at one focus.
2. A line from planet to Sun sweeps equal areas in equal times (faster when closer).
3. \(T^2 \propto r^3\) — the square of the orbital period is proportional to the cube of the semi-major axis.

Why an Orbit Is Continuous Free Fall

This is not a metaphor — it is literal physics. Consider a satellite in a circular orbit at radius \(r\) from Earth's center. At every instant:

1. Gravity pulls the satellite toward Earth's center. The gravitational force \(F_g = GMm/r^2\) provides the centripetal force: \(GMm/r^2 = mv^2/r\).

1. The satellite has tangential velocity \(v\). This sideways speed is what keeps it from falling straight down. In the time the satellite "falls" a distance \(h\) toward Earth, it also moves forward a distance \(vt\).

1. Earth curves away. Because Earth is a sphere, the surface drops by exactly \(h\) over that same horizontal distance. The satellite falls exactly as much as the ground curves away beneath it.

Newton's cannon thought experiment (from Principia, 1687): Imagine firing a cannonball horizontally from a mountain top. At low speed, it travels a short distance and hits the ground. At higher speed, it travels farther before impact (the ground curves away slightly). At the critical orbital speed (~7.8 km/s for low Earth orbit), the cannonball falls toward Earth at the exact rate that Earth's spherical surface recedes — it never hits the ground. That cannonball is in orbit.

This is why astronauts on the ISS experience "weightlessness." They are not beyond Earth's gravity — at the ISS altitude (~400 km), gravity is still about 89% of surface gravity (\(g \approx 8.7\ \text{m/s}^2\)). They feel weightless because they and their spacecraft are falling together at the same rate. This is the same principle behind the "vomit comet" aircraft that creates brief periods of free fall for astronaut training.

Worked Example: Orbital Speed of the ISS

Problem: The International Space Station orbits at an altitude of approximately 400 km above Earth's surface. What is its orbital speed?

Given: \(M_E = 5.97 \times 10^{24}\ \text{kg}\), \(R_E = 6.37 \times 10^6\ \text{m}\), \(G = 6.67 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2\).

Step 1 — Find the orbital radius. The radius is measured from Earth's center: \[ r = R_E + h = 6.37 \times 10^6 + 4.00 \times 10^5 = 6.77 \times 10^6\ \text{m} \]

Step 2 — Apply the circular orbit formula.

\[ v = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.77 \times 10^6}} \]

\[ v = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^6}} = \sqrt{5.88 \times 10^7} \approx 7,670\ \text{m/s} = 7.67\ \text{km/s} \]

This is roughly 27,600 km/h (17,100 mph). At this speed, the ISS completes one orbit around Earth in about 93 minutes.

Step 3 — Compute the orbital period using Kepler's third law.

For circular orbits: \(T^2 = \frac{4\pi^2}{GM_E}r^3\)

\[ T = 2\pi\sqrt{\frac{r^3}{GM_E}} = 2\pi\sqrt{\frac{(6.77 \times 10^6)^3}{3.98 \times 10^{14}}} = 2\pi\sqrt{\frac{3.10 \times 10^{20}}{3.98 \times 10^{14}}} = 2\pi\sqrt{7.79 \times 10^5} \]

\[ T \approx 2\pi \times 883 \approx 5,550\ \text{s} \approx 92.5\ \text{minutes} \]

Worked Example: Escape Velocity from Earth

Problem: What is the minimum speed a spacecraft must achieve at Earth's surface to escape Earth's gravity entirely (ignoring air resistance)?

Solution: Escape velocity is derived by setting total mechanical energy to zero at infinity:

\[ \frac{1}{2}mv_{\text{esc}}^2 - G\frac{M_E m}{R_E} = 0 \]

\[ v_{\text{esc}} = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{\frac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}} \]

\[ v_{\text{esc}} = \sqrt{\frac{7.96 \times 10^{14}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} \approx 11,200\ \text{m/s} = 11.2\ \text{km/s} \]

This is about 40,300 km/h (25,000 mph). Note that escape velocity is \(\sqrt{2} \approx 1.414\) times the circular orbital speed at the same radius:

\[ \frac{v{\text{esc}}}{v{\text{orb}}} = \sqrt{\frac{2GM/r}{GM/r}} = \sqrt{2} \]

This ratio is universal — it holds for any central body at any radius, whether Earth, the Moon, or the Sun.

ELI-10: Explain It Like I'm 10

An orbit is not floating — it is falling and missing. Throw a ball horizontally. It curves down and hits the ground. Throw it faster and it goes farther before hitting. Throw it fast enough (about 17,500 mph for Earth) and the ground curves away beneath it at the same rate the ball falls. The ball is now in orbit — forever falling, never landing. The Moon is doing exactly this around Earth.

