# The Force of gravity

**Mass** is a measure of an object's inertia. **Mass** also determines the strength of gravity.
Because of gravity all objects are attracted to each other, but we mostly notice the attraction towards the Earth because
it is so large and so close.

It's funny, we have a special word just for the direction of gravity, down.

# $$F_g=mg $$

\(F_g\) = the force of gravity, weight [N, Newtons, pounds]*vector*

\(m\) = mass [kg]

\(g\) = acceleration of gravity on the surface of the Earth = 9.8 [m/s²]

*vector*

The force of gravity depends on the mass of the object, but on the surface of the Earth **acceleration from gravity is the same for all objects.**

So why do feathers fall slower than bricks?

The acceleration of 9.8 m/s² on the surface of the Earth is just an approximation. The gravity of earth changes a bit depending on where you are.

What is the force of gravity in Los Angeles?

Why would the acceleration of gravity seem lower near the equator?

**Table:** Comparative gravities in various cities around the world

Location | Acceleration in m/s² | Acceleration in ft/s² |
---|---|---|

Amsterdam | 9.813 | 32.19 |

Athens | 9.800 | 32.15 |

Auckland | 9.799 | 32.15 |

Bangkok | 9.783 | 32.1 |

Brussels | 9.811 | 32.19 |

Buenos Aires | 9.797 | 32.14 |

Calcutta | 9.788 | 32.11 |

Cape Town | 9.796 | 32.14 |

Chicago | 9.803 | 32.16 |

Copenhagen | 9.815 | 32.2 |

Frankfurt | 9.810 | 32.19 |

Havana | 9.788 | 32.11 |

Helsinki | 9.819 | 32.21 |

Istanbul | 9.808 | 32.18 |

Jakarta | 9.781 | 32.09 |

Kuwait | 9.793 | 32.13 |

Lisbon | 9.801 | 32.16 |

London | 9.812 | 32.19 |

Los Angeles | 9.796 | 32.14 |

Madrid | 9.800 | 32.15 |

Manila | 9.784 | 32.1 |

Mexico City | 9.779 | 32.08 |

Montréal | 9.789 | 32.12 |

New York City | 9.802 | 32.16 |

Nicosia | 9.797 | 32.14 |

Oslo | 9.819 | 32.21 |

Ottawa | 9.806 | 32.17 |

Paris | 9.809 | 32.18 |

Rio de Janeiro | 9.788 | 32.11 |

Rome | 9.803 | 32.16 |

San Francisco | 9.800 | 32.15 |

Singapore | 9.781 | 32.09 |

Skopje | 9.804 | 32.17 |

Stockholm | 9.818 | 32.21 |

Sydney | 9.797 | 32.14 |

Taipei | 9.790 | 32.12 |

Tokyo | 9.798 | 32.15 |

Vancouver | 9.809 | 32.18 |

Washington, D.C. | 9.801 | 32.16 |

Wellington | 9.803 | 32.16 |

Zurich | 9.807 | 32.18 |

**Example:**Amazon says the book The Three-Body Problem by Cixin Liu has a mass of 0.44 kg. What force of gravity will it feel?

## solution

$$F_g=mg$$ $$F_g=(0.44)(9.8)$$ $$F_g=4.3N$$**Example:**How much mass does the book Cat's Cradle by Kurt Vonnegut have if it feels a force of gravity of 1.8N?

## solution

$$F_g=mg$$ $$\frac{F_g}{g}=m$$ $$\frac{1.8}{9.8}=m$$ $$0.18 kg=m$$### Weight and Mass

Another word for the force of gravity is **weight**. An object on the moon would weigh less than it does
on Earth because of the lower gravity, but it would still have the same mass.

Earth's gravity does extend into space, but it decreases with distance. It is about ~90% for astronauts in orbit around the earth, but they don't notice any gravity because they are in a freefall.

**Freefall** means that you are just letting gravity accelerate you without any opposing forces. To keep
from falling we are careful to always counter the force of gravity. This can be done with a parachute or a jetpack!
Mostly we counter gravity with a normal force from the ground.

**Example:**The book The Name of the Wind by Patrick Rothfuss has a mass of 0.34kg. How does its mass and weight change in a freefall on Earth?

## solution

The mass doesn't change from 0.34kg

The weight is the force of gravity.

It also doesn't change in a freefall

**Example:**The hardcover version of the book Seveneves by Neal Stephenson has a mass of 0.95 kg. What is the force of gravity felt by the book on Earth? What about on the Moon?

## solution

weight on the earth:

$$F_{g}=mg$$ $$F_{g}=0.95(9.8)$$ $$F_{g}=9.31N$$weight on the moon:

$$F_{g}=ma$$ $$F_{g}=0.95(1.6)$$ $$F_{g}=1.52N$$**Example:**The book Uprooted by Naomi Novik is resting on a table. Amazon says the shipping weight is 1.2 pounds.

What is the mass in kilograms of the book on Earth? On Mars?

What is the weight of the book in Newtons on Earth (a = 9.8 m/s²)?

On Mars (a = 3.711 m/s²)?

## solution

mass doesn't change

$$ 1.2lbs \frac{1kg}{2.2lbs} = 0. \overline{54}kg $$ $$ m = 0. \overline{54} kg $$weight on Earth

$$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(9.8)$$ $$F_{g}=5.35N$$weight on Mars

$$F_{g}=ma$$ $$F_{g}=(0.\overline{54})(3.711)$$ $$F_{g}=2.02N$$# Free Body Diagrams

A free body diagram (FBD) is a diagram that shows all the force vectors on a single mass. It helps visualize the directions and how different masses apply forces to each other.

