0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Define a right–handed coordinate system with the pivot as the origin.
Choose counter-clockwise (CCW) moments about the +z axis as positive. |
The stick rotates in the horizontal plane about a vertical axis, so the relevant component of the torque vector is its z-component. All forces are vertical (±y direction); their moment arms lie along ±x. Hence every torque vector points along ±z, making scalar bookkeeping convenient. |
| 2 | \[\vec\tau = \vec r \times \vec F\] | The torque produced by a force is the cross product of the position vector \(\vec r\) (from pivot to the point of application) and the force vector \(\vec F\). Because \(\vec r\) is horizontal and \(\vec F\) is vertical, \(\vec\tau\) is purely along the ±z direction. |
| 3 | Left mass (0.10 kg) at \(x=-0.50\,\text{m}\) : \[\tau_{\text L}=(-0.50\,\hat\imath)\times(-0.10g\,\hat\jmath)=+0.50(0.10g)\,\hat k\] \[\tau_{\text L}=+0.49\,\text{N·m}\] |
The cross product gives a +z (CCW) torque. Numerically, \(0.50\,\text{m}\times0.10\,\text{kg}\times9.8\,\text{m/s}^2=0.49\,\text{N·m}.\) |
| 4 | Right mass (0.15 kg) at \(x=+0.50\,\text{m}\) : \[\tau_{\text R}=(+0.50\,\hat\imath)\times(-0.15g\,\hat\jmath)=-0.50(0.15g)\,\hat k\] \[\tau_{\text R}=-0.735\,\text{N·m}\] |
The torque is −z (clockwise). Its magnitude is \(0.50\times0.15g=0.735\,\text{N·m}.\) |
| 5 | Tension \(T\) applied vertically at the left end: \[\tau_T=(-0.50\,\hat\imath)\times(+T\,\hat\jmath)=-0.50T\,\hat k\] |
We take the string to pull straight up (+y). Its lever arm is still 0.50 m, and the resulting torque is along −z (same sense as the heavier right mass). |
| 6 | Static rotational equilibrium about the vertical axis: \[\tau_{\text L}+\tau_{\text R}+\tau_T = 0\] |
The stick is not twisting about the pivot, so the algebraic sum of all z-components of torque must vanish. |
| 7 | \[ 0.49 – 0.735 – 0.50T = 0 \quad\Rightarrow\quad T = -0.49\,\text{N} \] |
Solving the equilibrium equation yields a negative sign, meaning the actual tension direction is downward (−y) at the left end. Its magnitude is 0.49 N. |
| 8 | \[|T| = 0.49\,\text{N}\quad\text{(directed downward at the left end)}\] | The string must pull downward with 0.49 N to balance the unequal weights and keep the stick from tilting while it is free to spin in the horizontal plane. |
Just ask: "Help me solve this problem."
A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a \( 30^\circ \) rough incline (\( \mu_k = 0.4 \)) of distance \( d \), with an initial speed of \( 5 \) \( \text{m/s} \), and then it slides back down.
A truck of mass 3500 kg hits the back of a small car of mass 1400 kg. Which car exerted more force on the other and why?
A box with a mass of \( M \) rests on a scale in an elevator that is moving downwards. The elevator slows with an acceleration of \( \dfrac{g}{4} \). Which of the following will give the reading of the scale?
A spring with spring constant \( k = 2.3 \) \( \text{N/m} \) is attached to an object of mass \( 10 \) \( \text{kg} \). If the object is hung from the ceiling by this spring, how much would the spring be stretched?
A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. No matter what the orientation of the force, the crate does not move. From least to greatest, rank the normal force on the crate.
\(0.49\,\text{N downward}\)
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.
