Free Body Diagrams

A free body diagram shows all the force vectors on a single mass. Typically you label the mass in the center of a box and draw force vectors going away from the box. A free body diagram is often drawn to visualize Newton's second law.

2.0 kg 200 N 500 N 300 N Example: Find the acceleration for the free body diagram above.
solution $$\sum F=ma$$ $$-200\,\mathrm{N}+300\,\mathrm{N}+500\,\mathrm{N}=ma$$ $$600\,\mathrm{N}=(2\,\mathrm{kg})a$$ $$300 \mathrm{\tfrac{m}{s^2}}=a$$
0.09 kg 0.34 N F= ? Example: This book The Martian by Andy Weir (mass = 0.09 kg) is resting on a table (acceleration = zero). The force of gravity produces a downwards force of 0.34 N. What force is required to keep the book at rest on the table.
solution $$\sum F=ma$$ $$F_{N} + F_{g} = ma$$ $$F_{N} - 0.34 = (0.09)(0)$$ $$F_{N} - 0.34 = 0$$ $$F_{N} = 0.34 \, \mathrm{N}$$
Example: Draw a free body diagram for a 83 kg person holding onto a rope in a game of "Tug of war". The rope is pulling them with a force of 520 N left. The person's feet are countering that with 540 N to the right.
solution 83 kg 520 N 540 N
Example: A red crate is falling, but a parachute is slowing its acceleration to only 2 m/s² down. The force of gravity is 833 N down. The parachute provides 663 N up. Draw a free body diagram and use it to find the mass of the crate.
solution m = ? 833 N 663 N $$\sum F=ma$$ $$F_1+F_2=ma$$ $$-833+663 = m(2)$$ $$\frac{-170}{2} = \frac{m(2)}{2}$$ $$-85 \, \mathrm{kg} = m$$

A negative mass doesn't make sense. Where did we mess up?
Oh, the acceleration is negative, because it points down.

$$\frac{-170}{-2} = \frac{m(-2)}{-2}$$ $$85 \, \mathrm{kg} = m$$

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.

mass F g

$$F_g=mg $$

\(F_g\) = the force of gravity, weight [N, Newtons, kg m/s²] vector
\(m\) = mass [kg]
\(g\) = acceleration of gravity on 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 does the book have?
solution $$F_g=mg$$ $$F_g=(0.44 \, \mathrm{kg} )(9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$F_g=4.3 \, \mathrm{N}$$
Example: How much mass does the book Cat's Cradle by Kurt Vonnegut have if it feels a force of gravity of 1.8 N?
solution $$F_g=mg$$ $$\frac{F_g}{g}=m$$ $$\frac{1.8 \,\mathrm{N}}{9.8\, \mathrm{\tfrac{m}{s^2}}}=m$$ $$0.18 \, \mathrm{kg}=m$$

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 0.34 kg. How does its mass and weight change in a freefall on Earth?
solution

The mass doesn't change as an object falls.

Weight just means force of gravity, which also doesn't change in a short freefall

$$F_{g}=mg$$ $$F_{g}=(0.34)(9.8)$$ $$F_{g}=3.33 \, \mathrm{N}$$
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
  • $$\text{weight on the earth}$$ $$F_{g}=mg$$ $$F_{g}=0.95(9.8)$$ $$F_{g}=9.31 \, \mathrm{N}$$
  • $$\text{weight on the moon}$$ $$F_{g}=ma$$ $$F_{g}=0.95(1.6)$$ $$F_{g}=1.52 \, \mathrm{N}$$

Weight changes on different planets, but mass doesn't change.

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?
solution $$\text{mass is the same everywhere}$$ $$ 1.2\mathrm{lbs} \frac{1\mathrm{kg}}{2.2\mathrm{lbs}} = 0. \overline{54}\mathrm{kg} $$ $$ m = 0. \overline{54} \, \mathrm{kg} $$

What is the weight of the book in Newtons on Earth (a = 9.8 m/s²)?
On Mars (a = 3.711 m/s²)?
solution
  • $$\text{weight on Earth}$$ $$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(9.8)$$ $$F_{g}=5.35\, \mathrm{N}$$
  • $$\text{weight on Mars}$$ $$F_{g}=ma$$ $$F_{g}=(0.\overline{54})(3.711)$$ $$F_{g}=2.02\, \mathrm{N}$$

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.

