0 attempts
0% avg
UBQ Credits
| 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. |
Just ask: "Help me solve this problem."
A \(5\)-meter long ladder is leaning against a wall, with the bottom of the ladder \(3\) meters from the wall. The ladder is uniform and has a mass of \(20 \, \text{kg}\). A person of mass \(80 \, \text{kg}\) is standing on the ladder at a distance of \(4\) meters from the bottom of the ladder. What is the force exerted by the wall on the ladder?
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 \(I = \tfrac{1}{3} m L^2\).
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of \( 5.0 \) \( \text{rev/s} \) in \( 8.0 \) \( \text{s} \). At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in \( 12.0 \) \( \text{s} \). Through how many revolutions does the tub turn during the entire \( 20 \)-s interval? Assume constant angular acceleration while it is starting and stopping.
A grinding wheel is in the form of a uniform solid disk of radius \( 7.00 \) \( \text{cm} \) and mass \( 2.00 \) \( \text{kg} \). It starts from rest and accelerates uniformly under the action of the constant torque of \( 0.600 \) \( \text{N m} \) that the motor exerts on the wheel.
A force of \(17 \, \text{N}\) is applied to the end of a \(0.63 \, \text{m}\) long torque wrench at an angle \(45^\circ\) from a line joining the pivot point to the handle. What is the magnitude of the torque about the pivot point produced by this force?
\(\boxed{\theta_{\min}\approx 51.3^\circ}\)
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
We crafted THE Ultimate A.P Physics 1 course so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
