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Objective: Determine the acceleration of the blocks and the tension in the cord given the masses, coefficients of friction, and incline angle given:
Part a: Calculate the acceleration of the blocks
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] F_{\text{gravity, parallel A}} = m_A g \sin(\theta) [/katex] | Parallel component of gravitational force for A. |
| 2 | [katex] F_{\text{gravity, parallel B}} = m_B g \sin(\theta) [/katex] | Parallel component of gravitational force for B. |
| 3 | [katex] F_{\text{friction A}} = \mu_A m_A g \cos(\theta) [/katex] | Frictional force on A. |
| 4 | [katex] F_{\text{friction B}} = \mu_B m_B g \cos(\theta) [/katex] | Frictional force on B. |
| 5 | [katex] F_{\text{net A}} = F_{\text{gravity, parallel A}} – F_{\text{friction A}} [/katex] | Net force on A. |
| 6 | [katex] F_{\text{net B}} = F_{\text{gravity, parallel B}} – F_{\text{friction B}} [/katex] | Net force on B. |
| 7 | [katex] F_{\text{net}} = F_{\text{net B}} – F_{\text{net A}} [/katex] | Total net force on the system. |
| 8 | [katex] a = \frac{F_{\text{net}}}{m_A + m_B} [/katex] | Acceleration of the system. |
Plug in the given values:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 9 | [katex] a = \frac{(m_B g \sin(\theta) – \mu_B m_B g \cos(\theta)) – (m_A g \sin(\theta) – \mu_A m_A g \cos(\theta))}{m_A + m_B} [/katex] | Substitute the net forces from steps 5 and 6. |
| 10 | [katex] a = \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{5 + 5} [/katex] | Substitute given values. |
| 11 | [katex] a = \frac{(5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.30 \cdot \cos(32^\circ))) – (5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.20 \cdot \cos(32^\circ)))}{10} [/katex] | Simplify the expression. |
| 12 | [katex] a = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} [/katex] | Combine like terms. |
| 13 | [katex] a = \frac{9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} [/katex] | Simplify further. |
Part b: Calculate the tension in the cord
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] T = m_A \cdot a + F_{\text{friction A}} [/katex] | Tension equals the force to accelerate block A plus frictional force on A. |
| 2 | [katex] F_{\text{friction A}} = \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] | Frictional force opposing the motion of block A. |
| 3 | [katex] T = m_A \cdot a + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] | Substitute the frictional force into the tension formula. |
| 4 | [katex] T = m_A \cdot \left( \frac{F_{\text{net}}}{m_A + m_B} \right) + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] | Substitute the expression for � from the acceleration calculation. |
| 5 | [katex] T = 5 \cdot \left( \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Insert given values for masses, coefficients of friction, gravitational acceleration, and angle. |
| 6 | [katex] T = 5 \cdot \left( \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Simplify the expression for the net force component. |
| 7 | [katex] T = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Further simplify the tension formula. |
| 8 | [katex] T = \frac{5 \cdot 9.8 \cdot 0.10 \cdot \cos(32^\circ)}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Combine like terms for the final tension calculation. |
| 9 | [katex] T \approx 17.66 , \text{N} [/katex] | Calculate the numeric value for tension. |
[katex] \boxed{T = 17.66 , \text{N}} [/katex]
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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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