Pro Tip – Draw an FBD to visualize the all forces and lever arms acting on the ladder. Note that you can split either the forces or the lever arm into components as long as the two are are perpendicular to each other.

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | h = L \sin(\theta) | Calculate the height h of the ladder against the wall using the sine function where \theta is the angle with the ground. |

2 | h = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} \approx 4.33 \, \text{m} | The angle \theta is given as 60^\circ . The \sin(60^\circ) = \frac{\sqrt{3}}{2} . |

3 | w_{\text{lad}} = mg w_{\text{lad}} = 20 \times 9.8 = 196 \, \text{N} |
Calculate the weight of the ladder using its mass m and gravitational acceleration g . |

4 | w_{\text{person}} = m_{\text{person}}g w_{\text{person}} = 80 \times 9.8 = 784 \, \text{N} |
Calculate the weight of the person using the person’s mass m_{\text{person}} and gravitational acceleration g . |

5 | \text{Moment at the bottom} = \text{Moment at the top} | The torque or moment due to the person and the ladder about the point where the bottom of the ladder contacts the ground must be balanced by the force exerted by the wall. |

6 | F_{\text{wall}} \times h = w_{\text{lad}} \times \frac{L}{2} \cos(\theta) + w_{\text{person}} \times d \cos(\theta) | The moment (or torque) at the top due to the force from the wall F_{\text{wall}} must counterbalance the moments generated by the weight of the ladder and person. L is the ladder length, d is the distance where the person stands from the bottom. |

7 | F_{\text{wall}} \times 4.33 = 196 \times \frac{5}{2} \times \frac{1}{2} + 784 \times 4 \times \frac{1}{2} | Substitute values for L = 5 \, \text{m}, d = 4 \, \text{m}, \cos(60^\circ) = \frac{1}{2}, h \approx 4.33 \, \text{m} . |

8 | F_{\text{wall}} \times 4.33 = 98 \times 2.5 + 784 \times 2 | Simplification of the equation to compute the force exerted by the wall. |

9 | F_{\text{wall}} \times 4.33 = 245 + 1568 | Total moments at the top due to the weight of both the ladder and person. |

10 | F_{\text{wall}} \times 4.33 = 1813 | Add the moments for the final calculation. |

11 | F_{\text{wall}} = \frac{1813}{4.33} \approx 418.71 \, \text{N} | Calculate the force exerted by the wall by dividing the total moment by the height h . |

12 | F_{\text{wall}} \approx 419 \, \text{N} |
Finding the final value and rounding off to the nearest whole number, providing the force in Newtons. |

Phy can also check your working. Just snap a picture!

- Statistics

Advanced

Mathematical

GQ

Two workers are holding a thin plate with length 5 m and height 2 m at rest by supporting the plate in the bottom corners. The workers are standing at rest on a slope of 10 degrees. Treat these supporting forces as vertical normal forces and calculate their magnitudes and state if both workers are sharing “the job” fairly.

- Rotational Motion, Torque

Beginner

Mathematical

GQ

- Rotational Kinematics

Beginner

Mathematical

GQ

- Rotational Kinematics

Advanced

Mathematical

MCQ

The figure shows a person’s foot. In that figure, the Achilles tendon exerts a force of magnitude F = 720 N. What is the magnitude of the torque that this force produces about the ankle joint?

- Rotational Motion, Torque

Advanced

Proportional Analysis

MCQ

A ice skater that is spinning in circles has an initial rotational inertia I_{i}. You can approximate her shape to be a cylinder. She is spinning with velocity ω_{i}. As she extends her arms she her rotational inertia changes by a factor of x and her angular velocity changes by a factor of y. Which one of the following options best describe x and y.

- Angular Momentum, Rotational Motion

419 N

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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_1m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

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 |

Momentum | Torque and Rotations |
---|---|

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients 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. |

k | Spring 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 |

Variable | SI 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)} |

Variable | Derived 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

Conversion Example

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

Start with the given measurement:

`\text{5 km}`

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}}`

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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - 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 .
- Bookmark questions that you can’t solve so you can come back to them later.
- 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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