Equilibrium of a particleEdexcel International A Level Maths: Revision notes

A particle is in equilibrium when it is at rest (or moving at constant velocity) under a set of forces. The resultant force is zero, so the sum of the components in any direction is zero. For coplanar forces, resolve in two perpendicular directions (usually horizontal and vertical, or parallel and perpendicular to a slope) and set the sum of components in each direction equal to zero. The forces then give two equations, enough to find two unknowns.

Weight $W=mg$ acts vertically downwards, with $g=9.8$ m s$^{-2}$ in this course. Normal reaction $R$ acts perpendicular to a surface, away from it. Tension $T$ acts along a string or rope, pulling on the particle; for a light string the tension is the same throughout. Thrust is the equivalent push along a rod. Friction $F$ acts along a rough surface, opposing (the tendency of) motion. A smooth surface has no friction. On a rough surface $F\le\mu R$, and in limiting equilibrium, when the particle is on the point of slipping, $F=\mu R$, where $\mu$ is the coefficient of friction.

Worked example. A crate of mass 20 kg is pulled by a rope with a force of 50 N at $30^\circ$ above the horizontal and stays at rest on a rough floor. Resolving vertically: $R+50\sin30^\circ=20g=196$, so $R=171$ N. Resolving horizontally: $F=50\cos30^\circ=43.3$ N. If the crate is on the point of slipping, $\mu=\frac{F}{R}=\frac{43.3}{171}=0.253$. Pulling at an angle upwards reduces the normal reaction, and pushing downwards increases it.

On a plane inclined at $\alpha$, resolve parallel and perpendicular to the plane. A particle of weight $W$ held by a force $P$ up a smooth slope: $P=W\sin\alpha$ and $R=W\cos\alpha$. If the holding force is horizontal (force $H$), its components are $H\cos\alpha$ up the slope and $H\sin\alpha$ into the slope. Parallel to the plane: $H\cos\alpha=W\sin\alpha$, so $H=W\tan\alpha$. Perpendicular: $R=W\cos\alpha+H\sin\alpha$. On a rough slope, add friction $F$ acting up the slope if the particle tends to slide down.

When a particle hangs from two light strings, each tension acts along its string. Resolve horizontally and vertically. Example. A weight of 80 N hangs from strings at $30^\circ$ and $60^\circ$ to the horizontal on opposite sides. Horizontally: $T_1\cos30^\circ=T_2\cos60^\circ$, so $T_2=\sqrt3\,T_1$. Vertically: $T_1\sin30^\circ+T_2\sin60^\circ=80$, so $2T_1=80$, $T_1=40$ N and $T_2=40\sqrt3\approx69.3$ N. Solve the simultaneous equations by substitution, and check by testing one of the equations with your values.