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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 \( \mu_s \). |
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A sphere of mass M and radius r, and rotational inertia I is released from the top of a inclined plane of height h. The surface has considerable friction. Using only the variable mentioned, derive an expression for the sphere’s center of mass velocity.
A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
The moment of inertia of a solid cylinder about its axis is given by [katex]I = \frac{1}{2}mR^2[/katex]. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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