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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 \tau_{boy} = \tau_{girl} This equation states the balancing condition where the torque (\tau) due to the boy must equal the torque due to the girl for the seesaw to be in equilibrium.
2 m_{boy} \cdot g \cdot d_1 = m_{girl} \cdot g \cdot d_2 Torque (\tau) is calculated by the formula \tau = F \cdot d where F is the force (here, the weight of the children, m \cdot g) and d is the distance from the pivot. Here, g is the acceleration due to gravity, m_{boy} and m_{girl} are the masses of the boy and girl respectively, and d_1 and d_2 are their respective distances from the fulcrum.
3 \frac{m_{boy}}{m_{girl}} = \frac{d_2}{d_1} Divide both sides of the equation by g \cdot d_1 \cdot d_2 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: m_{seesaw} \cdot g \cdot L = (m_{boy} + m_{girl}) \cdot g \cdot \frac{(d_2 – d_1)}{2} 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 (\frac{L}{2} 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, L is the total length of the seesaw.
5 m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L} Re-arranging the equation to solve for m_{seesaw}. This formula calculates the minimum mass of the seesaw required to achieve balance. Note that since d_1 > d_2, (d_1 – d_2) will be positive, ensuring a positive mass for the seesaw.
6 m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L} This is the final formula that yields the mass of the seesaw needed to balance with the boy and girl placed as described.

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m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L}

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##### Dark Mode Equation Sheet (Download)
KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \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}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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