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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."
A solid metal bar is at rest on a horizontal frictionless surface. It is free to rotate about a vertical axis at the left end. The figures below show forces of different magnitudes that are exerted on the bar at different locations. In which case does the bar’s angular speed about the axis increase at the fastest rate?
A mechanical wheel initially at rest on the floor begins rolling forward with an angular acceleration of \( 2\pi \, \text{rad/s}^2 \). If the wheel has a radius of \( 2 \, \text{m} \), what distance does the wheel travel in \( 3 \) seconds?
A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
A sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
A 5-meter long ladder is leaning against a wall, with the bottom of the ladder 3 meters from the wall. The ladder is uniform and has a mass of 20 kg. A person of mass 80 kg is standing on the ladder at a distance of 4 meters from the bottom of the ladder. The ladder makes an angle of 60 degrees with the ground. What is the force exerted by the wall on the ladder?
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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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