Q.A block of mass $m$ rests on a frictionless inclined plane of angle $\theta$ and is held in place by a light string running parallel to the incline. Draw the free body diagram of the block, name every force acting on it, and write the two equilibrium equations along and perpendicular to the incline.

The free body diagram of the block shows exactly three forces acting on it: its weight, $mg$, acting vertically downward; the normal reaction, $N$, acting perpendicular to the incline surface, away from it; and the tension, $T$, acting along the incline, directed up the slope (since the string is parallel to the incline and pulls the block toward the fixed support at the top). The incline being frictionless means there is no fourth (friction) force to include.

Choosing axes along and perpendicular to the incline (rather than purely horizontal and vertical) is the natural choice here, since two of the three forces ($N$ and $T$) already lie exactly along one of these two directions. The weight $mg$, which does not lie along either chosen axis, must be resolved into two components: $mg\sin\theta$ acting down the slope (along the incline), and $mg\cos\theta$ acting into the incline surface (perpendicular to it).

Since the block is held in equilibrium (at rest, zero …

List every external force on the isolated block (weight, normal, tension), resolve the weight into components along and perpendicular to the inc …

• Including the weight as a fourth separate force in addition to its own resolved components, effectively double-counting it. …