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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Define a right–handed coordinate system with the pivot as the origin.
Choose counter-clockwise (CCW) moments about the +z axis as positive. |
The stick rotates in the horizontal plane about a vertical axis, so the relevant component of the torque vector is its z-component. All forces are vertical (±y direction); their moment arms lie along ±x. Hence every torque vector points along ±z, making scalar bookkeeping convenient. |
| 2 | \[\vec\tau = \vec r \times \vec F\] | The torque produced by a force is the cross product of the position vector \(\vec r\) (from pivot to the point of application) and the force vector \(\vec F\). Because \(\vec r\) is horizontal and \(\vec F\) is vertical, \(\vec\tau\) is purely along the ±z direction. |
| 3 | Left mass (0.10 kg) at \(x=-0.50\,\text{m}\) : \[\tau_{\text L}=(-0.50\,\hat\imath)\times(-0.10g\,\hat\jmath)=+0.50(0.10g)\,\hat k\] \[\tau_{\text L}=+0.49\,\text{N·m}\] |
The cross product gives a +z (CCW) torque. Numerically, \(0.50\,\text{m}\times0.10\,\text{kg}\times9.8\,\text{m/s}^2=0.49\,\text{N·m}.\) |
| 4 | Right mass (0.15 kg) at \(x=+0.50\,\text{m}\) : \[\tau_{\text R}=(+0.50\,\hat\imath)\times(-0.15g\,\hat\jmath)=-0.50(0.15g)\,\hat k\] \[\tau_{\text R}=-0.735\,\text{N·m}\] |
The torque is −z (clockwise). Its magnitude is \(0.50\times0.15g=0.735\,\text{N·m}.\) |
| 5 | Tension \(T\) applied vertically at the left end: \[\tau_T=(-0.50\,\hat\imath)\times(+T\,\hat\jmath)=-0.50T\,\hat k\] |
We take the string to pull straight up (+y). Its lever arm is still 0.50 m, and the resulting torque is along −z (same sense as the heavier right mass). |
| 6 | Static rotational equilibrium about the vertical axis: \[\tau_{\text L}+\tau_{\text R}+\tau_T = 0\] |
The stick is not twisting about the pivot, so the algebraic sum of all z-components of torque must vanish. |
| 7 | \[ 0.49 – 0.735 – 0.50T = 0 \quad\Rightarrow\quad T = -0.49\,\text{N} \] |
Solving the equilibrium equation yields a negative sign, meaning the actual tension direction is downward (−y) at the left end. Its magnitude is 0.49 N. |
| 8 | \[|T| = 0.49\,\text{N}\quad\text{(directed downward at the left end)}\] | The string must pull downward with 0.49 N to balance the unequal weights and keep the stick from tilting while it is free to spin in the horizontal plane. |
Just ask: "Help me solve this problem."
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
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.
The elliptical orbit of a comet is shown above. Positions \(1\) and \(2\) are, respectively, the farthest and nearest positions to the Sun, and at position \(1\) the distance from the comet to the Sun is \(10\) times that at position \(2\). What is the ratio \(\dfrac{F_1}{F_2}\), the force on the comet at position \(1\) to the force on the comet at position \(2\)?
A uniform, solid, \( 100 \) \( \text{kg} \) cylinder with a diameter of \( 1.0 \) \( \text{m} \) is mounted so it is free to rotate about a fixed, horizontal, frictionless axis that passes through the centers of its circular ends. A \( 10 \) \( \text{kg} \) block is hung from a very light, thin cord wrapped around the cylinder’s circumference. When the block is released, the cord unwinds and the block accelerates downward. What is the acceleration of the block?
\(0.49\,\text{N downward}\)
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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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