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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 uniform solid cylinder of mass [katex] M [/katex] and radius [katex] R [/katex] is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force [katex] F [/katex] , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle [katex] \theta [/katex], what is the kinetic energy of the cylinder in terms of [katex] F [/katex] and [katex] \theta [/katex]?
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(5M\)that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will result in an increase in the total angular momentum of the child-platform system?
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)?
A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal, \( 2.3 \) \( \text{m} \)-long, \( 6.1 \) \( \text{kg} \) reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A \( 64 \) \( \text{kg} \) woman lies on the reaction board with her feet over the pivot. The scale reads \( 27 \) \( \text{kg} \). What is the distance from the woman’s feet to her center of mass? Express your answer with the appropriate units.
A mechanical wheel initially at rest on the floor begins rolling forward with an angular acceleration of \( 2\pi \, \text{rad/s}^2 \). If the wheel has a radius of \( 2 \, \text{m} \), what distance does the wheel travel in \( 3 \) seconds?
[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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