Lecture 14: Equilibrium and Elasticity

Course Outline

Center of Gravity
Equilibrium and Stability
Elasticity - Hooke's Law

For a rigid body to be in equilibrium, two conditions must be satisfied:

The vector sum of all forces acting on the body must be zero. $\sum \vec{F_d} = 0$
The vector sum of all torques acting on the body about any axis must be zero. $\sum \vec{\tau} = 0$

Center of Gravity

Average location of the weight of an object. Same as the center of mass for a uniform gravitational field.

\begin{equation} \vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \cdots + m_n\vec{r}_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum m_i\vec{r}_i}{\sum m_i} \end{equation}

Example

A uniform plank of length $L = 6.0\text{m}$ and mass $M = 90\text{kg}$ is supported by two sawhorses separated by $D = 1.5\text{m}$ and equidistant from the center of the plank

A person wants to stand on the right-hand end of the plank. If the plank is to remain in equilibrium, what is the maximum mass of the person?

⚠️

Hint: The person's weight acts at the center of gravity of the person-plank system.

Answer:

The center of gravity equation gives us:

x_{cg} = \frac{M(0) + m\left(\frac{L}{2}\right)}{M + m} = \frac{m}{M + m}\frac{L}{2}

we set $x_{cg} = x_S$ and solve for $m$ :

\begin{align*} \frac{m}{M + m}\frac{L}{2} &= \frac{D}{2}\\ mL &= (M + m)D\\ m &= \frac{MD}{L - D} = \frac{90\text{kg}\times 1.5\text{m}}{6.0\text{m} - 1.5\text{m}} = 30\text{ kg} \end{align*}

Elasticity

Spring Elasticity

A force $\rightarrow$ equilibrium $=$ restoring force
Object that exert a restoring force are elastic
Hooke's Law: $F = -kx$ $\rightarrow$ displaced by $x$ from equilibrium (linear slope)
$k$ is the spring constant (or force constant) $\rightarrow$ stiffness of the spring.
Minus sign indicates that the force is opposite to the displacement $\rightarrow$ restoring force

Example

A spring is attached to a $2.0 \text{ kg}$ block. The other end of the spring is pulled by a motorized toy train that moves forward at $5.0 \text{cm/s}$ .

The spring constant is $50 \text{ N/m}$ and the coefficient of static friction between the block and the surface is $0.60$ .

The spring is at equilibrium length at $t = 0 \text{ s}$ . How far does the spring stretch before the block starts to move? And what the time it takes for the block to start moving?

Answer:

The force of the spring when the toy moves as far as $x$ :

F_s = kx

When the block almost moves, this spring force is same as the static friction between the block and the floor: $(F_s = F_f)$

\begin{align*} F_f = F_s = \mu_s F_N &= \mu_s mg\\ kx &= \mu_s mg\\ x &= \frac{\mu_s mg}{k} = \frac{(0.60)(2.0\text{ kg})(9.8\text{ m/s}^2)}{50\text{ N/m}} = 0.235\text{ m} \end{align*}

Then to calculate the time it takes for the block to start moving, we use the equation of motion:

t = \frac{x}{v} = \frac{0.235\text{ m}}{0.050\text{ m/s}} = 4.7\text{ s}

Work Done by a Spring

Work done by a spring is the area under the force-displacement curve.

\begin{align} W &= \int F ds\nonumber\\ W &= -\int_{s_i}^{s_f} k(s-s_{eq})^2 ds\nonumber\\ W &= -\frac{1}{2}k(s_f^2 - s_i^2) \Big\rvert_{s_i}^{s_f}\nonumber\\ W &= \frac{1}{2}k\Delta s_{i}^2 - \frac{1}{2}k\Delta s^2_{f} \end{align}

Example

A block with mass $m$ is pushed to a spring until a displacement of $\Delta x$ then released. What is the speed of the block when it passes through the equilibrium position?

Answer:

Using conservation of energy:

\begin{align} \underbrace{\frac{1}{2}k\Delta x^2}_{\text{spring}} + 0 &= 0 + \underbrace{\frac{1}{2}mv^2}_{\text{kinetic}}\nonumber\\ v &= \sqrt{\frac{k}{m}}\Delta x \end{align}

What if the mass is hanging from the spring at equilibrium $y$ and spring constant $k$ ?

using equation above we can use static equilibrium formula:

\begin{align} F_s = F_g &= k\Delta y = mg\nonumber\\ k &= \frac{mg}{\Delta y} \end{align}

⚠️

Elasticity is not the sole property of a spring!

Medium Elasticity

Any solid will be deformed by a force
The graph of force vs. displacement is not linear
- Elastic limit: the point beyond which the material is permanently deformed
- As long as the strech is less than the elastic limit, the material will return to its original shape when the force is removed.
A strech beyond the elastic limit is called plastic deformation and it is permanent.
For most materials, the force is proportional to the displacement up to the elastic limit.

\begin{equation} F = k\Delta L \end{equation}

Tensile Stress and Young Modulus

The elasticity is directly related to the spring constant.
The force pulling each bond is related to the strech:

\begin{equation} \frac{F}{A} = \frac{k\Delta L}{A} = \frac{Y\Delta L}{L} \rightarrow Y = \frac{\text{tensile stress}}{\text{tensile strain}} \end{equation}

Volume Stress and Bulk Modulus

The force per unit area $F/A$ is applied to all surfaces.
The volume strain is defined as the change in volume divided by the original volume: $\Delta V/V$
The bulk modulus is defined as the ratio of the volume stress to the volume strain:

\begin{equation} B = \frac{F/A}{\Delta V/V} \end{equation}

Shear Stress

The shear strain is defined as the change in angle $\theta$ divided by the original angle: $\Delta \theta$

S = \frac{F_{||}/A}{x/h}

Example

A $2.0 \text{ m}$ long, $1.0 \text{ mm}$ diameter steel wire is suspended from the ceiling. Hanging a $4.5 \text{ kg}$ mass from the wire causes it to stretch by $1.0 \text{ mm}$ .

What is the Young's modulus of the wire? Can you identify the material?

Answer:

The force pulling on the wire is the weight of the mass:

\frac{F}{A} = \frac{mg}{A} = \frac{(4.5\text{ kg})(9.8\text{ m/s}^2)}{\pi(0.0005\text{ m})^2} = 5.6\times 10^7\text{ N/m}^2

The resulting strech of $1.0 \text{ mm}$ is the change in length divided by the original length:

Y = \frac{F/A}{\Delta L/L} = \frac{5.6\times 10^7\text{ N/m}^2}{(1.0\times 10^{-3}\text{ m})/(2.0\text{ m})} = 11\times 10^{10}\text{ N/m}^2

This is the Young's modulus of the medium $\rightarrow$ the wire is made of copper.

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