Multiple-body systems

We can use Newton’s second law to analyze systems involving multiple bodies in much the same way that we use it for single bodies. We will usually consider bodies that are connected together with massless strings that do not stretch. This allows us to assume that each body has the same acceleration. The equation we use for multiple-body systems is

`vec F_"net" = m_"tot"vec a_"sys"`,

where `vec F_"net"` is the vector sum of all forces acting on the system (not internal tension forces), `m_"tot"` is sum of the masses of all bodies, and `vec a_"sys"` is the acceleration of the system.

This is very straightforward when all bodies are in a line. Once we find the acceleration, we can determine the force of tension acting on individual bodies using Newton’s second law again, but this time considering the body individually (draw a new free body diagram) and having acceleration as a known variable.

Pulley systems look two-dimensional, but we can turn them into one-dimensional problems by making one direction of the pulley negative and the other positive. This means, for example, that the force of gravity could be positive for one body and negative for another if they are on opposite sides of the pulley.

Example

In a pulley system where mass A (21 kg) is on a 18º frictionless ramp and mass B (12 kg) is suspended, find the acceleration of the system and the tension in the string.

As always, we start with a sketch:

18ºAB[bwd][fwd]
A pulley system of two bodies

There are three forces acting on the system:

Adding the first two gives us the parallel component of `vec F_"g,A"`:

`vec F_"g,A" + vec F_"N,A" = vec F_"g,ramp" = |vec F_"g,A"|cos(90º-theta) = |vec F_"g,A"|sintheta`.

Now we can use Newton’s second law to find acceleration.

`vec F_"net" = |vec F_"g,A"|sintheta + vec F_"g,B" = m_"tot"vec a_"sys"`.

If we let [fwd] be positive, then `vec F_"g,ramp"` will be negative. Since `vec F_"g" = mg`,

`vec F_"net" = -m_"A"gsintheta + m_"B"g = m_"tot"vec a_"sys"`.

Rearranging to solve for acceleration gives us

`vec a_"sys" = (-m_"A"gsintheta + m_"B"g)/m_"tot"`,

and after substituting we have

`vec a_"sys" = (-(21\ "kg")(9.80\ "m/s"^2)sin18º + (12\ "kg")(9.80\ "m/s"^2))/(21\ "kg" + 12\ "kg") = 1.64\ "m/s"^2`,

therefore the acceleration of the system is 1.6 m/s2 [fwd].

Now we can use the acceleration to find the tension in the string. We can do this using either A or B. It’s slightly easier with B. There are two forces acting on B:

We can write our net force equation again:

`vec F_"net" = vec F_"g,B" + vec F_"T" = m_"B"g + vec F_"T" = m_"B"vec a`.

Solving for the force of tension, we have

`vec F_"T" = m_"B"vec a - m_"B"g = m_"B"(vec a - g)`,

which gives us the solution

`vec F_"T" = (12\ "kg")(1.64\ "m/s"^2 - 9.80\ "m/s"^2) = -97.92\ "N"`,

therefore the magnitude of the force of tension is 98 N.