| Derivation or Formula | Reasoning |
|---|---|
| \[\text{Given points: }A=(0,0),\;F_1\text{ at }(0,-2),\;F_2\text{ at }(2,-2),\;F_3\text{ at }(3,-1)\] | Set the pivot at \(A\) and list each force’s application point. |
| \[\text{Let each force magnitude be }F\] | All three forces have equal magnitude \(F\). |
| \[\text{Use }\tau_z = r_x F_y – r_y F_x\] | In 2D, the torque about the pivot (out of page \(z\)-component) is given by \(\tau_z\). Positive corresponds to counterclockwise (CCW). |
| \[\vec r_1=(0,-2)-(0,0)=(0,-2)\] | Position vector from \(A\) to where \(F_1\) acts. |
| \[\vec F_1=\left(-\frac{F}{\sqrt2},-\frac{F}{\sqrt2}\right)\] | “SW at \(45^\circ\) to vertical” means left and down with equal components: \(F_x<0,\;F_y<0\), each of magnitude \(F/\sqrt2\). |
| \[\tau_{1A}=r_{1x}F_{1y}-r_{1y}F_{1x}=(0)\left(-\frac{F}{\sqrt2}\right)-(-2)\left(-\frac{F}{\sqrt2}\right)=-\frac{2F}{\sqrt2}=-\sqrt2F\] | Compute torque from \(F_1\) about \(A\). Negative means clockwise (CW). |
| \[\vec r_2=(2,-2)-(0,0)=(2,-2)\] | Position vector from \(A\) to where \(F_2\) acts. |
| \[\vec F_2=\left(-\frac{F}{\sqrt2},\frac{F}{\sqrt2}\right)\] | “NW at \(45^\circ\) to vertical” means left and up with equal components: \(F_x0\). |
| \[\tau_{2A}=r_{2x}F_{2y}-r_{2y}F_{2x}=2\left(\frac{F}{\sqrt2}\right)-(-2)\left(-\frac{F}{\sqrt2}\right)=\frac{2F}{\sqrt2}-\frac{2F}{\sqrt2}=0\] | Torque from \(F_2\) cancels because the moment from its vertical component is canceled by the moment from its horizontal component about \(A\). |
| \[\vec r_3=(3,-1)-(0,0)=(3,-1)\] | Position vector from \(A\) to where \(F_3\) acts. |
| \[\vec F_3=(F,0)\] | \(F_3\) acts horizontally east: \(F_x=F,\;F_y=0\). |
| \[\tau_{3A}=r_{3x}F_{3y}-r_{3y}F_{3x}=3(0)-(-1)(F)=F\] | The line of action is above/below pivot by \(1\,\text{m}\) (since \(r_y=-1\)), producing a CCW (positive) torque. |
| \[\tau_{\text{net},A}=\tau_{1A}+\tau_{2A}+\tau_{3A}=-\sqrt2F+0+F=(1-\sqrt2)F\] | Sum the three torques about \(A\). |
| \[\boxed{\tau_{\text{net},A}=(1-\sqrt2)F\;\text{Nm}\;<0\;\Rightarrow\;\text{CW}}\] | Since \(\sqrt2\approx1.414\), \((1-\sqrt2)F<0\), so the net torque about \(A\) is clockwise. |
| Derivation or Formula | Reasoning |
|---|---|
| \[B=(0,-1)\] | Set the pivot at \(B\). |
| \[\tau_z = r_x F_y – r_y F_x\] | Same torque relation about the new pivot. |
| \[\vec r_1=(0,-2)-(0,-1)=(0,-1)\] | Position vector from \(B\) to where \(F_1\) acts. |
| \[\vec F_1=\left(-\frac{F}{\sqrt2},-\frac{F}{\sqrt2}\right)\] | Same force components as before. |
| \[\tau_{1B}=0\left(-\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=-\frac{F}{\sqrt2}\] | Torque from \(F_1\) about \(B\) is negative (CW). |
| \[\vec r_2=(2,-2)-(0,-1)=(2,-1)\] | Position vector from \(B\) to where \(F_2\) acts. |
| \[\vec F_2=\left(-\frac{F}{\sqrt2},\frac{F}{\sqrt2}\right)\] | Same \(F_2\) components as before. |
| \[\tau_{2B}=2\left(\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=\frac{2F}{\sqrt2}-\frac{F}{\sqrt2}=\frac{F}{\sqrt2}\] | Net torque from \(F_2\) about \(B\) is positive (CCW). |
| \[\vec r_3=(3,-1)-(0,-1)=(3,0)\] | Position vector from \(B\) to where \(F_3\) acts. |
| \[\vec F_3=(F,0)\] | Same horizontal force. |
| \[\tau_{3B}=3(0)-0(F)=0\] | Since \(\vec r_3\) is purely horizontal and \(\vec F_3\) is also horizontal, the line of action passes through the pivot height, giving zero moment arm about \(B\). |
