Note \(T_a \) represents tension in the angled wire. \(T_h \) represents tension in the horizontal wire.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[W = 600\,\text{N}\] | Define the gymnast’s weight \(W\). |
| 2 | \[\sum F_x :\; T_a \cos 37^\circ – T_h = 0\] | Horizontal equilibrium: the horizontal component of \(T_a\) balances \(T_h\). |
| 3 | \[\sum F_y :\; T_a \sin 37^\circ – W = 0\] | Vertical equilibrium: the upward component of \(T_a\) supports the weight. |
| 4 | \[T_a = \frac{W}{\sin 37^\circ}\] | Solve the vertical equilibrium equation for the angled-cable tension \(T_a\). |
| 5 | \[T_a = \frac{600}{\sin 37^\circ} \approx 9.97\times10^2\,\text{N}\] | Insert \(W = 600\,\text{N}\) and \(\sin 37^\circ \approx 0.602\). |
| 6 | \[T_h = T_a \cos 37^\circ\] | Use the horizontal equilibrium equation to solve for the horizontal-cable tension \(T_h\). |
| 7 | \[T_h = 9.97\times10^2 \times \cos 37^\circ \approx 7.96\times10^2\,\text{N}\] | Insert \(T_a\) and \(\cos 37^\circ \approx 0.799\). |
| 8 | \[\boxed{T_a \approx 1.0 \text{ kN}},\;\boxed{T_h \approx 8.0\times10^2 \text{ N}}\] | Final tensions in the angled and horizontal cables. |
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Which of the following best explains why astronauts experience weightlessness while orbiting the earth?
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
A \( 60 \ \text{kg} \) person is riding in an elevator. At time \( t_1 \), the elevator is accelerating downward with a magnitude of \( 2 \ \text{m/s}^2 \). A short time later, at time \( t_2 \), the elevator is accelerating upward with a magnitude of \( 2 \ \text{m/s}^2 \). The ratio of the normal force exerted by the elevator on the person at time \( t_1 \) to that at time \( t_2 \) is most nearly
Three blocks are stacked on top of one another. The top block has a mass of \( 4.6 \, \text{kg} \), the middle one has a mass of \( 1.2 \, \text{kg} \), and the bottom one has a mass of \( 3.7 \, \text{kg} \).
Identify and calculate any normal forces between the objects.
A conical pendulum is formed by attaching a ball of mass \( m \) to a string of length \( \ell \), then allowing the ball to move in a horizontal circle of radius \( r \). The following figure shows that the string traces out the surface of a cone, hence the name.
A block hangs from the ceiling by a massless rope. A \( 3.0 \, \text{kg} \) block is attached to the first block and hangs below it on another piece of massless rope. The tension in the top rope is \( 63.0 \, \text{N} \).
If an object is moving, is it possible for the net force acting on it to be zero? Explain.
When a horizontal force of \( 4.5 \, \text{N} \) acts on a block on a resistance-free surface, it produces an acceleration of \( 2.5 \, \text{m/s}^2 \). Suppose a second \( 4.0 \, \text{kg} \) block is dropped onto the first. What is the magnitude of the acceleration of the combination if the same force continues to act? Assume that the second block does not slide on the first block.
In the diagram shown, a \(20 \, \text{N}\) force is applied to block \(B\) (\(7 \, \text{kg}\)). Block \(A\) has a mass of \(3 \, \text{kg}\). Assume frictionless conditions.
A point \( P \) is subjected to three simultaneous forces of magnitudes \( F_A > F_B > F_C \). Point \( P \) is in equilibrium. Which of the following statements is not always true about the magnitudes of the forces?
\(T_a \approx 1.0 \times 10^3 \\text{ N}\)
\(T_h \approx 8.0 \times 10^2 \\text{ N}\)
Note \(T_a \) represents tension in the angled wire. \(T_h \) represents tension in the horizontal wire.
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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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