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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."
A meterstick is supported at its center, which is aligned with the center of a cradle located at position \( x = 0 \) \( \text{m} \). Two identical objects of mass \( 1.0 \) \( \text{kg} \) are suspended from the meterstick. One object hangs \( 0.25 \) \( \text{m} \) to the left of the support point, and the other object hangs \( 0.50 \) \( \text{m} \) to the right of the support point. The system is released from rest and is free to rotate. Which of the following claims correctly describes the subsequent motion of the system containing the meterstick, cradle, and the two objects?
The figure above shows a uniform beam of length \( L \) and mass \( M \) that hangs horizontally and is attached to a vertical wall. A block of mass \( M \) is suspended from the far end of the beam by a cable. A support cable runs from the wall to the outer edge of the beam. Both cables are of negligible mass. The wall exerts a force \( F_w \) on the left end of the beam. For which of the following actions is the magnitude of the vertical component of \( F_w \) smallest?
A \( 0.72 \) \( \text{m} \)-diameter solid sphere can be rotated about an axis through its center by a torque of \( 10.8 \) \( \text{Nm} \) which accelerates it uniformly from rest through a total of \( 160 \) revolutions in \( 15.0 \) \( \text{s} \). What is the mass of the sphere?
A disk of radius \( R = 0.5 \) \( \text{cm} \) rests on a flat, horizontal surface such that frictional forces are considered to be negligible. Three forces of unknown magnitude are exerted on the edge of the disk, as shown in the figure. Which of the following lists the essential measuring devices that, when used together, are needed to determine the change in angular momentum of the disk after a known time of \( 5.0 \) \( \text{s} \)?
The figure shows a person’s foot. In that figure, the Achilles tendon exerts a force of magnitude F = 720 N. What is the magnitude of the torque that this force produces about the ankle joint?
\(\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}\)
\(f = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\)
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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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