| 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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Four systems are in rotational motion. Which of the following combinations of rotational inertia and angular speed for each of the systems corresponds to the greatest rotational kinetic energy?
| System | Rotational Inertia | Angular Speed |
|---|---|---|
| A | \( I_0 \) | \( \omega_0 \) |
| B | \( I_0 \) | \( 4\, \omega_0 \) |
| C | \( 2 I_0 \) | \( 2\, \omega_0 \) |
| D | \( 6 I_0 \) | \( \omega_0 \) |
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at either end of the rod?
Which of the following must be zero if an object is spinning at a constant rate? There may be more than one right answer.
In both cases, a massless rod is supported by a fulcrum, and a \(200 \, \text{kg}\) hanging mass is suspended from the left end of the rod by a cable. A downward force \(F\) keeps the rod in rest. The rod in Case A is \(50 \, \text{cm}\) long, and the rod in Case B is \(40 \, \text{cm}\) long (each rod is marked at \(10 \, \text{cm}\) intervals). The magnitude of each vertical force \(F\) exerted on the rod will be
A \(25 \, \text{g}\) steel ball is attached to the top of a \(24 \, \text{cm}\)-diameter vertical wheel of negligible mass. Starting from rest, the wheel accelerates at \(470 \, \text{rad/s}^2\). The ball is released after \(\frac{3}{4}\) of a revolution. How high does it go above the center of the wheel?
A light string is attached to a massive pulley of known rotational inertia \( I_P \), as shown in the figure. A student must determine the relationship between the torque exerted on the pulley and the change in the pulley’s angular velocity when the torque is applied for \( 2.0 \) \( \text{s} \). In addition to a stopwatch to measure the time interval, what two measurements could the student make in order to determine the relationship? Select two answers.
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end (assume it does not slip). The other end of the beam is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the (1) tension force in the cable and (2) the total force the wall exerts on the beam.
The driver of a car traveling at \( 30.0 \) \( \text{m/s} \) applies the brakes and undergoes a constant negative acceleration of \( 2.00 \) \( \text{m/s}^2 \). How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of \( 0.300 \) \( \text{m} \)?
Which of the following situations will increase the moment of inertia of a solid cylinder \( I = \tfrac{1}{2} M R^{2} \) by the same amount?
A net torque is applied to the edge of a spinning object as it rotates about its internal axis. The table shows the net torque exerted on the object at different instants in time. How can a student use the data table to determine the change in angular momentum of the object from \( 0 \) to \( 6 \) \( \text{s} \)? Justify your selection.
| Time \( (\text{s}) \) | Net Torque \( (\text{N} \cdot \text{m}) \) |
|---|---|
| 0 | 0 |
| 2 | 1.5 |
| 4 | 3.0 |
| 6 | 4.5 |
\(T \approx 6.63\times 10^3\,\text{N},\quad H_x \approx 5.74\times 10^3\,\text{N},\quad H_y \approx 5.96\times 10^3\,\text{N}\)
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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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