9 Connected Bodies and Basic Mechanical Systems

Learn how to analyze connected bodies using free-body diagrams, force and moment equations, and cable-length constraints, with fixed- and movable-pulley examples.

Choose the model and identify the bodies

Start by deciding which bodies to include and which idealizations apply. These choices determine what forces and constraints belong in the analysis.

A has constant length and carries along its length; it pulls but does not push. A changes the cable’s direction. For one continuous ideal cable, has the same magnitude throughout. A is supported by multiple cable segments, so forces from those segments act on it.

A smooth contact exerts a normal force but no friction force. A rough contact may also exert friction. These are modeling assumptions rather than universal properties: if a pulley has significant mass or friction, or a cable has mass, tensions on its two sides may differ.

Draw body and system diagrams

Draw a separate for each body whose force or motion is needed. Show only forces and moments acting on that body, such as its weight, applied loads, support reactions, contact forces, and cable tensions. Draw arrows along the cable and away from the body.

For a rigid body in a plane, use both force and moment equations; for a particle, use force equations. You can also draw a system FBD that treats several connected parts as one system. Forces internal to the selected system cancel and are omitted, while forces from the surroundings remain.

Apply equilibrium and motion equations

Use a consistent sign convention and write the appropriate equations for each body. For a body in static equilibrium, the resultant force and the moment about point $O$ are zero:

\sum \mathbf F=\mathbf 0,\qquad \sum M_O=0.

For a translating body that accelerates, apply Newton’s second law:

\sum \mathbf F=m\mathbf a.

Forces between separate bodies form equal-and-opposite interaction pairs, but each force appears on a different body’s FBD. A cable is an unknown force, not automatically equal to the load’s weight. It equals the weight only when the body’s force balance makes that true; for an accelerating body, use Newton’s second law.

Use cable-length constraints

A taut, inextensible cable imposes a relationship between the positions of the bodies it connects. Express the total variable cable length as a constant, counting every segment whose length changes. Differentiating the length relation gives the velocity relation, and differentiating again gives the acceleration relation.

For a single cable passing over one fixed pulley and connecting two bodies, let each coordinate be positive in the direction that increases its corresponding cable segment. The constant-length relation is

x_1+x_2=\text{constant}.

Therefore,

v_1+v_2=0,\qquad a_1+a_2=0.

The bodies have equal acceleration magnitudes in opposite directions. More complicated cable paths require counting every changing segment; connected bodies do not necessarily have the same displacement or acceleration.

Analyze a fixed pulley with two masses

For an ideal fixed pulley with hanging masses $m_1$ and $m_2$ , assume $m_2>m_1$ and take the downward direction of $m_2$ as positive. Separate FBDs give

T-m_1g=m_1a,\qquad m_2g-T=m_2a.

The cable constraint means the masses have equal acceleration magnitudes in opposite directions. Solving the two body equations together with the constraint gives

a=\frac{m_2-m_1}{m_1+m_2}g,\qquad T=\frac{2m_1m_2}{m_1+m_2}g.

For $m_1=2\,\text{kg}$ , $m_2=3\,\text{kg}$ , and $g=9.81\,\text{m/s}^2$ , the acceleration is $a=1.96\,\text{m/s}^2$ and the is $T=23.5\,\text{N}$ . The heavier mass descends. Both body equations and the cable constraint are needed to solve the system.

Analyze a

Consider an ideal attached to a load and supported by two vertical segments of the same cable. Each segment pulls upward with $T$ . For a stationary load of weight $W$ , the FBD of the pulley and load gives

2T-W=0,\qquad T=\frac{W}{2}.

Thus, a $200\,\text{N}$ load requires a $100\,\text{N}$ pull in this idealized arrangement. The two supporting cable segments share the load. If the pulley’s mass is not negligible, include the pulley itself in the FBD.

Let $y$ be the downward displacement of the and $x$ the downward displacement of the free end. The changing cable length is $x+2y$ , so

x+2y=\text{constant},\qquad a_x+2a_y=0.

The free end moves twice as far as the load, in the opposite direction. The force advantage is accompanied by a corresponding increase in the pulling distance.

Follow a reliable solution procedure

Isolate the bodies. Decide whether to draw individual FBDs, a system FBD, or both.
Draw and label forces. Include weights, tensions, applied forces, contact forces, and support reactions; show relevant angles and dimensions.
Set coordinates and signs. Choose positive directions for each body and keep them consistent.
Write constraints. Express each inextensible cable’s length in terms of its changing segments, then differentiate if velocity or acceleration is required.
Write equations. Use equilibrium for bodies at rest or Newton’s second law for accelerating bodies. Include moment equilibrium for rigid bodies when needed.
Solve and check. Verify units and signs, and check whether the predicted motion agrees with the constraint. A negative result means the actual direction is opposite the assumed positive direction.

The analysis combines body-by-body FBDs, equations of equilibrium or motion, and cable-length constraints. Ideal continuous cables carry equal , while a can be supported by several forces. The cable geometry determines how connected bodies’ displacements and accelerations are related, so counting changing cable segments carefully is essential.