| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \tau = F r \] | Torque is defined as the product of the applied force \(F\) and the perpendicular distance \(r\) from the pivot (the hinges). |
| 2 | \[ r_{\text{edge}} > r_{\text{hinge}} \] | The distance from the hinge to the knob on the far side, \(r_{\text{edge}}\), is greater than any distance closer to the hinges, \(r_{\text{hinge}}\). |
| 3 | \[ \tau_{\text{edge}} = F r_{\text{edge}} \] | Applying the same force \(F\) at the knob (far side) yields a torque proportional to the larger lever arm \(r_{\text{edge}}\). |
| 4 | \[ \tau_{\text{hinge}} = F r_{\text{hinge}} \] | If the knob were near the hinges, the lever arm \(r_{\text{hinge}}\) would be much smaller, producing smaller torque. |
| 5 | \[ \tau_{\text{edge}} > \tau_{\text{hinge}} \] | Because \(r_{\text{edge}} > r_{\text{hinge}}\), the torque produced at the knob’s usual position is greater, making the door easier to rotate open for the same applied force. |
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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?
An object weighing 120 N is set on a rigid beam of negligible mass at a distance of 3 m from a pivot, as shown above. A vertical force is to be applied to the other end of the beam a distance of 4 m from the pivot to keep the beam at rest and horizontal. What is the magnitude F of the force required?
A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal, \( 2.3 \) \( \text{m} \)-long, \( 6.1 \) \( \text{kg} \) reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A \( 64 \) \( \text{kg} \) woman lies on the reaction board with her feet over the pivot. The scale reads \( 27 \) \( \text{kg} \). What is the distance from the woman’s feet to her center of mass? Express your answer with the appropriate units.
One end of a string is wrapped around a pulley that is free to rotate with negligible friction about an axle at its center. The other end of the string is attached to a block. The block is released from rest and moves downward with constant acceleration. Which of the following correctly indicates whether the amount of work done on the pulley by the string during each successive complete rotation remains constant or increases, and provides a valid justification?
The system in the Figure is in equilibrium. A concrete block of mass \(225 \, \text{kg}\) hangs from the end of a uniform strut whose mass is \(45.0 \, \text{kg}\).
A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 1.00 N·m? Assume that the moment of inertia of a rectangle is \( I = \frac{1}{12}M(a^2+b^2) \)
If a constant net torque is applied to an object it will (select all that applies):
Two masses, \( m_1 \) and \( m_2 \), are suspended on either side of a pulley with a radius \( R \), as shown. The heavier mass, \( m_2 \), is initially held at rest above the ground by a distance \( h \) before being released. An student measures that it takes an amount of time \( t \) for the heavier mass to hit the ground after being released.
A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.
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?
\(\text{To maximize torque by using the largest possible lever arm so less force is needed to open the door}\)
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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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