Systems, Tension & Connected Objects
- Treat connected objects as one system to find the shared acceleration
- Isolate a single body to find an internal force such as tension
- Analyze Atwood machines and objects on inclines with a rotated axis
Two objects, one acceleration
When two objects are joined by an inextensible string or are in contact and pushed together, they must move with the same magnitude of acceleration — if one sped up relative to the other, the string would stretch or the blocks would overlap. That single fact turns a two-body problem into a one-body problem: add the masses, apply only the external forces, and solve for a.
Why internal forces vanish from the system equation
The tension the string exerts on block A and the tension it exerts on block B are a Newton's third-law pair. They are equal in magnitude and opposite in direction, so when you draw a boundary around both blocks and sum forces, they cancel exactly. This is why the system approach never mentions tension. The price is that the system approach also cannot find tension — for that you must go back and isolate one block.
A 3.0 kg block on a frictionless table is connected by a light string over an ideal pulley to a 2.0 kg block hanging off the edge. Find the acceleration and the tension.
- 1.System: the only external force along the direction of motion is the hanging block's weight, (2.0)(10) = 20 N.
- 2.Total mass = 3.0 + 2.0 = 5.0 kg, so a = 20 N ÷ 5.0 kg = 4.0 m/s².
- 3.Isolate the 3.0 kg block on the table: the only horizontal force on it is the tension, so T = m a = (3.0)(4.0) = 12 N.
- 4.Check with the hanging block: 20 N − T = (2.0)(4.0) = 8 N, so T = 12 N. Consistent.
Tension is not the hanging weight
In that example the tension came out as 12 N, not the 20 N weight of the hanging block. It has to be less: if the tension equaled the weight, the hanging block would have zero net force and would not accelerate. Tension equals weight only when the acceleration is zero. Students who write T = mg for an accelerating system lose the point every time.
Inclines: rotate the axes, split the weight
On a ramp of angle θ the motion is along the surface, so tilt your coordinate system to match: x along the incline, y perpendicular to it. Only the weight needs resolving. The component along the incline is mg sin θ; the component into the surface is mg cos θ, which is what the normal force must balance. Note that N = mg cos θ, not mg — this is why friction on a ramp is weaker than friction on level ground for the same block.
The Atwood machine
Two masses hang from opposite sides of one pulley. The heavier side falls, the lighter side rises, and the shared acceleration is the net weight divided by the total mass: a = (m₁ − m₂)g / (m₁ + m₂). When the masses are nearly equal the acceleration is small, which is precisely why Atwood built the device — it slows gravity down enough to time by hand.
An "ideal" pulley is massless and frictionless, which is what lets you say the tension is the same on both sides of the string. In Unit 5 the pulley gets rotational inertia and that stops being true — the tensions differ, and the difference is what supplies the torque that spins it.
A 4.0 kg and a 6.0 kg block sit in contact on a frictionless floor. A 20 N horizontal force pushes on the 4.0 kg block. What is the contact force between the blocks?
A 5.0 kg mass and a 3.0 kg mass hang from an ideal Atwood machine. What is the magnitude of the acceleration? (g = 10 m/s²)
A 2.0 kg block is released on a frictionless 30° incline. What is its acceleration down the slope? (g = 10 m/s², sin 30° = 0.50)
Draw a separate free-body diagram for each object, even when you plan to use the system shortcut. Rubrics award the diagrams independently of the algebra, and a labeled diagram with no numbers still earns points.
Answer the 3 checkpoints as you read.
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