Tension in inextensible strings

WHAT is tension?

WHY is tension the same throughout a massless string?

This is derived, not assumed.

HOW the inextensibility constraint works (the 80/20 idea)

Derivation for two masses over a pulley: Let $x_1, x_2$ be lengths of string from pulley down to each mass. Total length: $L = x_1 + x_2 + (\text{wrap over pulley}) = \text{constant}$ Differentiate w.r.t. time: $\dot{x}_1 + \dot{x}_2 = 0 \quad\Rightarrow\quad \ddot{x}_1 = -\ddot{x}_2$

So magnitudes are equal, $|a_1| = |a_2| = a$. When one mass goes down, the other goes up by the same amount. That's the whole secret.

Worked Example 1 — Two masses, frictionless pulley (Atwood machine)

Masses $m_1 > m_2$ hang over an ideal pulley. Find $a$ and $T$.

Step 1 — FBD each mass. Why? Newton's 2nd law applies to each body separately; tension is an external force on each.

For $m_1$ (accelerates down with $a$): $m_1 g - T = m_1 a$ Why this sign? Net force = (down weight) − (up tension), and motion is downward, so we call down positive for this mass.

For $m_2$ (accelerates up with the same $a$, by inextensibility): $T - m_2 g = m_2 a$ Why same $a$? The string is inextensible ⟹ equal acceleration magnitudes.

Step 2 — Add the equations to eliminate $T$: $m_1 g - m_2 g = (m_1 + m_2) a \;\Rightarrow\; \boxed{a = \frac{(m_1 - m_2)g}{m_1 + m_2}}$

Step 3 — Back-substitute for $T$: $\boxed{T = \frac{2 m_1 m_2}{m_1 + m_2}\,g}$ Why this step? Plug $a$ into either equation; this is the actual answer the question wants.

Worked Example 2 — Block on table connected to hanging block

$m_1$ on a frictionless table, string over a pulley at the edge, hanging mass $m_2$.

Step 1 — FBD. $m_2$ (falls): $\; m_2 g - T = m_2 a$ $m_1$ (slides horizontally): $\; T = m_1 a$ Why? On the table, only $T$ acts horizontally; weight & normal cancel vertically.

Step 2 — Add: $m_2 g = (m_1 + m_2)a \Rightarrow \boxed{a = \frac{m_2 g}{m_1 + m_2}}, \quad T = \frac{m_1 m_2 g}{m_1 + m_2}$

Worked Example 3 — Connected blocks pulled by a force

Two blocks $m_1, m_2$ on a frictionless floor, string between them, force $F$ pulls $m_2$ (so $m_1$ is dragged behind).

Step 1 — System first (find $a$): $\;F = (m_1+m_2)a \Rightarrow a = \frac{F}{m_1+m_2}$. Why treat as one system? They share $a$; internal tension cancels for the whole system.

Step 2 — Isolate $m_1$ to get $T$: only $T$ pulls $m_1$, so $T = m_1 a = \boxed{\frac{m_1 F}{m_1 + m_2}}$ Why $m_1$ not $m_2$? Tension is the only force accelerating $m_1$, so it directly equals $m_1 a$.

Common Mistakes (Steel-manned)

Active Recall

Connections

• Newton's Second Law — every tension equation is $F=ma$ applied to one body.
• Free Body Diagrams — the tool that makes tension visible.
• Constraint Relations — generalization of "inextensible ⟹ same $a$".
• Atwood Machine — canonical tension problem.
• Frictionless Pulleys vs Pulleys with Inertia — when $T$ is not the same on both sides.
• Normal Force — the other contact-force cousin (pushes instead of pulls).

Concept Map

Hinglish (regional understanding)