| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \sum \vec{F} = 0 \] | Translational equilibrium means that the net force acting on the object is zero. |
| 2 | \[ \vec{a} = \frac{\sum \vec{F}}{m} = 0 \] | According to Newton’s second law, if the net force is zero, the acceleration of the object must also be zero. |
| 3 | \text{Selected: (b) The net force on the object is 0 and (c) The acceleration of the object is 0} | For translational equilibrium, both the net force and the resulting acceleration must be zero. |
| 4 | \text{Explanation for other options:} | Option (a): An object may have nonzero kinetic energy even while moving at a constant velocity. Option (d): Net torque being zero is a condition for rotational equilibrium, not necessarily translational equilibrium. |
| 5 | \[\boxed{\text{(b) and (c)}}\] | This is the final answer, satisfying the conditions required for translational equilibrium. |
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There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
If an elephant were chasing you, its enormous mass would be most threatening. But if you zigzagged, its mass would be to your advantage. Why?
A boy and a girl are balanced on a massless seesaw. The boy has a mass of \(60 \, \text{kg}\) and the girl’s mass is \(50 \, \text{kg}\). If the boy sits \(1.5 \, \text{m}\) from the pivot point on one side of the seesaw, where must the girl sit on the other side for equilibrium?
A spring with a spring constant of \( 600. \) \( \text{N/m} \) is used for a scale to weigh fish. What is the mass of a fish that would stretch the spring by \( 7.5 \) \( \text{cm} \) from its normal length?
Young David experimented with slings before tackling Goliath. He found that he could develop an angular speed of \( 8.0 \) \( \text{rev/s} \) in a sling \( 0.60 \) \( \text{m} \) long. If he increased the length to \( 0.90 \) \( \text{m} \), he could revolve the sling only \( 6.0 \) times per second.
A meter stick of mass 200 grams is balanced at the 40-cm mark when a 100-gram mass is suspended from the 10-cm mark. What is the distance from the pivot point to the center of mass of the meter stick? Give your answer in centimeters.
A \( 4 \)-\( \text{kg} \) ball and a \( 1 \)-\( \text{kg} \) ball are positioned a distance \( L \) apart on a bar of negligible mass. How far from the \( 4 \)-\( \text{kg} \) mass should the fulcrum be placed to balance the bar?
If I weigh \( 741 \) \( \text{N} \) on Earth at a place where \( g = 9.80 \) \( \text{m/s}^2 \) and \( 5320 \) \( \text{N} \) on the surface of another planet, what is the acceleration due to gravity on that planet?
Find the tension in each cable supporting the gymnast who weighs \( 600 \) \( \text{N} \). The gymnast is at rest, holding a junction point where two cables are attached: one cable is horizontal, and the second cable is attached to the ceiling making an angle of \( 37^{\circ} \) above the horizontal, as shown in the diagram.
A forward horizontal force of \(12 \, \text{N}\) is used to pull a \(240 \, \text{N}\) crate at constant velocity across a horizontal floor. The coefficient of friction is
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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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