| Derivation / Formula | Reasoning |
|---|---|
| \[T = m_B g\] | The hanging block B is in equilibrium (no motion), so its weight \(m_B g\) is balanced by the string tension \(T\). |
| \[T = m_A g \sin\theta + f_s\] | For block A on the incline, tension pulls it up the plane, opposed by the downslope components of weight \(m_A g \sin\theta\) and static friction \(f_s\). |
| \[f_s = \mu_s N\] | Maximum static friction equals the normal force \(N\) times the coefficient of static friction \(\mu_s\). |
| \[N = m_A g \cos\theta\] | The normal force on block A equals the perpendicular component of its weight, \(m_A g \cos\theta\). |
| \[T = m_A g \sin\theta + \mu_s m_A g \cos\theta\] | Substitute \(f_s\) and \(N\) into the force balance for block A. |
| \[\mu_s = \frac{m_B – m_A \sin\theta}{m_A \cos\theta}\] | Insert \(T = m_B g\) from block B and solve algebraically for \(\mu_s\). Note that \(g\) cancels. |
| \[\mu_s = \frac{17\,\text{kg} – 10\,\text{kg} \,(\sin45^\circ)}{10\,\text{kg}\,(\cos45^\circ)}\] | Plug in given masses and \(\theta = 45^\circ\) (where \(\sin45^\circ = \cos45^\circ = \frac{\sqrt{2}}{2} \approx 0.707\)). |
| \[\mu_s = \frac{17 – 10(0.707)}{10(0.707)} \approx 1.4\] | Compute numerator and denominator to obtain the minimum static-friction coefficient. |
| \[\boxed{\mu_s \approx 1.4}\] | This value ensures the system remains in static equilibrium; any smaller \(\mu_s\) lets block B descend. |
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The steepest street in the world is Baldwin Street in Dunedin, New Zealand. It has an inclination angle of \( 38.0^\circ \) with respect to the horizontal. Suppose a wooden crate with a mass of \( 25.0 \) \( \text{kg} \) is placed on Baldwin Street. An additional force of \( 59 \) \( \text{N} \) must be applied to the crate perpendicular to the pavement in order to hold the crate in place. If the coefficient of static friction between the crate and the pavement is \( 0.599 \), what is the magnitude of the frictional force?
A student presses a \( 0.5 \) \( \text{kg} \) book against the wall. If the \( \mu_s \) between the book and the wall is \( 0.2 \), what force must the student apply to hold the book in place?
In the diagram shown, a \(20 \, \text{N}\) force is applied to block \(B\) (\(7 \, \text{kg}\)). Block \(A\) has a mass of \(3 \, \text{kg}\). Assume frictionless conditions.
Describe two examples in which the force of friction exerted on an object is in the direction of motion of the object.
A child slides down a slide with a \( 34^\circ \) incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.
A \(5.5 \, \text{kg}\) block slides down a \(30^\circ\) incline that is \(2.2 \, \text{m}\) long. If \(\mu = 0.20\), what is the acceleration of the block?
A child on a sled reaches the bottom of a hill with a velocity of \( 10.0 \, \text{m/s} \) and travels \( 25.0 \, \text{m} \) along a horizontal straightaway to a stop. If the child and sled together have a mass of \( 60.0 \, \text{kg} \), what is the average retarding force on the sled on the horizontal straightaway?
A block rests on a flat plane inclined at an angle of \(30^\circ\) with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
In the diagram above, block \(A\) has a mass of \(3.2 \, \text{kg}\) and block \(B\) a mass of \(2.4 \, \text{kg}\). The pulley is frictionless and has no mass.
The coefficient of static friction between hard rubber and normal street pavement is about \(0.85\). On how steep a hill (maximum angle) can you leave a car parked?
\(\mu_s \approx 1.4\)
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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] |
Metric Prefixes
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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