| Derivation or Formula | Reasoning |
|---|---|
| \[\text{Forces on the ladder:}\] | We identify all external forces acting on the uniform ladder in static equilibrium, with a smooth (frictionless) wall and a rough ground. |
| \[W=50\,\text{N}\ \text{acting at the center (at }L/2\text{ from either end)}\] | The ladder is uniform, so its weight 0\(W\) acts downward at its center of mass located at \(L/2\). |
| \[N_g\ \text{at ground, upward}\] | The ground exerts a normal reaction \(N_g\) on the ladder, perpendicular to the ground (vertical upward). |
| \[f_s\ \text{at ground, horizontal toward the wall}\] | Static friction at the ground prevents the bottom from sliding outward (away from the wall). The impending motion would make the bottom slip outward, so friction must act toward the wall. |
| \[N_w\ \text{at wall, horizontal away from the wall}\] | The wall is smooth, so it exerts only a normal force \(N_w\), perpendicular to the wall (horizontal). It pushes the ladder away from the wall. |
| \[f_w=0\] | Because the wall is smooth, there is no friction force at the wall. |
| Derivation or Formula | Reasoning |
|---|---|
| \[\sum F_y=0:\quad N_g-W=0\] | Static equilibrium requires the net vertical force to be zero. Only \(N_g\) (up) and \(W\) (down) act vertically. |
| \[N_g=W\] | From \(N_g-W=0\), the ground normal equals the weight. |
| \[\sum F_x=0:\quad f_s-N_w=0\] | Static equilibrium requires the net horizontal force to be zero. Horizontally we have friction \(f_s\) (toward wall) and wall normal \(N_w\) (away from wall). |
| \[f_s=N_w\] | From \(f_s-N_w=0\), the friction force must balance the wall normal force. |
| \[\sum \tau_{\text{about bottom}}=0\] | Take torques about the bottom contact point to eliminate \(N_g\) and \(f_s\) from the torque equation (they act at the pivot, producing no moment arm). |
| \[\tau_{N_w}=N_w(L\sin\theta)\] | The top contact point is at height \(L\sin\theta\). The force \(N_w\) is horizontal, so its perpendicular moment arm about the bottom is \(L\sin\theta\). |
| \[\tau_W=W\left(\frac{L}{2}\cos\theta\right)\] | The weight acts at the midpoint. Its horizontal distance from the bottom is \((L/2)\cos\theta\). Since \(W\) is vertical, that horizontal distance is the perpendicular lever arm. |
| \[\sum \tau=0:\quad N_w(L\sin\theta)-W\left(\frac{L}{2}\cos\theta\right)=0\] | Choose counterclockwise positive. \(N_w\) tends to rotate the ladder counterclockwise; \(W\) tends to rotate it clockwise. |
| \[N_w\sin\theta=\frac{W}{2}\cos\theta\] | Cancel \(L\) from both terms and rearrange. |
| \[N_w=\frac{W}{2}\cot\theta\] | Solve the torque balance for \(N_w\) in terms of \(\theta\). |
| \[f_s=N_w=\frac{W}{2}\cot\theta\] | Use \(f_s=N_w\) from horizontal equilibrium. |
| \[f_s\le \mu N_g\] | Static friction can take any value up to its maximum \(\mu N_g\). The minimum angle occurs at impending slip, where friction is at maximum. |
| \[\frac{W}{2}\cot\theta=\mu W\] | At the threshold of slipping, set \(f_s=\mu N_g\) and use \(N_g=W\). |
| \[\frac{1}{2}\cot\theta=\mu\] | Cancel \(W\) (nonzero) from both sides. |
| \[\cot\theta=2\mu\] | Multiply both sides by \(2\). |
| \[\tan\theta=\frac{1}{2\mu}\] | Take the reciprocal to solve for \(\theta\) using \(\tan\theta\). |
| \[\tan\theta=\frac{1}{2(0.4)}=\frac{1}{0.8}=1.25\] | Substitute \(\mu=0.4\) and compute. |
| \[\theta=\arctan(1.25)\approx 51.3^\circ\] | Compute the angle whose tangent is \(1.25\). |
| \[\boxed{\theta_{\min}\approx 51.3^\circ}\] | This is the minimum angle so that the required friction does not exceed the maximum static friction. |
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A uniform rod of length \( L \) and mass \( M \) is free to rotate about one end, as shown in the diagram. The free end is released from rest at a horizontal position, as shown. The pivot point is supported by a stand so that only the free end can move. The moment of inertia of a rod about its end is \(\tfrac{1}{3} M L^{2}\).
A ball of radius \( r \) rolls on the inside of a circular track of radius \( R \). If the ball starts from rest at the left vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping? For a solid spherical ball, the moment of inertia is \(\frac{2}{5} m r^2\).
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To increase the moment of inertia of a body about an axis, you must
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\(\boxed{\theta_{\min}\approx 51.3^\circ}\)
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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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