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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1. Sum of Forces in the Horizontal Direction (x-axis) | \[ \sum F_x = 0 \implies N_1 – f = 0 \implies N_1 = f \] | Horizontal forces balance; friction equals wall’s normal force. |
| 2. Sum of Forces in the Vertical Direction (y-axis) | \[ \sum F_y = 0 \implies N_2 – m_1 g – m_2 g = 0 \implies N_2 = (m_1 + m_2) g \] | Vertical forces balance; ground’s normal force equals total weight. |
| 3. Sum of Torques About the Bottom of the Ladder | \[ N_1 L \sin \theta – m_1 g d \cos \theta – m_2 g \left( \dfrac{L}{2} \cos \theta \right) = 0 \] \[ N_1 = \dfrac{m_1 g d + \dfrac{1}{2} m_2 g L}{L \tan \theta} \] | Set net torque to zero; solve for \( N_1 \). |
| 4. Frictional Force at the Point of Slipping | \[ f_{\text{max}} = \mu_{\text{min}} N_2 = \mu_{\text{min}} (m_1 + m_2) g \] \[ N_1 = f = \mu_{\text{min}} (m_1 + m_2) g \] | Relate friction to normal force and coefficient of friction. |
| 5. Solving for \( \mu_{\text{min}} \) | \[ \mu_{\text{min}} (m_1 + m_2) g = \dfrac{m_1 g d + \dfrac{1}{2} m_2 g L}{L \tan \theta} \] \[ \mu_{\text{min}} = \dfrac{m_1 d + \dfrac{1}{2} m_2 L}{(m_1 + m_2) L \tan \theta} \] | Equate expressions for \( N_1 \) and solve for \( \mu_{\text{min}} \). |
| 6. Answer to First Part | \[ \mu_{\text{min}} = \dfrac{m_1 d + \dfrac{1}{2} m_2 L}{(m_1 + m_2) L \tan \theta} \] | Minimum coefficient of static friction required. |
| 7. Given \( \mu_s = \dfrac{3}{2} \mu_{\text{min}} \) | \[ \mu_s = \dfrac{3}{2} \mu_{\text{min}} \] | Actual coefficient of friction is 1.5 times \( \mu_{\text{min}} \). |
| 8. Calculating Frictional Force \( f \) | \[ f = N_1 = \mu_{\text{min}} (m_1 + m_2) g \] | Frictional force required for equilibrium. |
| 9. Answer to Second Part | \[ f = \mu_{\text{min}} (m_1 + m_2) g \] | Magnitude of frictional force with increased \( \mu_s \). |
Just ask: "Help me solve this problem."
A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?
A pulley has an initial angular speed of \( 12.5 \) \( \text{rad/s} \) and a constant angular acceleration of \( 3.41 \) \( \text{rad/s}^2 \). Through what angle does the pulley turn in \( 5.26 \) \( \text{s} \)?
Young David experimented with slings before tackling Goliath. He found that he could develop an angular speed of \( 8.0 \) \( \text{rev/s} \) in a sling \( 0.60 \) \( \text{m} \) long. If he increased the length to \( 0.90 \) \( \text{m} \), he could revolve the sling only \( 6.0 \) times per second.
Consider a solid uniform sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, translational or rotational?
A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal, \( 2.3 \) \( \text{m} \)-long, \( 6.1 \) \( \text{kg} \) reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A \( 64 \) \( \text{kg} \) woman lies on the reaction board with her feet over the pivot. The scale reads \( 27 \) \( \text{kg} \). What is the distance from the woman’s feet to her center of mass? Express your answer with the appropriate units.
1. \( \mu_{\text{min}} = \dfrac{m_1 d + \dfrac{1}{2} m_2 L}{(m_1 + m_2) L \tan \theta} \)
2. \( f = \mu_{\text{min}} (m_1 + m_2) g \)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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