Free body diagrams are a technique that helps visualize the force vectors used to solve Newton's second law.
A free body diagram shows all the force vectors on an object. Mass is typically written inside a box, with force vectors pointed away from the box.
solution
solution
$$\sum F=ma$$ $$200\,\mathrm{N}+300\,\mathrm{N}+500\,\mathrm{N}=ma$$ $$600\,\mathrm{N}=(2\,\mathrm{kg})a$$ $$300 \mathrm{\tfrac{m}{s^2}}=a$$solution
$$\sum F=ma$$ $$F_{N} + F_{g} = ma$$ $$F_{N}  0.34\, \mathrm{N} = (0.09\, \mathrm{kg})(0)$$ $$F_{N}  0.34\,\mathrm{N} = 0$$ $$F_{N} = 0.34 \, \mathrm{N}$$solution
$$ \text{vertical}$$ $$\sum F=ma$$ $$22 \, \mathrm{N}  85 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$63 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$3.9 \, \mathrm{\tfrac{m}{s^2}} = a$$$$ \text{horizontal}$$ $$\sum F=ma$$ $$78 \, \mathrm{N}+55 \, \mathrm{N}140 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$7 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$0.43 \, \mathrm{\tfrac{m}{s^2}} = a$$
solution
$$\sum F=ma$$ $$F_1+F_2=ma$$ $$833+663 = m(2)$$ $$\frac{170}{2} = \frac{m(2)}{2}$$ $$85 \, \mathrm{kg} = m$$A negative mass doesn't make sense. Where did we mess up?
Oh, the acceleration is negative, because it points down.
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.
There is a special word to describe the direction of gravity, down.
$$F_g=mg $$
\(F_g\) = 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 only on the mass of the object because on the surface of the Earth acceleration from gravity is the same for all objects. If you aren't on the surface of the Earth, there is a different way to calculate gravity.
solution
Air friction produces a force that opposes motion. Feathers have a large surface area compared to their small mass so they have more air friction.
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.
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 
Question: What factors might explain why the measured acceleration of gravity changes in different locations on the surface of Earth?
solution
solution
$$F_g=mg$$ $$F_g=(0.44 \, \mathrm{kg} )(9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$F_g=4.3 \, \mathrm{N}$$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.
$$F_g = \mathrm{weight}$$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 jet pack, or just the ground.
answer
An object's mass doesn't change when it is falling.
Weight just means the force of gravity, which also doesn't change in a short freefall.
Local Massive Objects Surface Gravity
name  g (m/s²)  


Sun  275 
Mercury  3.7  
Venus  8.9  
Earth  9.8  
Moon  1.6  
Mars  3.7  
Jupiter  25.8  

Saturn  10.4 

Uranus  8.7 
Neptune  11.2 
solution
Weight and force of gravity mean the same thing. A planet's gravity field determines your weight, but not your mass.

$$\text{Earth}$$ $$F_{g}=mg$$ $$F_{g}=(0.95)(9.8)$$ $$F_{g}=9.31 \, \mathrm{N}$$

$$\text{Moon}$$ $$F_{g}=mg$$ $$F_{g}=(0.95)(1.6)$$ $$F_{g}=1.52 \, \mathrm{N}$$
Converting kilograms (kg) into pounds (lbs)
$$1 \, \mathrm{kg} = 2.2\, \mathrm{lbs} \quad \scriptsize \text{(On Earth)}$$ $$1 \, \mathrm{N} = 0.2248\, \mathrm{lbs}$$Pounds are a unit of force and kilograms are a unit of mass. You can't convert directly between them because they are different concepts, but you can use the force of gravity equation to find a conversion that works for only Earth's surface.
solution
Mass doesn't depend on gravity, so it's the same everywhere.
$$ 1.2\,\mathrm{lbs} \left( \frac{1\,\mathrm{kg}}{2.2\,\mathrm{lbs}} \right)= 0.\overline{54}\,\mathrm{kg}$$ $$ m = 0. \overline{54} \, \mathrm{kg} $$What is the weight of the book in Newtons on Earth? On Mars?
solution

$$\text{weight on Earth}$$ $$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(9.8)$$ $$F_{g}=5.35\, \mathrm{N}$$

$$\text{weight on Mars}$$ $$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(3.711)$$ $$F_{g}=2.02\, \mathrm{N}$$
The Normal Force
Typically a normal force will balance the force of gravity to keep an object from accelerating up or down.
A normal force occurs when two objects are in contact. It is perpendicular to the point of contact. A normal force prevents objects from passing through each other.
A normal force will scale to a value that will keep the net force and acceleration zero.
Normal forces come from the combined effect of electromagnetic forces and the Pauli exclusion principle.
The electromagnetic force allows chemical bonds to form. These bonds give solid matter its rigid structure, which is required for normal forces.
At the atomic scale, particles can't pass through each other primarily because of a quantum mechanical effect called the Pauli exclusion principle . The Pauli exclusion principle is mostly responsible for keeping particles, like electrons, separate.
solution
$$F_{g}=mg$$ $$F_{g}=(100)(9.8)$$ $$F_{g}=980\, \mathrm{N}$$$$\sum F=ma$$ $$F_{N}F_{g}=ma$$ $$F_{N}  980=(100)(2)$$ $$F_{N}=1180\, \mathrm{N}$$
solution
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.
$$F_{g}=mg$$ $$F_{g}=(20)(9.8)$$ $$F_{g}=196\, \mathrm{N}$$$$\sum F=ma$$ $$F_{N}F_{g}=ma$$ $$F_{N}  196=(20)(0) $$ $$F_{N}=196 \, \mathrm{N}$$
solution