6.6 Key Assumptions

All three topics in this chapter rest on simplifying assumptions. Recognizing them is essential for knowing when the models apply — and when they break down.

Static Equilibrium Assumptions

1. Rigid body: The object does not deform under load. In reality, every object flexes slightly, but for most engineering problems (beams, bridges, ladders), the deformation is negligible for force and torque calculations.
2. Point of application: Forces act at well-defined points (e.g., weight at center of mass). In reality, distributed loads (like snow on a roof) require integration, but we approximate them as point forces for tractability.
3. Static: The object is genuinely at rest. If it accelerates, \(\sum F = ma\) replaces \(\sum F = 0\), and \(\sum \tau = I\alpha\) replaces \(\sum \tau = 0\).

Elasticity Assumptions

1. Small deformations (Hookean regime): Stress is proportional to strain (\(\sigma = Y\epsilon\)). This linear relationship holds only below the material's elastic limit. Beyond that, permanent (plastic) deformation or fracture occurs.
2. Uniform cross-section and homogeneous material: The equations assume the material is the same throughout and the cross-sectional area does not change along the length.
3. Isothermal conditions: Temperature changes cause thermal expansion/contraction, which is treated separately. The moduli \(Y, S, B\) are temperature-dependent.

Gravitation Assumptions

1. Point masses: Newton's law \(F = Gm_1m_2/r^2\) is exact for point masses. For extended bodies (like planets), it is exact for the force between them only if they are spherically symmetric — the mass can be treated as concentrated at the center (shell theorem, proved by Newton).
2. Inverse-square law holds in classical regime: General relativity modifies gravitation for very strong fields (near black holes) or very high precision (GPS satellites require relativistic corrections).
3. Two-body problem: Orbital formulas like \(v = \sqrt{GM/r}\) assume the central body is much more massive than the orbiting body. When masses are comparable, both bodies orbit their common center of mass.

Topic Summary

• Static equilibrium requires \(\sum\mathbf{F}=0\) AND \(\sum\tau=0\) (about any axis). The "smart pivot" trick — placing the pivot at an unknown force — is the most powerful problem-solving strategy.
• Stress is force/area; strain is fractional deformation. Elastic moduli (\(Y, S, B\)) relate them. Young's modulus for steel is ~\(2 \times 10^{11}\) Pa; for rubber it is ~\(10^7\) Pa.
• Universal gravitation: \(F_g = Gm_1m_2/r^2\). Inverse-square law. Near Earth's surface, this reduces to \(F = mg\) with \(g = GM_E/R_E^2\) — not a coincidence, but a derived result.
• Proportional reasoning with \(1/r^2\): doubling distance quarters the force; tripling distance reduces it to one-ninth.
• Gravitational PE: \(U = -Gm_1m_2/r\). Negative by convention (zero at infinity).
• Orbits are continuous free fall — objects falling at the same rate the ground curves away. \(v{\text{orb}} = \sqrt{GM/r}\), \(v{\text{esc}} = \sqrt{2GM/r}\). Escape velocity is always \(\sqrt{2}\) times orbital speed.
• Kepler's laws describe elliptical orbits and the \(T^2 \propto r^3\) relationship.
• Assumptions matter: Rigid bodies, small deformations, point masses — each model has limits, and knowing them prevents misapplication.

Essential Equations

Concept Check

1. Why can you choose any point as the pivot for torque calculations in equilibrium problems?
2. A steel wire and a rubber band of identical dimensions are stretched by the same force. Which stretches more? Why?
3. If the distance between Earth and the Sun doubled, how would the gravitational force change?
4. Why does \(U = -GMm/r\) have a negative sign? What would happen if \(U\) were zero at Earth's surface?
5. If a satellite's orbital radius increases, does its speed increase or decrease? Why?
6. A 4.0 m uniform beam (mass 30 kg) is supported at both ends. A 25 kg weight sits 1.5 m from the left end. Which support bears more weight, and by how much?
7. A copper wire (Young's modulus \(1.1 \times 10^{11}\) Pa) of diameter 1.5 mm and length 2.0 m supports a 20 kg mass. How much does it stretch?
8. The ISS orbits at ~400 km altitude. If it were moved to an altitude of 1,600 km (4× higher), would its orbital period increase by a factor of 2, 4, or 8? (Hint: use Kepler's third law.)

Open Educational References

• OpenStax, College Physics, Chapters 6, 9: Gravitation, Statics and Elasticity
• OpenStax, University Physics, Volume 1, Chapters 12–13: Equilibrium, Elasticity, Gravitation