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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."
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A car is moving up the side of a circular roller coaster loop of radius \( 12 \) \( \text{m} \). The angular velocity is \( 1.8 \) \( \text{rad/s} \) and angular acceleration is \( -0.82 \) \( \text{rad/s}^2 \). The car is at the same elevation as the center of the loop. Find the magnitude and direction (relative to the horizontal) of the acceleration.
In an experiment, an external torque is applied to the edge of a disk of radius \( 0.5 \) \( \text{m} \) such that the edge of the disk speeds up as it continues to rotate. The tangential speed as a function of time is shown for the edge of the disk. The rotational inertia of the disk is \( 0.125 \) \( \text{kg} \cdot \text{m}^2 \). Can a student use the graph and the known information to calculate the net torque exerted on the edge of the disk?
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 \( \frac{1}{2} m r^2 \).
Two workers are holding a thin plate with length \(5 \, \text{m}\) and height \(2 \, \text{m}\) at rest by supporting the plate in the bottom corners. The workers are standing at rest on a slope of \(10^\circ\). Treat these supporting forces as vertical normal forces and calculate their magnitudes and state if both workers are sharing “the job” fairly.
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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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