| 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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A massless rigid rod of length \(3d\) is pivoted at a fixed point \(W\), and two forces each of magnitude \(F\) are applied vertically upward as shown above. A third vertical force of magnitude \(F\) may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude \(F\) is applied at point?
How long does it take for a rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s2?
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at a point \( \frac{L}{4} \) away from the center?
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 \( 0.300 \) \( \text{m} \)?
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.
A sphere of mass \( M \) and radius \( r \), and rotational inertia \( I \) is released from the top of an inclined plane of height \( h \). The surface has considerable friction. Using only the variables mentioned, derive an expression for the sphere’s center of mass velocity.
An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his angular momentum about the axis of rotation?
A meterstick is supported at its center, which is aligned with the center of a cradle located at position \( x = 0 \) \( \text{m} \). Two identical objects of mass \( 1.0 \) \( \text{kg} \) are suspended from the meterstick. One object hangs \( 0.25 \) \( \text{m} \) to the left of the support point, and the other object hangs \( 0.50 \) \( \text{m} \) to the right of the support point. The system is released from rest and is free to rotate. Which of the following claims correctly describes the subsequent motion of the system containing the meterstick, cradle, and the two objects?
What is the ratio of the moment of inertia of a cylinder of mass \( m \) and radius \( r \) to the moment of inertia of a hoop of the same mass and same radius?
A grinding wheel is in the form of a uniform solid disk of radius \( 7.00 \) \( \text{cm} \) and mass \( 2.00 \) \( \text{kg} \). It starts from rest and accelerates uniformly under the action of the constant torque of \( 0.600 \) \( \text{N m} \) that the motor exerts on the wheel.
\(T \approx 6.63\times 10^3\,\text{N},\quad H_x \approx 5.74\times 10^3\,\text{N},\quad H_y \approx 5.96\times 10^3\,\text{N}\)
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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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