| 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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A disk increases from 2 complete revolutions in 2 seconds to 5 complete revolutions in 2 seconds. What is its average angular acceleration?
The figure above shows a uniform beam of length \( L \) and mass \( M \) that hangs horizontally and is attached to a vertical wall. A block of mass \( M \) is suspended from the far end of the beam by a cable. A support cable runs from the wall to the outer edge of the beam. Both cables are of negligible mass. The wall exerts a force \( F_w \) on the left end of the beam. For which of the following actions is the magnitude of the vertical component of \( F_w \) smallest?
A \( 50 \, \text{kg} \) person is sitting on a seesaw \( 1.2 \, \text{m} \) from the balance point. On the other side, a \( 70 \, \text{kg} \) person is balanced. How far from the balance point is the second person sitting?
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
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?
| Wagon | Wheel Structure | Moment of Inertia | Wheel Mass | Wheel Radius |
|---|---|---|---|---|
| Wagon \(A\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.5 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(B\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.2 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(C\) | Hollow hoop | \[M R^2\] | \[ 0.1 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
Three wagons have identical total mass (including their wheels) and each has four wheels. However, the wheels on each wagon have different designs with varying mass distributions and radii as shown in a reference chart. When accelerating each wagon from a standstill to \( 10 \) \( \text{m/s} \), which wagon requires the most energy input?
A mechanical wheel initially at rest on the floor begins rolling forward with an angular acceleration of \( 2\pi \, \text{rad/s}^2 \). If the wheel has a radius of \( 2 \, \text{m} \), what distance does the wheel travel in \( 3 \) seconds?
A motorcycle has tires with a diameter of \( 44.0 \) \( \text{cm} \). Cruising down the highway, they are rotating at \( 1150 \) \( \text{rpm} \) (revolutions per minute).
Car A of mass \( m_A \) is moving to the east along a straight road. Car B of mass \( m_B \) is moving to the north along another straight road. At the instant \( t = 0 \) shown in the figure, both cars are at their closest point to a flagpole, with Car A a distance \( d_A \) from the flagpole and Car B a distance \( d_B \) from the flagpole. The cars continue, each moving with constant speed, and reach the intersection of the two roads at time \( t = t_f \). Which of the following correctly expresses the magnitude of the total angular momentum of the two-car system about the flagpole at time \( t = 0 \)?
A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it, as shown in the figure above. The outside edge of the disk is moving at a linear speed \( v \), and the clay is moving at speed \( \frac{v}{2} \). The clay sticks to the outside edge of the disk. How does the angular momentum of the system after the clay sticks compare to the angular momentum of the system before the clay sticks, and what is an explanation for the comparison?
\(\tau_{\text{net},A}=(1-\sqrt2)F\;\text{Nm}\;\text{(clockwise)},\;\tau_{\text{net},B}=0\;\text{Nm},\;\tau_{\text{net},C}=0\;\text{Nm}\)
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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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