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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\tau_{\text{beam}} = F_{\text{beam}} \cdot r_{\text{beam}} \cdot \sin(\theta)\) | Torque due to the beam. \(F_{\text{beam}} = m_{\text{beam}} \cdot g\) and \(r_{\text{beam}} = \frac{L}{2}\). |
| 2 | \(F_{\text{beam}} = m_{\text{beam}} \cdot g\) | Force from the beam’s weight. \(m_{\text{beam}} = 500 \, \text{kg}\), \(g = 9.8 \, \text{m/s}^2\). |
| 3 | \(F_{\text{beam}} = 500 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 4900 \, \text{N}\) | Weight of the beam. |
| 4 | \(r_{\text{beam}} = \frac{L}{2} = \frac{6.00 \, \text{m}}{2} = 3.00 \, \text{m}\) | Distance from bolt to the center of mass of the beam. |
| 5 | \(\tau_{\text{beam}} = 4900 \, \text{N} \cdot 3.00 \, \text{m}\) | Torque due to the beam. |
| 6 | \(\tau_{\text{beam}} = 14700 \, \text{Nm}\) | Torque exerted by the beam. |
| 7 | \(\tau_{\text{worker}} = F_{\text{worker}} \cdot r_{\text{worker}} \cdot \sin(\theta)\) | Torque due to the worker. \(F_{\text{worker}} = m_{\text{worker}} \cdot g\), \(r_{\text{worker}} = L\). |
| 8 | \(F_{\text{worker}} = m_{\text{worker}} \cdot g\) | Force due to the worker’s weight. \(m_{\text{worker}} = 70 \, \text{kg}\). |
| 9 | \(F_{\text{worker}} = 70 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 686 \, \text{N}\) | Weight of the worker. |
| 10 | \(r_{\text{worker}} = L = 6.00 \, \text{m}\) | Distance from the bolt to the worker. |
| 11 | \(\tau_{\text{worker}} = 686 \, \text{N} \cdot 6.00 \, \text{m}\) | Torque due to the worker. |
| 12 | \(\tau_{\text{worker}} = 4116 \, \text{Nm}\) | Torque exerted by the worker. |
| 13 | \(\tau_{\text{total}} = \tau_{\text{beam}} + \tau_{\text{worker}} = 14700 \, \text{Nm} + 4116 \, \text{Nm}\) | Total torque about the bolt. |
| 14 | \(\tau_{\text{total}} = 18816 \, \text{Nm}\) | Final total torque about the bolt. |
Just ask: "Help me solve this problem."
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A meter stick with a uniformly distributed mass of 0.5 kg is supported by a pivot placed at the 0.25 m mark from the left. At the left end, a small object of mass 1.0 kg is placed at the zero mark, and a second small object of mass 0.5 kg is placed at the 0.5 m mark. The meter stick is supported so that it remains horizontal, and then it is released from rest. Find the change in the angular momentum of the meter stick, one second after it is released,.
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\(\tau_{\text{total}} = 18816 \, \text{Nm}\)
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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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