| 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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What is the net torque acting on the pivot supporting a \(10 \, \text{kilogram}\) beam \(2 \, \text{meters}\) long as shown above? Assume that the positive direction is clockwise.
An isolated spherical star of radius \( R_o \), rotates about an axis that passes through its center with an angular velocity of \( \omega_o \). Gravitational forces within the star cause the star’s radius to collapse and decrease to a value \( r_o < R_o \), but the mass of the star remains constant. A graph of the star’s angular velocity as a function of time as it collapses is shown. Which of the following predictions is correct about the angular momentum \( L \) of the star immediately after the collapse?
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of \( 5.0 \) \( \text{rev/s} \) in \( 8.0 \) \( \text{s} \). At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in \( 12.0 \) \( \text{s} \). Through how many revolutions does the tub turn during the entire \( 20 \)-s interval? Assume constant angular acceleration while it is starting and stopping.
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?
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Two spheres of equal size and equal mass are rotated with an equal amount of torque. One of the spheres is solid with its mass evenly distributed throughout its volume, and the other is hollow with all of its mass concentrated at the edges. Which sphere would rotate faster if the same amount of torque is applied for the same period of time for both?
A meter stick of mass \( .2 \) kg is pivoted at one end and supported horizontally. A force of \( 3 \) N downwards is applied to the free end, perpendicular to the length of the meter stick. What is the net torque about the pivot point?
\(T \approx 6.63\times 10^3\,\text{N},\quad H_x \approx 5.74\times 10^3\,\text{N},\quad H_y \approx 5.96\times 10^3\,\text{N}\)
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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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