Typically you label the mass in the center of the box and draw force vectors going away from the box.

**Example:**Find the acceleration for the free body diagram above.

## solution

$$\sum F=ma$$ $$-200+300+500=2a$$ $$600=2a$$ $$300 \tfrac{m}{s^2}=a$$**Example:**This book The Martian by Andy Weir (mass = 0.09kg) is resting on a table (acceleration = zero). The force of gravity produces a downwards force of 0.34N. What force is required to keep the book at rest on the table.

## solution

$$\sum F=ma$$ $$F_{up} + F_{g} = ma$$ $$F_{up} - 0.34 = (0.09)(0)$$ $$F_{up} - 0.34 = 0$$ $$F_{up} = 0.34N$$**Example:**Draw a free body diagram for a 83kg person holding onto a rope in a game of "Tug of war". The rope is pulling them with a force of 520N left. The person's feet are countering that with 540N to the right.

## solution

**Example:**A red crate is falling, but a parachute is slowing its acceleration to only 2 m/s² down. The force of gravity is 833N down. The parachute provides 663N up. Draw a free body diagram and use it to find the mass of the crate.

## solution

$$\sum F=ma$$ $$F_1+F_2=ma$$ $$-833+663 = m(2)$$ $$\frac{-170}{2} = \frac{m(2)}{2}$$ $$-85 kg = m$$Wait, a negative mass doesn't make sense. Oh, the acceleration is negative, because it points down.

$$\frac{-170}{-2} = \frac{m(-2)}{-2}$$ $$85 kg = m$$### The Normal Force

A **normal force** occurs when two objects are in contact. It is perpendicular to the point of contact.

Typically a normal force will balance the force of gravity to keep an object from accelerating up or down.

**Example:**You are accelerating up in an elevator at 2m/s². If your mass is 100kg, what is the normal force you feel from the elevator?

## solution

$$F_{g}=mg$$ $$F_{g}=(100)(-9.8)$$ $$F_{g}=-980N$$$$\sum F=ma$$ $$F_{g}+F_{N}=ma$$ $$F_{N}=ma-F_{g}$$ $$F_{N}=(100)(2)-(-980)$$ $$F_{N}=1180N$$

**Example:**A 20kg box is at rest on a horizontal sidewalk. Find the force of gravity and the normal force on the box.

## solution

$$F_{g}=mg$$ $$F_{g}=(20)(-9.8)$$ $$F_{g}=-196N$$In the simple case of a flat horizontal surface and no vertical acceleration the force of gravity will be equal and opposite to the normal force.

$$\sum F=ma$$ $$F_{g}+F_{N}=ma$$ $$F_{N}=ma-F_{g}$$ $$F_{N}=(20)(0)-(-196)$$ $$F_{N}=196N$$**Example:**A 20kg box is at rest on a steep sidewalk. The sidewalk is at an angle 20 degrees from horizontal. Find the force of gravity and the normal force on the box. What is the acceleration of the box? (assume no friction)

## solution

$$F_{g}=mg$$ $$F_{g}=(20)(-9.8)$$ $$F_{g}=196N$$Separate the gravity vector into components parallel and perpendicular to the ground

**perpendicular to ground**

$$\sum F_{\perp}=ma$$ $$F_{g\perp}+F_{N}=ma$$ $$F_{N}=ma-F_{g\perp}$$ $$F_{N}=(20)(0)-(-184)$$ $$F_{N}=184N$$

**parallel to ground**

$$\sum F_{\parallel}=ma$$ $$F_{g\parallel}=ma$$ $$(67)=(20)a$$ $$3.35\small\frac{m}{s^{2}}=\normalsize a$$

### Tension Forces

Tension is the pulling force from a string or rope. Ropes are useful for transferring a force over a distance. If you wanted to apply a force on your dog while going for a walk you could use the tension on the leash to transfer that force.**Example:**A helium balloon is attached to a 0.5g paperclip. If the balloon and paperclip are accelerating up at 0.023m/s², what is the tension on the paper clip from the balloon in Newtons?

## solution

$$a = 0.023\small\frac{m}{s^{2}}$$ $$m = 0.5g = 0.0005kg$$ $$\sum F=ma$$ $$F_{g} + F_{T} = ma$$ $$F_{T} = ma - F_{g}$$ $$F_{T} = ma - mg$$ $$F_{T} = 0.0005*0.023 - 0.0005*9.8$$ $$F_{T} = 0.0049115N$$**Example:**A 100 kg person is pulling a 10kg crate with a rope (ignore the mass of the rope). Both the person and the crate are accelerating to the left at 0.1 m/s². What force is the person producing in order to accelerate to the left?

## solution

**crate**

**person**

**Example:**You are walking your dog with the leash at a 45 degree angle down towards the dog. Neither your nor the dog are accelerating. The dog is pulling on the leash forward with a force of 100N. Calculate the x part of the tension force on the leash. Then calculate the total tension force.

## solution

**dog**

The 100N is only the x-part of the tension force vector.

$$F_t^2 = 100^2+100^2$$ $$F_t^2 = 20000$$ $$F_t = 141N$$