$$F_{g}=mg$$ $$F_{g}=(20)(9.8)$$ $$F_{g}=196 \, \mathrm{N}$$

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

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

$$\text{parallel to ground}$$
$$F_{g\parallel}=F_{g}\sin(20)$$ $$F_{g\parallel}=(196)\sin(20)$$ $$F_{g\parallel}=67 \, \mathrm{N}$$
$$\sum F_{\parallel}=ma$$ $$F_{g\parallel}=ma$$ $$(67)=(20)a$$ $$3.35 \, \mathrm{\tfrac{m}{s^{2}}}=a$$
Tension Forces
When a person is walking a dog they are able to apply a force on the dog with a leash. They use the tension on the leash to transfer that force from their hand to the dog.
Tension is the pulling force from a chain, string, or rope. Tension is useful for transferring a force over a distance. In most situations, the tension is the same for both ends.
solution
$$\sum F=ma$$ $$F_{g} + T = ma$$ $$T = ma + F_{g}$$ $$T = ma + mg$$ $$T = (0.0005\, \mathrm{kg})(0.023\, \mathrm{\tfrac{m}{s^2}})+ (0.0005\, \mathrm{kg}) (9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$T = 0.0049115\, \mathrm{N}$$solution
The tension force is equal and opposite for the person and crate.
$$\text{crate on right}$$ $$\sum F=ma$$ $$T = 10 (0.1)$$ $$T = \color{#f05}1 \, \mathrm{N}$$$$\text{person on left}$$ $$\sum F=ma$$ $$F_{\mathrm{person}} + T = ma$$ $$F_{\mathrm{person}} {\color{#f05}+ 1 \, \mathrm{N}} = (100 \, \mathrm{kg})(0.1\, \mathrm{\tfrac{m}{s^2}})$$ $$F_{\mathrm{person}} + 1\, \mathrm{N} = 10\, \mathrm{N}$$ $$F_{\mathrm{person}} = 11\,\mathrm{N}$$
solution
$$\text{dog: horizontal}$$ $$\sum F=ma$$ $$T_x + F_{\mathrm{dog}} = 0$$ $$T_x + 100 \, \mathrm{N} = 0$$ $$T_x= 100 \,\mathrm{N}$$The 100 N is only the xpart 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.
$$T^2 = 100^2+100^2$$ $$T^2 = 20\,000$$ $$T = 141 \, \mathrm{N}$$solution
$$ \text{box}$$ $$\sum F=ma$$ $$F_{\mathrm{tension}}=(10)(0.5)$$ $$F_{\mathrm{tension}}=5\, \mathrm{N}$$The tension force on the string is equal but opposite for the squirrel and box.
$$ \text{squirrel}$$ $$\sum F=ma$$ $$F_{\mathrm{squirrel}} + F_{\mathrm{tension}}=ma$$ $$F_{\mathrm{squirrel}} + 5=(0.4)(0.5)$$ $$F_{\mathrm{squirrel}}=0.25$$ $$F_{\mathrm{squirrel}}=5.2 \, \mathrm{N}$$The negative sign tells us that the squirrel's force is pointed left.
$$F_{\mathrm{squirrel}}=5.2 \, \mathrm{N} \text{ left}$$Assume no friction, no air resistance, Earth gravity, and a massless rope.
hint
Both free body diagrams share the same tension and acceleration. Build two equations with Newton's second law and solve a system of equations for T and a. While T and a are the same magnitude they have different directions, so watch the sign of acceleration in particular.
This Khan academy video covers this problem in more depth.
solution
$$\text{20kg box: horizontal}$$ $$\sum F=ma_x$$ $$T = ma_x$$ $$T = \color{#d3a}20a_x$$$$\text{10kg box: vertical}$$ $$\sum F=ma_y$$ $$T  F_g = ma_y$$ $$T  mg = ma_y$$ $$T  (10)(9.8) = 10a_y$$
$${\color{#d3a}20a_x}  98 = 10a_y$$
$$a_x = a_y$$
$${\color{#d3a}20a_y}  98 = 10a_y$$ $$98 = 10a_y + 20a_y$$ $$98 = 30a_y$$ $$3.2\overline{6} \, \mathrm{\tfrac{m}{s^2}} = a_y$$
solution (treating system as one mass)
$$\text{gravity only pulls on the 10kg part}$$ $$F_g = m g$$ $$F_g = (10 \, \mathrm{kg})(9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$F_g = \color{#d3a}98 \, \mathrm{N}$$$$\text{use both masses to find acceleration in the vertical}$$ $$\sum F=ma$$ $$F_g = (10\, \mathrm{kg}+20\, \mathrm{kg}) a$$ $$F_g = (30\, \mathrm{kg}) a$$ $$98\, \mathrm{N} = (30\,\mathrm{kg}) a$$ $$3.2\overline{6} \, \mathrm{\tfrac{m}{s^2}}= a$$
solution
We have two variables and two equations, this means we can plug one equation into the other. The Tension is the same for each body. The accelerations are the same, but in opposite directions, so we need to make one acceleration negative.
The 25kg object is falling down at 4.2 m/s². The 10kg is rising up at 4.2 m/s².