| 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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Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?
The rotating systems, shown in the figure above, differ only in that the two identical movable masses are positioned a distance r from the axis of rotation (left), or a distance r/2 from the axis of rotation (right). What happens if you release the hanging blocks simultaneously from rest?
A wheel of radius \( R \) and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses \( m \), \( M \), and \( 2M \), respectively, are mounted on the rim of the wheel, as shown above. If the system is in static equilibrium, what is the value of \( m \) in terms of \( M \)?
Two uniform solid balls, one of radius \( R \) and mass \( M \), the other of radius \( 2R \) and mass \( 8M \), roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
Suppose just two external forces act on a stationary, rigid object and the two forces are equal in magnitude and opposite in direction. Under what condition does the object start to rotate?
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?
An ice skater that is spinning in circles has an initial rotational inertia \(I_i\). You can approximate her shape to be a cylinder. She is spinning with velocity \(\omega_i\). As she extends her arms, her rotational inertia changes by a factor of \(x\) and her angular velocity changes by a factor of \(y\). Which one of the following options best describe \(x\) and \(y\)?
A disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
A \(350\ \text{g}\) ball is attached to the end of a thin, uniform rod of mass \(500\ \text{g}\) and length \(1.2\ \text{m}\). The system is rotated in a horizontal circle about the opposite end of the rod. Calculate the moment of inertia of the system about the axis of rotation. Hint: the moment of inertia of a thin rod about the end of the rod is \(I = \tfrac{1}{3} m L^2\).
A massless rigid rod of length \(3d\) is pivoted at a fixed point \(W\), and two forces each of magnitude \(F\) are applied vertically upward as shown above. A third vertical force of magnitude \(F\) may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude \(F\) is applied at point?
\(\boxed{\theta_{\min}\approx 51.3^\circ}\)
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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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