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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \sum F_y = T_1 \sin \theta_1 + T_2 \sin \theta_2 – mg = 0 \) | In vertical equilibrium, the vertical components of the tensions must balance the weight of the ball. \( \theta_2 \) is smaller as \( T_2 \) is more vertical. |
| 2 | \( T_1 \cos \theta_1 = T_2 \cos \theta_2 \) | In horizontal equilibrium, the horizontal components of the tensions must be equal. |
| 3 | \( \tan \theta_1 = \frac{T_1 \sin \theta_1}{T_1 \cos \theta_1} \) | Definition of tangent in terms of sine and cosine for \( T_1 \). |
| 4 | \( \tan \theta_2 = \frac{T_2 \sin \theta_2}{T_2 \cos \theta_2} \) | Definition of tangent in terms of sine and cosine for \( T_2 \). |
| 5 | \( \theta_2 < \theta_1 \) implies \( \tan \theta_2 < \tan \theta_1 \) | Since \( T_2 \) is more vertical, it has a smaller angle with the horizontal than \( T_1 \). |
| 6 | \( T_1 < T_2 \) | \( T_2 \) must provide a larger portion of the weight, so it must be greater than \( T_1 \). |
| 7 | \( \boxed{T_1 < T_2} \) | Final conclusion based on the analysis of tension and angles. |
Just ask: "Help me solve this problem."
Two students push a 1750 kg car with a force of 758 N along a perfectly level road at a constant velocity of 4.00 m/s. Find the force of friction.
Three students are pulling on a bag of skittles. Each is pulling with a horizontal force. If student 1 pulls Eastward with [katex]170 \, \text{N}[/katex], student 2 pulls Southward with [katex]100 \, \text{N}[/katex] and student 3 pulls with [katex]200 \, \text{N}[/katex] at an angle of [katex]20^\circ [/katex] west of north, what is the net force caused by the three students on the bag of skittles?
The escape speed of an object of mass \( m \) from a planet of mass \( M \) and radius \( r \) depends on the gravitational constant and
Four blocks of masses \( 20 \, \text{kg}, \, 30 \, \text{kg}, \, 40 \, \text{kg}, \, \text{and} \, 50 \, \text{kg} \) are stacked on top of one another in an elevator in order of decreasing mass with the lightest mass on the top of the stack. The elevator moves downward with an acceleration of \( 3.2 \, \text{m/s}^2 \). Find the contact force between the \( 30 \, \text{kg} \) and \( 40 \, \text{kg} \) masses.
The heaviest train ever pulled by a single engine was over [katex] 2 \, \text{km} [/katex] long. A force of [katex] 1.13 \times 10^5 \, \text{N} [/katex] is needed to get the train to start moving. If the coefficient of static friction is [katex] 0.741 [/katex] and the coefficient of kinetic friction is [katex] .592 [/katex], what is the train’s mass?
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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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