| \[\tau_{\text{net},B}=\left(-\frac{F}{\sqrt2}\right)+\left(\frac{F}{\sqrt2}\right)+0=0\] | Sum of torques cancels exactly. |
| \[\boxed{\tau_{\text{net},B}=0\;\text{Nm}\;\Rightarrow\;\text{zero}}\] | Equal and opposite torques from \(F_1\) and \(F_2\) cancel, and \(F_3\) produces no torque about \(B\). |
| Derivation or Formula | Reasoning |
|---|---|
| \[C=(1,-1)\] | Set the pivot at \(C\). |
| \[\tau_z = r_x F_y – r_y F_x\] | Same torque formula. |
| \[\vec r_1=(0,-2)-(1,-1)=(-1,-1)\] | Position vector from \(C\) to where \(F_1\) acts. |
| \[\vec F_1=\left(-\frac{F}{\sqrt2},-\frac{F}{\sqrt2}\right)\] | Same \(F_1\) components. |
| \[\tau_{1C}=(-1)\left(-\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=\frac{F}{\sqrt2}-\frac{F}{\sqrt2}=0\] | About \(C\), the perpendicular effects of \(F_1\)’s components cancel, giving zero net torque from \(F_1\). |
| \[\vec r_2=(2,-2)-(1,-1)=(1,-1)\] | Position vector from \(C\) to where \(F_2\) acts. |
| \[\vec F_2=\left(-\frac{F}{\sqrt2},\frac{F}{\sqrt2}\right)\] | Same \(F_2\) components. |
| \[\tau_{2C}=1\left(\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=\frac{F}{\sqrt2}-\frac{F}{\sqrt2}=0\] | Similarly, \(F_2\)’s component torques cancel about \(C\). |
| \[\vec r_3=(3,-1)-(1,-1)=(2,0)\] | Position vector from \(C\) to where \(F_3\) acts. |
| \[\vec F_3=(F,0)\] | Horizontal eastward force. |
| \[\tau_{3C}=2(0)-0(F)=0\] | \(\vec r_3\) is horizontal and \(\vec F_3\) is horizontal, so there is no moment arm about \(C\). |
| \[\tau_{\text{net},C}=0+0+0=0\] | All three individual torques are zero about \(C\). |
| \[\boxed{\tau_{\text{net},C}=0\;\text{Nm}\;\Rightarrow\;\text{zero}}\] | The lines of action produce no net rotational tendency about \(C\); each force has either zero moment arm or cancels via its components. |
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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.
Four identical lead balls with large mass are connected by rigid but very light rods in the square configuration shown in the preceding figure. The balls are rotated about the three labeled axes. Which of the following correctly ranks the rotational inertia \(I\) of the balls about each axis?
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
Two points, A and B, are on a disk that rotates about an axis. Point A is \( 3 \) times as far from the axis as point B. If the speed of point B is \( v \), then what is the speed of point A?
A ladder at rest is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on the ladder by the floor?
Angular momentum cannot be conserved if
A disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end (assume it does not slip). The other end of the beam is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the (1) tension force in the cable and (2) the total force the wall exerts on the beam.
The rotating systems, shown in the figure above, differ only in that the two identical movable masses are positioned a distance r from the axis of rotation (left), or a distance r/2 from the axis of rotation (right). What happens if you release the hanging blocks simultaneously from rest?
When is the angular momentum of a system constant?
\(\tau_{\text{net},A}=(1-\sqrt2)F\;\text{Nm}\;\text{(clockwise)},\;\tau_{\text{net},B}=0\;\text{Nm},\;\tau_{\text{net},C}=0\;\text{Nm}\)
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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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