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. |
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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 \( M \)?
What is the net torque acting on the pivot supporting a \(10 \, \text{kilogram}\) beam \(2 \, \text{meters}\) long as shown above? Assume that the positive direction is clockwise.
Consider a rigid body that is rotating. Which of the following is an accurate statement?
A solid sphere \( I = 0.06 \, \text{kg} \cdot \text{m}^2 \) spins freely around an axis through its center at an angular speed of \( 20 \, \text{rad/s} \). It is desired to bring the sphere to rest by applying a friction force of magnitude \( 2.0 \, \text{N} \) to the sphere’s outer surface, a distance of \( 0.30 \, \text{m} \) from the sphere’s center. How much time will it take the sphere to come to rest?
The figure shows scale drawings of four objects, each of the same mass and uniform thickness, with the mass distributed uniformly. Which one has the greatest moment of inertia when rotated about an axis perpendicular to the plane of the drawing at point P?
Two workers are holding a thin plate with length \(5 \, \text{m}\) and height \(2 \, \text{m}\) at rest by supporting the plate in the bottom corners. The workers are standing at rest on a slope of \(10^\circ\). Treat these supporting forces as vertical normal forces and calculate their magnitudes and state if both workers are sharing “the job” fairly.
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?
Two masses, \( m_y = 32 \) \( \text{kg} \) and \( m_z = 38 \) \( \text{kg} \), are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius \( R = 0.311 \) \( \text{m} \) and mass \( 3.1 \) \( \text{kg} \). Initially, \( m_y \) is on the ground and \( m_z \) rests \( 2.5 \) \( \text{m} \) above the ground.
A uniform solid sphere of mass M and radius R is placed on a frictionless horizontal surface. A massless string is wrapped around the sphere and is pulled with a force F. The string makes an angle of θ with the horizontal. What is the minimum value of the coefficient of static friction between the sphere and the surface required for the sphere to start rolling without slipping?
\(\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}\)
\(f = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\)
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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] |
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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