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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."
A 25 g steel ball is attached to the top of a 24-cm-diameter vertical wheel. Starting from rest, the wheel accelerates at [katex] 470 \, \frac{rad}{s^2}[/katex]. The ball is released after [katex]\frac{3}{4} [/katex] of a revolution. How high does it go above the center of the wheel?
A force of 17 N is applied to the end of a 0.63-m long torque wrench at an angle 45° from a line joining the pivot point to the handle. What is the magnitude of the torque about the pivot point produced by this force?
A uniform ladder of length L and weight [katex] W = 50 N [/katex] rests against a smooth vertical wall. If the coefficient of static friction between the ladder and the ground is [katex] \mu = .4 [/katex].
An object moves at a constant speed of [katex] 9.0 \frac{m}{s} [/katex] in a circular path of radius of 1.5 m. What is the angular acceleration of the object?
Select all that applies. If a constant net torque is applied to an object, that object will
\(\tau_{\text{total}} = 18816 \, \text{Nm}\)
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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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