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.
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
$$\sum F=ma$$ $$F_{N} + F_{g} = ma$$ $$F_{N}  0.34 = (0.09)(0)$$ $$F_{N}  0.34 = 0$$ $$F_{N} = 0.34 \, \mathrm{N}$$solution
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.
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, 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 
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.
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.
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
$$F_{g}=mg$$ $$F_{g}=(0.34)(9.8)$$ $$F_{g}=3.33 \, \mathrm{N}$$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.
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.
solution

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

$$F_{g}=mg$$ $$F_{g}=(20)(9.8)$$ $$F_{g}=196\, \mathrm{N}$$
$$\sum F=ma$$ $$F_{g}+F_{N}=ma$$ $$F_{N}=ma+F_{g}$$ $$F_{N}=(20)(0) + 196$$ $$F_{N}=196 \, \mathrm{N}$$ 
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.
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
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.solution
$$a = 0.023 \, \mathrm{\tfrac{m}{s^{2}}}$$ $$m = 0.5\, \mathrm{g} = 0.0005 \, \mathrm{kg}$$$$\sum F=ma$$ $$F_{g} + F_{\mathrm{Ten}} = ma$$ $$F_{\mathrm{Ten}} = ma + F_{g}$$ $$F_{\mathrm{Ten}} = ma + mg$$ $$F_{\mathrm{Ten}} = (0.0005)(0.023) + (0.0005) (9.8)$$ $$F_{\mathrm{Ten}} = 0.0049115\, \mathrm{N}$$
solution
The tension force is equal and opposite for the person and crate.
$$\text{crate}$$ $$\sum F=ma$$ $$F_\mathrm{Ten} = 10 (0.1)$$ $$F_\mathrm{Ten} = 1 \, \mathrm{N}$$$$\text{person}$$ $$\sum F=ma$$ $$F_{\mathrm{person}} + F_{\mathrm{Ten}} = ma$$ $$F_{\mathrm{person}} + 1 = (100)(0.1)$$ $$F_{\mathrm{person}} + 1 = 10$$ $$F_{\mathrm{person}} = 11\,\mathrm{N}$$
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 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.
$$F_\mathrm{tension}^2 = 100^2+100^2$$ $$F_\mathrm{tension}^2 = 20000$$ $$F_\mathrm{tension} = 141 \, \mathrm{N}$$