Equilibrium & resolving forces
Forces in Action - OCR A-Level Physics
Key Definition
Equilibrium
An object is in equilibrium when (1) the resultant force on it is zero and (2) the resultant moment about any point is zero. The object is either stationary or moving with constant velocity.
An object is in equilibrium when (1) the resultant force on it is zero and (2) the resultant moment about any point is zero. The object is either stationary or moving with constant velocity.
$$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M = 0$$
Resolving a force
Resolving a force
Splitting a single force into two perpendicular components, usually horizontal and vertical. For a force $F$ at angle $\theta$ to the horizontal: $F_x = F\cos\theta$ and $F_y = F\sin\theta$.
Splitting a single force into two perpendicular components, usually horizontal and vertical. For a force $F$ at angle $\theta$ to the horizontal: $F_x = F\cos\theta$ and $F_y = F\sin\theta$.
$$F_x = F\cos\theta, \qquad F_y = F\sin\theta$$
- For a child on a swing pulled back by a horizontal force $P$, with rope tension $T$ at angle $\theta$ to the vertical: vertical balance $W = T\cos\theta$; horizontal balance $P = T\sin\theta$.
- On a slope of angle $\theta$, it is simpler to resolve parallel and perpendicular to the slope: parallel component of weight $= W\sin\theta$, perpendicular component $= W\cos\theta$.
- For an object stationary on a slope, friction balances the parallel component: $F_f = W\sin\theta$. The normal contact force balances the perpendicular component: $N = W\cos\theta$.
- Three coplanar forces in equilibrium form a closed triangle when drawn tip-to-tail. This is the triangle-of-forces method.
Diagram pending
Block on an inclined plane at angle $\theta$. Weight $W$ vertically downward, resolved into $W\sin\theta$ down the slope and $W\cos\theta$ into the slope. Normal contact force $N = W\cos\theta$ perpendicular to slope; friction $F_f$ up the slope.
Will be replaced with a GeoGebra SVG in stream 2.
Diagram pending
Triangle of forces for the child on a swing: weight $W$ down, tension $T$ at angle $\theta$ to the vertical, horizontal pull $P$. Drawn tip-to-tail to form a closed triangle showing equilibrium.
Will be replaced with a GeoGebra SVG in stream 2.
Common Mistake
HIGH
Students often: swap $\sin$ and $\cos$ when resolving, e.g. write the parallel component on a slope as $W\cos\theta$.
Instead: draw a clear right-angled triangle and identify which side is adjacent to $\theta$. The component along the angle's adjacent direction uses $\cos$; the opposite direction uses $\sin$. On a slope of angle $\theta$, the parallel component is $W\sin\theta$ because the slope is opposite to $\theta$ measured from the vertical weight.
Instead: draw a clear right-angled triangle and identify which side is adjacent to $\theta$. The component along the angle's adjacent direction uses $\cos$; the opposite direction uses $\sin$. On a slope of angle $\theta$, the parallel component is $W\sin\theta$ because the slope is opposite to $\theta$ measured from the vertical weight.
Examiner Tips and Tricks
- State BOTH equilibrium conditions before solving: $\sum F = 0$ AND $\sum M = 0$. Markers award marks for stating the principle.
- When using the triangle-of-forces method, label each side with the force it represents and confirm it closes.
- Choose your axes to make the maths simple: on a slope, resolve parallel and perpendicular to the slope, not horizontal and vertical.