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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[W_b = 1450 \times 9.8,\quad W_{block} = 80 \times 9.8\] | Compute the weights of the steel beam and the block using \(g=9.8\,\text{m/s}^2\). |
| 2 | \[d_b = \frac{6.6}{2} = 3.3\,\text{m},\quad d_{block} = 6.6 – 2 = 4.6\,\text{m}\] | The beam’s weight acts at its midpoint, and the block is placed 2 m from the free (cable) end so its distance from the wall is \(6.6-2=4.6\,\text{m}\). |
| 3 | \[T\sin(30^\circ)\times 6.6 = W_b\times3.3 + W_{block}\times4.6\] | Take moments about the wall. Only the vertical component of the cable’s tension produces a moment. The beam’s and block’s weights (acting downward) tend to rotate the beam clockwise, while the vertical component of the cable (acting upward) produces a counterclockwise moment. |
| 4 | \[T = \frac{W_b\times3.3 + W_{block}\times4.6}{6.6\sin(30^\circ)}\] | Solve the moment equation for the cable tension \(T\). |
| 5 | \[T = \frac{1450(9.8)(3.3) + 80(9.8)(4.6)}{6.6\times0.5}\] | Substitute the numerical values and use \(\sin(30^\circ)=0.5\). |
| 6 | \[T = \frac{(1450\times9.8\times3.3)+(80\times9.8\times4.6)}{3.3}\] | Simplify the denominator since \(6.6\times0.5 = 3.3\). |
| 7 | \[T \approx \frac{46893+3606.4}{3.3} \approx \frac{50499.4}{3.3} \approx 15303\,\text{N}\] | The numerator comes from \(1450\times9.8\times3.3\approx46893\,\text{N}\cdot\text{m}\) and \(80\times9.8\times4.6\approx3606.4\,\text{N}\cdot\text{m}\). Dividing by 3.3 gives the cable tension. |
| 8 | \[\boxed{T \approx 1.53 \times 10^{4}\,\text{N} \text{ (directed along the cable, }30^\circ\text{ above horizontal)}}\] | This is the force between the cable and the wall; the cable pulls on the wall along its length, at \(30^\circ\) above the horizontal. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[R_x = T\cos(30^\circ)\] | For horizontal equilibrium of the beam, the cable’s horizontal pull must be balanced by the wall’s horizontal reaction. |
| 2 | \[R_x \approx 15303\times0.866 \approx 13245\,\text{N}\] | Using \(\cos(30^\circ)\approx0.866\), calculate the horizontal component of the reaction force. |
| 3 | \[R_y = (W_b + W_{block}) – T\sin(30^\circ)\] | For vertical equilibrium the wall’s vertical reaction plus the cable’s vertical component must support the weights of the beam and block. |
| 4 | \[R_y = (1450\times9.8+80\times9.8) – 15303\times0.5\] | Substitute the values, recalling that \(1450+80=1530\) and \(\sin(30^\circ)=0.5\). |
| 5 | \[R_y \approx 1530\times9.8 – 7651.5 \approx 14994 – 7651.5 \approx 7342.5\,\text{N}\] | Compute the total weight, then subtract the cable’s vertical contribution to find the vertical reaction at the wall. |
| 6 | \[R = \sqrt{R_x^2+R_y^2} \approx \sqrt{(13245)^2+(7342.5)^2} \approx 15152\,\text{N}\] | Combine the horizontal and vertical components to get the magnitude of the reaction force at the beam–wall contact. |
| 7 | \[\theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right) \approx \tan^{-1}\left(\frac{7342.5}{13245}\right) \approx 29^\circ\] | Determine the angle of the reaction force relative to the horizontal. |
| 8 | \[\boxed{R \approx 1.52 \times 10^{4}\,\text{N} \text{ at }29^\circ \text{ above the horizontal}}\] | This is the net force between the steel beam and the wall. Its horizontal component balances the cable’s pull and its vertical component supports the beam and block. |
Just ask: "Help me solve this problem."
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?
A centrifuge rotor rotating at \( 9200 \) \( \text{rpm} \) is shut off and is eventually brought uniformly to rest by a frictional torque of \( 1.20 \) \( \text{N} \cdot \text{m} \). If the mass of the rotor is \( 3.10 \) \( \text{kg} \) and it can be approximated as a solid cylinder of radius \( 0.0710 \) \( \text{m} \), through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is given by \( \frac{1}{2} m r^2 \).
A windmill blade with a rotational inertia of \( 6.0 \) \( \text{kg} \cdot \text{m}^2 \) has an initial angular velocity of \( 8 \) \( \text{rad/s} \) in the clockwise direction. It is then given an angular acceleration of \( 4 \) \( \text{rad/s}^2 \) in the clockwise direction for \( 10 \) seconds. What is the change in rotational kinetic energy of the blade over this time interval?
A seesaw is balanced on a fulcrum, with a boy of mass [katex] M_1 [/katex] sitting on one end and a girl of mass [katex] M_2 [/katex] sitting on the other end. The seesaw is a uniform plank of length [katex]L[/katex] and mass [katex] M[/katex]. The fulcrum is located at the midpoint of the plank. Does [katex] M_1 = M_2 [/katex]. Justify your working.
A hoop with a mass [katex]m[/katex] and unknown radius is rolling without slipping on a flat surface with an angular speed [katex]\omega[/katex]. The hoop encounters a hill and continues to roll without slipping until it reaches a maximum height [katex]h[/katex].
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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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