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To solve the problem related to the balanced seesaw with a boy and a girl sitting on it, we adhere to the principles of torque and leverage. Here, the seesaw must balance so the torques due to the boy and girl must be equal in magnitude but opposite in direction.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\tau_{boy} = \tau_{girl}[/katex] | This equation states the balancing condition where the torque ([katex]\tau[/katex]) due to the boy must equal the torque due to the girl for the seesaw to be in equilibrium. |
| 2 | [katex]m_{boy} \cdot g \cdot d_1 = m_{girl} \cdot g \cdot d_2[/katex] | Torque ([katex]\tau[/katex]) is calculated by the formula [katex]\tau = F \cdot d[/katex] where [katex]F[/katex] is the force (here, the weight of the children, [katex]m \cdot g[/katex]) and [katex]d[/katex] is the distance from the pivot. Here, [katex]g[/katex] is the acceleration due to gravity, [katex]m_{boy}[/katex] and [katex]m_{girl}[/katex] are the masses of the boy and girl respectively, and [katex]d_1[/katex] and [katex]d_2[/katex] are their respective distances from the fulcrum. |
| 3 | [katex]\frac{m_{boy}}{m_{girl}} = \frac{d_2}{d_1}[/katex] | Divide both sides of the equation by [katex]g \cdot d_1 \cdot d_2[/katex] to isolate the ratio of masses, which shows that the ratio of the boy’s mass to the girl’s mass is the inverse of their distances from the fulcrum. This ratio will ensure that their torques balance each other. |
| 4 | Mass of seesaw needed: [katex]m_{seesaw} \cdot g \cdot L = (m_{boy} + m_{girl}) \cdot g \cdot \frac{(d_2 – d_1)}{2}[/katex] | We need to add the minimum mass of the seesaw to keep it balanced at the pivot point itself. Assuming the mass is evenly distributed, its leverage point would be at the center ([katex]\frac{L}{2}[/katex] from the pivot). The seesaw’s mass should counteract any net torque resultant from the boy and girl’s differing distances from the pivot. Here, [katex]L[/katex] is the total length of the seesaw. |
| 5 | [katex]m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L}[/katex] | Re-arranging the equation to solve for [katex]m_{seesaw}[/katex]. This formula calculates the minimum mass of the seesaw required to achieve balance. Note that since [katex]d_1 > d_2[/katex], [katex](d_1 – d_2)[/katex] will be positive, ensuring a positive mass for the seesaw. |
| 6 | [katex]m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L}[/katex] | This is the final formula that yields the mass of the seesaw needed to balance with the boy and girl placed as described. |
Just ask: "Help me solve this problem."
A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.
A Christmas ornament made from a thin hollow glass sphere hangs from a thin wire of negligible mass. It is observed to oscillates with a frequency of \( 2.50 \) \( \text{Hz} \) in a city where \( g = 9.80 \) \( \text{m/s}^2 \). What is the radius of the ornament? The moment of inertia of the ornament is given by \( I = \frac{5}{3} mr^2 \).
Which of the following must be true for an object at translational equilibrium?
A wheel of radius \( R \) and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses \( m \), \( M \), and \( 2M \), respectively, are mounted on the rim of the wheel, as shown above. If the system is in static equilibrium, what is the value of \( m \) in terms of \( M \)?
Four forces are exerted on a disk of radius \( R \) that is free to spin about its center, as shown above. The magnitudes are proportional to the length of the force vectors, where \( F_1 = F_4 \), \( F_2 = F_3 \), and \( F_1 = 2F_2 \). Which two forces combine to exert zero net torque on the disk?
[katex]m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L}[/katex]
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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] |
General Metric Conversion Chart
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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