Laws of Motion · Lesson 10 of 14

Connected Bodies, Strings and Pulleys

“Use combined-system equations and separate force diagrams to determine acceleration and internal tension.”

Learning Objectives

• Use ideal-string assumptions and acceleration constraints. • Find a system acceleration before solving for internal forces. • Solve two connected horizontal bodies and an Atwood machine. • Include kinetic friction in a trolley-and-hanging-mass arrangement.

A combined system can simplify the acceleration

When two blocks are connected by an ideal light inextensible string, their motions are constrained. On a straight horizontal line they share one acceleration. Over a single fixed ideal pulley, one side rises by the same distance that the other falls, so their acceleration magnitudes are equal. A combined-system equation can eliminate internal string forces, but the acceleration signs must still respect the geometry.

After finding acceleration, isolate one body to determine tension. Tension does not have to equal the applied force or a hanging body’s weight. Those equalities hold only in particular force balances. Draw a separate diagram for each body, and use a common tension only when the string and pulley assumptions justify it.

Two bodies on a smooth horizontal surface

The pulled end changes the tension

Problem
Blocks A = 10 kg and B = 20 kg are connected on a smooth horizontal surface. A 600 N horizontal force pulls the outer end of one block away from the other. Find acceleration and tension for each pulled end.

1.For the combined 30 kg system, external horizontal force is 600 N. Acceleration is 600/30 = 20 m/s² in either case.
2.If the 10 kg block is pulled, the tension is the only horizontal force on the 20 kg block. Thus T = 20 × 20 = 400 N.
3.If the 20 kg block is pulled, tension accelerates the 10 kg block. Thus T = 10 × 20 = 200 N.
4.The acceleration is unchanged, but the tension differs because it must accelerate a different trailing mass. The physical arrangement must keep the string taut.

Trailing horizontal block— For a trailing block on a smooth horizontal surface, tension is the only horizontal external force, so T = ma. The force on the other connected body belongs on its own diagram. Arrow lengths are schematic unless stated otherwise.

Two hanging masses over a fixed pulley

Let m₂ > m₁ and take m₂ downward and m₁ upward as positive for their respective equations. The heavier mass has m₂g − T = m₂a; the lighter has T − m₁g = m₁a. Adding cancels tension and gives the common acceleration magnitude. Substitute into either individual equation to recover T.

Ideal Atwood machineLaTeX

a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}}, T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}

The string is light and inextensible; the fixed pulley is ideal with negligible inertia and friction. Positive a here means the heavier mass descends.

Eight and twelve kilograms

Problem
An 8 kg mass and a 12 kg mass hang over an ideal pulley. Use g = 10 m/s². Find acceleration and tension.

1.a = (12 − 8) × 10/(12 + 8) = 2 m/s². The 12 kg mass goes down and the 8 kg mass goes up.
2.For the 8 kg mass, T − 80 = 8 × 2, giving T = 96 N.
3.Check with the other mass: 120 − 96 = 24 N = 12 × 2. The tension lies between the two weights and is not equal to either.

A trolley pulled by a hanging mass

For a trolley of mass M moving horizontally and a hanging mass m moving downward, the trolley equation is T − f_k = Ma. The hanging-mass equation is mg − T = ma. With horizontal string force on the trolley and no vertical trolley acceleration, N = Mg and f_k = μ_kMg. Add the equations to obtain (M + m)a = mg − μ_kMg, then solve for tension.

Sliding trolley and hanging massLaTeX

a = \frac{m g - μ _{k} M g}{M + m}, T = m (g - a)

This form assumes the hanging mass is moving downward and the trolley slides toward the pulley. If the system starts at rest, first test whether static friction can hold it.

A 20 kg trolley and 3 kg hanging block

Problem
The trolley slides on a horizontal surface with μ_k = 0.04. Use g = 10 m/s² and ideal string and pulley assumptions.

1.Trolley normal force is 200 N and kinetic friction is 0.04 × 200 = 8 N.
2.Add 30 − T = 3a and T − 8 = 20a. This gives a = 22/23 = 0.9565 m/s².
3.Tension is 30 − 3(22/23) = 27.1304 N, about 27.1 N.
4.The coefficient supplied is kinetic, so this calculation describes the stated sliding phase. A coefficient of static friction would be needed to decide whether an initially resting arrangement begins to move by itself.

Check the ideal assumptions

If a pulley has significant rotational inertia, different tensions on the two sides can supply its turning effect. A massive string can also require a tension gradient to accelerate its own parts. Movable pulleys can give unequal acceleration magnitudes linked by a length constraint. The simple formulas here are not replacements for drawing and analysing the actual arrangement.

Internal cancellation is useful but limited

The combined diagram removes string forces internal to the selected collection. It does not tell you their magnitudes. Once an internal force is requested, isolate a component and restore that interaction as an external force on the smaller system. This is the same boundary reasoning used in momentum conservation and third-law force pairs.

Quick check

Two connected blocks accelerate on a smooth horizontal surface. To determine their connecting tension after finding acceleration, what should you do?

Key Takeaways

• Use string length to establish acceleration constraints. • Combined equations often eliminate internal tension. • Separate component diagrams reveal the internal forces. • Tension equals weight only when the relevant acceleration and other forces make that balance valid. • Static and kinetic friction describe different phases.

Previous · Lesson 9

Sliding Friction, Inclines and Rolling Resistance

Next · Lesson 11

Forces in Circular Motion