| 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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Pulleys \( X \) and \( Y \) are each attached to a block by a string that wraps around the pulley. Both blocks are released and have the same linear acceleration \( a \). As the blocks fall, the pulleys rotate about their centers. Pulley \( Y \) has a larger radius than Pulley \( X \). How does the angular acceleration \( \alpha_X \) of Pulley \( X \) compare to the angular acceleration \( \alpha_Y \) of Pulley \( Y \)?
A disk of radius 35 cm rotates at a constant angular velocity of 10 rad/s. How fast does a point on the rim of the disk travel (in m/s)?
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?
Two identical solid disks, each of mass \( M \) and radius \( R \), are welded together so that they touch at exactly one point on their rims. Determine the moment of inertia of the combined object about an axis that is perpendicular to the plane of the disks and passes through their point of contact. Hint: The moment of inertia of a solid disk about its center is \(\frac{1}{2} M R^{2}\).
A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
During the experiment, students collect data about the angular momentum of a rigid, uniform spinning wheel about an axle as a function of time, which was used to create the graph that is shown. A frictional torque is exerted on the wheel. A student makes the following statement about the data. “The frictional torque exerted on the wheel is independent of the wheel’s angular speed.” Does the data from the graph support the student’s statement? Justify your selection.
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?
An object is experiencing a nonzero net force. Which of the following statements is most accurate?
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at either end of the rod?
A rod may freely rotate about an axis that is perpendicular to the rod and is along the plane of the page. The rod is divided into four sections of equal length of 0.2 m each, and four forces are exerted on the rod, as shown in the figure. Frictional forces are considered negligible. Which of the following describes an additional torque that must be applied in order to keep the rod from rotating?
\(\boxed{\theta_{\min}\approx 51.3^\circ}\)
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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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