Everything from the 3 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
Newton's second law: resultant force \(= ma\), resolved in the direction of the acceleration. Perpendicular to the motion the forces still balance. Weight is \(W = mg\) with \(g = 10\ \text{m s}^{-2}\), and a sliding particle has friction \(F = \mu R\).
\(F = ma\) uses the RESULTANT force. Writing \(50 = 8a\), or \(50\cos 30^\circ = 8a\), ignores friction and is the most common lost mark.
Keep signs consistent. If the particle is slowing down, \(a\) is negative in your positive direction, and so is the resultant force.
"Moving upwards" does not mean "accelerating upwards". A lift moving up while slowing down has a DOWNWARD acceleration, so the reaction is less than the weight.
Along the plane: resultant force \(= ma\). Perpendicular: \(R = mg\cos\alpha\) (plus or minus any other perpendicular components). Moving up a rough plane, friction acts down it, so \(mg\sin\alpha + \mu R\) slows the particle; moving down, friction acts up, so the driving force is \(mg\sin\alpha - \mu R\).
\(a = g\sin\alpha\), not \(g\cos\alpha\). Check with a limiting case: on a vertical "slope" (\(\alpha = 90^\circ\)) the particle falls freely with \(a = g\), and \(\sin 90^\circ = 1\).
After a particle stops at the top, do not assume it slides back. Compare the weight component \(mg\sin\alpha\) with the maximum friction \(\mu mg\cos\alpha\): it stays put if \(\mu \ge \tan\alpha\).
Using \(R = mg\cos\alpha\) when the applied force has a component perpendicular to the plane. In the second example that gives friction 5.20 N instead of 9.20 N.
A light inextensible string has the same tension all along it (also over a smooth pulley), and the particles it joins have accelerations of equal size. Write \(F = ma\) for EACH particle in its own direction of motion, then solve the equations together. A tow-bar can be in tension or in thrust.
A rope can only pull. A tow-bar can push, and when the car brakes hard it does. If your tension comes out negative, the connector is in thrust (for a rope, that would mean it goes slack).
The tension is NOT the weight of the hanging particle. If it were, B would have no resultant force and could not accelerate. In Fig. 1, \(T = 14.4\) N, less than B's 20 N weight.
After the string goes slack, the rising particle does NOT stop at once; it keeps moving up and decelerates at \(g\). Forgetting this second stage gives 1.2 m instead of 1.5 m.