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Part (a): Minimum coefficient
| Derivation/Formula | Reasoning |
|---|---|
| \[x_2 = \tfrac{L}{2}\cos\theta,\quad x_1 = d\cos\theta,\quad y_{\text{top}} = L\sin\theta\] | Set geometry: horizontal lever arms to the weights are \( \tfrac{L}{2}\cos\theta \) for the ladder’s center (mass \( m_2 \)) and \( d\cos\theta \) for the person (mass \( m_1 \)); the top contact is at height \( L\sin\theta \). |
| \[N_1(L\sin\theta) – m_2 g\left(\tfrac{L}{2}\cos\theta\right) – m_1 g(d\cos\theta) = 0\] | Torque about the bottom with \( \)counterclockwise positive: wall normal \( N_1 \) gives a positive moment \( N_1(L\sin\theta) \); weights \( m_2 g \) and \( m_1 g \) at offsets \( \tfrac{L}{2}\cos\theta \) and \( d\cos\theta \) give clockwise (negative) moments. |
| \[N_1 = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\] | Solve the torque equation algebraically for \( N_1 \). |
| \[f – N_1 = 0\] | Horizontal force balance: the floor friction \( f \) (to the right) balances the wall’s leftward normal \( N_1 \), so \( f = N_1 \). |
| \[N_2 – (m_1 + m_2)g = 0\] | Vertical force balance: the ground normal \( N_2 \) supports the total weight \( (m_1+m_2)g \). |
| \[f = \mu_{\min} N_2\] | Impending slip condition at the threshold of motion defines \( \mu_{\min} \) via \( f = \mu_{\min} N_2 \). With \( f = N_1 \), one has \( \mu_{\min} = \tfrac{N_1}{N_2} \). |
| \[\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}\] | Substitute \( N_1 = \tfrac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta} \) and \( N_2 = (m_1+m_2)g \); the \( g \) cancels, yielding the simplified ratio. |
| \[\boxed{\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}}\] | Final expression for the minimum coefficient ensuring no slip. |
Part (b): Friction magnitude
| Derivation/Formula | Reasoning |
|---|---|
| \[\mu_s = \tfrac{3}{2}\,\mu_{\min}\] | Given that the available static friction coefficient \( \mu_s \) exceeds the minimum \( \mu_{\min} \) by a factor of \( \tfrac{3}{2} \). |
| \[f = N_1\] | In static equilibrium, the actual friction adjusts to balance horizontal forces; thus \( f \) equals the wall normal \( N_1 \), not \( \mu_s N_2 \) unless at the threshold. |
| \[N_1 = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\] | Use the same torque result from part (a); it does not depend on \( \mu_s \). |
| \[\boxed{f = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}}\] | Substitute to obtain the friction magnitude; since \( \mu_s > \mu_{\min} \), this satisfies \( f \le \mu_s N_2 \) with margin. |
Just ask: "Help me solve this problem."
Consider a uniform hoop of radius R and mass M rolling without slipping. Which is larger, its translational kinetic energy or its rotational kinetic energy?
A miniature, solid globe with mass \( 0.25 \) \( \text{kg} \) and radius \( 0.10 \) \( \text{m} \) is spinning in place about a vertical axis with the equator horizontal, as shown. A point on the globe’s equator, represented by the dot in the figure, has a linear speed of \( 4.0 \) \( \text{m/s} \). The rotational inertia of a solid sphere of mass \( m \) and radius \( r \) is \( \tfrac{2}{5}mr^{2} \). The rotational kinetic energy of the globe is most nearly
Two thin coins are made from identically the same metal, but one coin has triple the diameter of the other. What is the ratio of the moment of inertia of the large coin compared to the small coin? Take the axis of rotation to be perpendicular to the coin and through its center; assume that the coins have the same thickness. Hint: The moment of inertia of a solid disk about its center is \(\frac{1}{2} M R^{2}\).
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
A point P is at a distance \( R \) from the axis of rotation of a rigid body whose angular velocity and angular acceleration are \( \omega \) and \( \alpha \) respectively. The linear speed, centripetal acceleration, and tangential acceleration of the point can be expressed as:
| Linear speed | Centripetal acceleration | Tangential acceleration | |
|---|---|---|---|
| \( (a) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\alpha \) |
| \( (b) \) | \( R\omega \) | \( R\alpha \) | \( R\omega^{2} \) |
| \( (c) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega \) |
| \( (d) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\omega \) |
| \( (e) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega^{2} \) |
\(\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}\)
\(f = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\)
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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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