| 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 pulley has an initial angular speed of \( 12.5 \) \( \text{rad/s} \) and a constant angular acceleration of \( 3.41 \) \( \text{rad/s}^2 \). Through what angle does the pulley turn in \( 5.26 \) \( \text{s} \)?
A disk of radius 35 cm rotates at a constant angular velocity of 10 rad/s. How fast does a point on the rim of the disk travel (in m/s)?
Two uniform solid balls, one of radius \( R \) and mass \( M \), the other of radius \( 2R \) and mass \( 8M \), roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting?
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously from the top of an inclined plane and do not slip, which one will reach the bottom first?
The moment of inertia of a solid cylinder about its axis is given by \( 0.5MR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A uniform ladder of length \(L\) and weight \(W = 50 \, \text{N}\) rests against a smooth vertical wall. If the coefficient of static friction between the ladder and the ground is \(\mu = 0.4\).
An object weighing 120 N is set on a rigid beam of negligible mass at a distance of 3 m from a pivot, as shown above. A vertical force is to be applied to the other end of the beam a distance of 4 m from the pivot to keep the beam at rest and horizontal. What is the magnitude F of the force required?
A disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
To increase the moment of inertia of a body about an axis, you must
\(T \approx 6.63\times 10^3\,\text{N},\quad H_x \approx 5.74\times 10^3\,\text{N},\quad H_y \approx 5.96\times 10^3\,\text{N}\)
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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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