In one line
Draw an FBD of each body with its inertia force, link them through the common string tension and acceleration, and solve the equations together.
Explained simply
Two blocks tied by a rope over a pulley move together like train coaches: same speed, same acceleration. Each block's own equation has the rope tension, so solving them together gives both the acceleration and the tension.
Technically, for an inextensible string over a light, frictionless pulley, the tension is the same on both sides and the bodies have equal magnitudes of acceleration.
How it works
- Decide the direction of motion (the heavier side or the larger driving force).
- Draw an FBD for each body: weight, normal reaction, friction (\( \mu_k N \)), tension T, and inertia force \( m_i a \) opposite to its acceleration.
- Write the dynamic-equilibrium equation along the motion for each body.
- Add the equations to eliminate T and get a; substitute back for T.
- For a movable pulley, use the string-length relation (one body's acceleration may be twice the other's).
Key formula
Block A (\( m_A \)) on a rough table, block B (\( m_B \)) hanging: \( a = \dfrac{(m_B - \mu m_A)g}{m_A + m_B} \), \( T = m_A(a + \mu g) \). Masses in kg, a in m/s\(^2\), T in N.
Quick example
A = 10 kg on a table (\( \mu = 0.2 \)), B = 5 kg hanging. B: \( 49.05 - T - 5a = 0 \). A: \( T - 19.62 - 10a = 0 \). Adding: \( 29.43 = 15a \), so a = 1.962 m/s\(^2\), T = 39.24 N. Check B: \( 49.05 - 39.24 = 9.81 = 5 \times 1.962 \).
Common mistakes
- Using different tensions on the two sides of a light frictionless pulley.
- Forgetting that a movable pulley halves or doubles accelerations.
- Taking friction as \( \mu m g \) on an incline (it is \( \mu m g\cos\theta \)).
How MU asks it
6 to 8-mark numericals: two blocks on inclined planes over a pulley, block on a table with a hanging mass, systems released from rest ('find velocity after 2 s'). Draw separate FBDs, write the equations clearly, and finish with the kinematics step.