When two or more bodies are connected by a cord and move in tandem, then they are considered to be a system of bodies. When working problems involving systems of bodies,

the first step is to draw
freebody diagrams
for each object. Remember, that the principle forces that we are now considering are: normals, weight, friction, tensions and generic applied forces.
 since both masses are attached, they share the same kinematics properties: displacement, velocity, acceleration and time.
 when writing equations for net F = ma, the direction of motion is considered to be the positive direction for the acceleration of the objects comprising the system.
Table Surface
Let's consider as our first example, two identical objects being dragged across the surface of a frictionless table by two cords. The two freebody diagrams would look like:
left mass


right mass




In order to determine the acceleration of the system and the tension in each cord, we will need to write the equations of motion for each mass: net F_{x} = ma_{x} and net F_{y} = ma_{y}.

left mass

right mass

net F_{x} = ma_{x} 
T_{1} = ma

T_{2}  T_{1} = ma

net F_{y} = ma_{y} 
 mg = 0

 mg = 0

Solving the two equations for net F_{x} = ma_{x} simultaneously yields the equation:
T_{1} = ma
T_{2}  T_{1} = ma

T_{2} = 2ma
If the numerical values for T_{2} and m are given, then this equation will allow you to solve for the acceleration of the system. Once the acceleration is known, then the tension in cord #1 can also be calculated.
Hanging Masses
Now suppose that the right mass is hanging off a frictionless table and the cord connecting it to the left mass passes over a "massless, frictionless" pulley. How would this change the freebody diagrams for each mass and their equations of motion?

mass on table

hanging mass




net F_{x} = ma_{x} 
T = ma



net F_{y} = ma_{y} 
 mg = 0

mg  T = ma

Solving these equations simultaneously for the acceleration yields the equation: T = ma mg  T = ma  mg = 2ma a = ½g
Once again, if the numerical value for m is given then this equation will allow you to solve for the acceleration of the system. Remember that "g" represents the acceleration due to gravity, 9.8 m/sec^{2}. Once the acceleration is known, then the tension in the cord can be calculated.
