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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."
During the experiment, students collect data about the angular momentum of a rigid, uniform spinning wheel about an axle as a function of time, which was used to create the graph that is shown. A frictional torque is exerted on the wheel. A student makes the following statement about the data. “The frictional torque exerted on the wheel is independent of the wheel’s angular speed.” Does the data from the graph support the student’s statement? Justify your selection.
A solid ball of mass \( M \) and radius \( R \) has rotational inertia \( \frac{2}{5} M R^{2} \) about its center. It rolls without slipping along a level surface at speed \( v \) just before it begins rolling up an inclined plane. Which of the following expressions correctly represents the maximum vertical height the solid ball can ascend to when it rolls up the incline without slipping?
A ladder at rest is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on the ladder by the floor?
Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?
The system above is NOT balanced since \(m_2\) is twice the mass of \(m_1\). Which of the following changes would NOT balance the system so that there is 0 net torque? Assume the plank has no mass of its own.
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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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