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UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Coordinate System: } +x \text{ (East), } +y \text{ (North)}\] | Define the coordinate system with East as the positive \(x\) direction and North as the positive \(y\) direction. |
| 2 | \[F_{1x} = 170 \;\text{N}, \quad F_{1y} = 0 \;\text{N}\] | Student 1 pulls Eastward with \(170\,\text{N}\); hence, all force is in the \(x\) direction. |
| 3 | \[F_{2x} = 0 \;\text{N}, \quad F_{2y} = -100 \;\text{N}\] | Student 2 pulls Southward with \(100\,\text{N}\); therefore, the \(y\) component is negative. |
| 4 | \[F_{3x} = -200\sin(20^\circ), \quad F_{3y} = 200\cos(20^\circ)\] | Student 3 pulls with \(200\,\text{N}\) at \(20^\circ\) west of north. The \(y\) component is \(200\cos(20^\circ)\) (northward) and the \(x\) component is \(-200\sin(20^\circ)\) (westward). |
| 5 | \[F_{\text{net},x} = 170 – 200\sin(20^\circ)\] | Sum the \(x\) components: Student 1 contributes \(170\,\text{N}\) east, and Student 3 contributes \(-200\sin(20^\circ)\,\text{N}\) (west). |
| 6 | \[F_{\text{net},y} = -100 + 200\cos(20^\circ)\] | Sum the \(y\) components: Student 2 gives \(-100\,\text{N}\) (south) and Student 3 gives \(200\cos(20^\circ)\,\text{N}\) (north). |
| 7 | \(200\sin(20^\circ) \approx 68.4 \;\text{N}, \quad 200\cos(20^\circ) \approx 187.9 \;\text{N}\) | Calculate the approximate numerical values of the components for Student 3. |
| 8 | \[F_{\text{net},x} \approx 170 – 68.4 = 101.6 \;\text{N}\] | Compute the net \(x\) component using the approximated value. |
| 9 | \[F_{\text{net},y} \approx -100 + 187.9 = 87.9 \;\text{N}\] | Compute the net \(y\) component using the approximated value. |
| 10 | \[F_{\text{net}} = \sqrt{(101.6)^2 + (87.9)^2} \approx 134.4 \;\text{N}\] | Find the magnitude of the net force using the Pythagorean theorem. |
| 11 | \[\theta = \tan^{-1}\left(\frac{87.9}{101.6}\right) \approx 40.9^\circ\]\] | Determine the direction of the net force measured as the angle north of east. |
| 12 | \[\boxed{134.4 \;\text{N},\; 40.9^\circ \; \text{north of east}}\] | State the final net force magnitude and its direction. |
Just ask: "Help me solve this problem."
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
If the coefficient of static friction is \( \mu_s = 0.5 \), how much force must be applied to a spring (spring constant of \( 0.8 \) \( \text{N/m} \)) which is attached to a block of wood (mass \( 4.0 \) \( \text{kg} \)) in order to just begin to move the block?
A simple Atwood’s machine remains motionless when equal masses \(M\) are placed on each end of the chord. When a small mass \(m\) is added to one side, the masses have an acceleration \(a\). What is \(M\)? You may neglect friction and the mass of the cord and pulley.
Two balls have their centers \( 2.0 \) \( \text{m} \) apart. One ball has a mass of \( 8.0 \) \( \text{kg} \). The other has a mass of \( 6.0 \) \( \text{kg} \). What is the gravitational force between them?
The heaviest train ever pulled by a single engine was over [katex] 2 \, \text{km} [/katex] long. A force of [katex] 1.13 \times 10^5 \, \text{N} [/katex] is needed to get the train to start moving. If the coefficient of static friction is [katex] 0.741 [/katex] and the coefficient of kinetic friction is [katex] .592 [/katex], what is the train’s mass?
\(\boxed{134.4\,\text{N}\text{ at }40.9^\circ\text{ north of east}}\)
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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] |
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] |
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