| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[m_b = 225\,\text{kg},\quad m_s = 45.0\,\text{kg},\quad g = 9.80\,\text{m/s}^2\] | Identify given masses: block \(m_b\), uniform strut \(m_s\). Use \(g\) for weight calculations. |
| 2 | \[W_b = m_b g = (225)(9.80)=2205\,\text{N}\] | Weight of the hanging block acts downward at the strut end. |
| 3 | \[W_s = m_s g = (45.0)(9.80)=441\,\text{N}\] | Weight of the uniform strut acts downward at its center of mass (midpoint). |
| 4 | \[\theta_s = 45^\circ,\quad \theta_T = 30^\circ\] | Strut makes \(45^\circ\) with the horizontal; cable makes \(30^\circ\) with the horizontal (as shown). |
| 5 | \[\Delta x_{\text{end}} = L\cos 45^\circ,\quad \Delta x_{\text{mid}} = \tfrac{L}{2}\cos 45^\circ\] | For torques about the hinge, the lever arm for vertical weights equals the horizontal distance to their lines of action. |
| 6 | \[\phi = |45^\circ-30^\circ|=15^\circ\] | The angle between the strut (position vector) and the cable force is \(15^\circ\), needed for the cable torque magnitude. |
| 7 | \[\tau_T = T\,L\sin 15^\circ\] | Torque magnitude from cable about hinge: \(\tau = rF\sin\phi\) with \(r=L\). |
| 8 | \[\tau_{W_b} = W_b\,(L\cos 45^\circ)\] | Block’s weight produces clockwise torque about hinge with lever arm \(L\cos45^\circ\). |
| 9 | \[\tau_{W_s} = W_s\,\Big(\tfrac{L}{2}\cos 45^\circ\Big)\] | Strut’s weight acts at midpoint, so lever arm is \(\tfrac{L}{2}\cos45^\circ\), also clockwise. |
| 10 | \[\sum \tau_{\text{hinge}}=0:\quad T L\sin 15^\circ – W_b(L\cos45^\circ) – W_s\Big(\tfrac{L}{2}\cos45^\circ\Big)=0\] | Equilibrium requires net torque about hinge equals zero. Cable torque counterclockwise balances clockwise torques from weights. |
| 11 | \[T\sin 15^\circ = \cos45^\circ\Big(W_b+\tfrac{1}{2}W_s\Big)\] | Cancel \(L\) and factor \(\cos45^\circ\) to solve for \(T\). |
| 12 | \[T = \dfrac{\cos45^\circ\,(W_b+\tfrac{1}{2}W_s)}{\sin15^\circ}\] | Algebraic isolation of \(T\). |
| 13 | \[T = \dfrac{(0.7071)\,(2205+0.5\cdot441)}{0.2588} = \dfrac{(0.7071)(2425.5)}{0.2588} \approx 6626\,\text{N}\] | Substitute numeric values: \(\cos45^\circ\approx0.7071\), \(\sin15^\circ\approx0.2588\). |
| 14 | \[\boxed{T \approx 6.63\times 10^3\,\text{N}}\] | Final cable tension. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\sum F_x=0:\quad H_x + T_x = 0\] | Only horizontal forces are the hinge reaction \(H_x\) and the cable’s horizontal component \(T_x\) (weights are vertical). |
| 2 | \[T_x = -T\cos30^\circ\] | The cable pulls the strut toward the left along the cable, so its horizontal component on the strut is negative (left). |
| 3 | \[H_x – T\cos30^\circ=0\] | Substitute \(T_x\) into equilibrium equation. |
| 4 | \[H_x = T\cos30^\circ\] | Solve for hinge horizontal component. |
| 5 | \[H_x = (6626)(0.8660)\approx 5739\,\text{N}\] | Compute with \(\cos30^\circ\approx0.8660\). |
| 6 | \[\boxed{H_x \approx 5.74\times 10^3\,\text{N}\ \text{(to the right)}}\] | Direction is to the right to balance the cable’s leftward pull. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\sum F_y=0:\quad H_y + T_y – W_b – W_s = 0\] | Vertical forces: hinge reaction \(H_y\) (unknown), cable vertical component \(T_y\) (downward), and the two weights (downward). |
| 2 | \[T_y = -T\sin30^\circ\] | Cable pulls the strut down along the cable direction at the strut end, so the vertical component is negative (down). |
| 3 | \[H_y – T\sin30^\circ – W_b – W_s = 0\] | Substitute \(T_y\) into the vertical equilibrium equation. |
| 4 | \[H_y = T\sin30^\circ + W_b + W_s\] | Algebraic solution for \(H_y\). |
| 5 | \[H_y = (6626)(0.5) + 2205 + 441 = 3313 + 2646 = 5959\,\text{N}\] | Use \(\sin30^\circ=0.5\) and add total downward loads the hinge must support. |
| 6 | \[\boxed{H_y \approx 5.96\times 10^3\,\text{N}\ \text{(upward)}}\] | Upward hinge force balances all downward forces. |
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\(T \approx 6.63\times 10^3\,\text{N},\quad H_x \approx 5.74\times 10^3\,\text{N},\quad H_y \approx 5.96\times 10^3\,\text{N}\)
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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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