mass Normal Force Force of gravity
Example: You are accelerating up in an elevator at 2 m/s². If your mass is 100 kg, what is the normal force you feel from the elevator?
solution
  • $$F_{g}=mg$$ $$F_{g}=(100)(-9.8)$$ $$F_{g}=-980\, \mathrm{N}$$
    $$\sum F=ma$$ $$F_{g}+F_{N}=ma$$ $$F_{N}=ma-F_{g}$$ $$F_{N}=(100)(2)-(-980)$$ $$F_{N}=1180\, \mathrm{N}$$









  • 100 kg F g = 980 N F N = 1180 N
Example: A 20 kg box is at rest on a horizontal sidewalk. Find the force of gravity and the normal force on the box.
solution
  • $$F_{g}=mg$$ $$F_{g}=(20)(9.8)$$ $$F_{g}=196\, \mathrm{N}$$
    $$\sum F=ma$$ $$-F_{g}+F_{N}=ma$$ $$F_{N}=ma+F_{g}$$ $$F_{N}=(20)(0) + 196$$ $$F_{N}=196 \, \mathrm{N}$$
  • 20 kg F N = 196 N Fg = 196 N

    In the simple case of a flat horizontal surface with no vertical acceleration the force of gravity will always be equal and opposite to the normal force.

20 kg F N Fg Example: A 20 kg 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
  • $$F_{g}=mg$$ $$F_{g}=(20)(9.8)$$ $$F_{g}=196 \, \mathrm{N}$$

  • Fg Fg⊥ Fg∥

Separate the gravity vector into components parallel and perpendicular to the ground. Acceleration is zero in the perpendicular direction.

  • $$\text{perpendicular to ground}$$

    $$F_{g\perp}=F_{g}\cos(20)$$ $$F_{g\perp}=(196)\cos(20)$$ $$F_{g\perp}=184 \, \mathrm{N}$$
    $$\sum F_{\perp}=ma$$ $$-F_{g\perp}+F_{N}=ma$$ $$F_{N}=ma+F_{g\perp}$$ $$F_{N}=(20)(0) + 184$$ $$F_{N}=184 \, \mathrm{N}$$
  • $$\text{parallel to ground}$$

    $$F_{g\parallel}=F_{g}\sin(20)$$ $$F_{g\parallel}=(196)\sin(20)$$ $$F_{g\parallel}=67 \, \mathrm{N}$$
    $$\sum F_{\parallel}=ma$$ $$F_{g\parallel}=ma$$ $$(67)=(20)a$$ $$3.35 \, \mathrm{\tfrac{m}{s^{2}}}= a$$
20 kg F N = 184 N Fg = 196 N

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 0.5 g paper clip. If the balloon and paper clip are accelerating up at 0.023 m/s², what is the tension on the paper clip from the balloon in Newtons?
solution $$a = 0.023 \, \mathrm{\tfrac{m}{s^{2}}}$$ $$m = 0.5\, \mathrm{g} = 0.0005 \, \mathrm{kg}$$
$$\sum F=ma$$ $$-F_{g} + F_{\mathrm{Ten}} = ma$$ $$F_{\mathrm{Ten}} = ma + F_{g}$$ $$F_{\mathrm{Ten}} = ma + mg$$ $$F_{\mathrm{Ten}} = (0.0005)(0.023) + (0.0005) (9.8)$$ $$F_{\mathrm{Ten}} = 0.0049115\, \mathrm{N}$$
Example: A 100 kg person is pulling a 10 kg 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 100 kg F person F Ten 10 kg F Ten

The tension force is equal and opposite for the person and crate.

$$\text{crate}$$ $$\sum F=ma$$ $$F_\mathrm{Ten} = 10 (-0.1)$$ $$F_\mathrm{Ten} = -1 \, \mathrm{N}$$
$$\text{person}$$ $$\sum F=ma$$ $$F_{\mathrm{person}} + F_{\mathrm{Ten}} = ma$$ $$F_{\mathrm{person}} + 1 = (100)(-0.1)$$ $$F_{\mathrm{person}} + 1 = -10$$ $$F_{\mathrm{person}} = -11\,\mathrm{N}$$
F Tx 10 kg F Tension F dog 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 100 N. Calculate the x part of the tension force on the leash. Then calculate the total tension force.
solution $$\text{dog: horizontal}$$ $$\sum F=ma$$ $$-F_{tx} + F_{\mathrm{dog}} = 0$$ $$-F_{tx} + 100 = 0$$ $$F_{tx} = 100 \, \mathrm{N}$$

The 100 N is only the x-part of the tension force vector. We can find the total tension force with the Pythagorean theorem. At 45° the x and y parts of the force are the same.

$$F_\mathrm{tension}^2 = 100^2+100^2$$ $$F_\mathrm{tension}^2 = 20000$$ $$F_\mathrm{tension} = 141 \, \mathrm{N}$$
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