use a vector triangle to represent coplanar forces in equilibrium

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Equilibrium of Forces

Equilibrium of Forces

In mechanics, a system of forces is in equilibrium when the resultant force is zero. For coplanar forces this condition can be visualised using a vector (force) triangle. The triangle provides a convenient graphical method to check equilibrium and to determine an unknown force.

Key Principle

For any set of coplanar forces to be in equilibrium:

\$\sum \vec{F}=0\$

which is equivalent to saying that the forces can be arranged head‑to‑tail to form a closed polygon – for three forces this polygon is a triangle.

Constructing the \cdot ector Triangle

  1. Identify all the forces acting on the body and resolve them into their components in the plane of interest.

  2. Represent each force as a vector arrow whose length is proportional to the magnitude of the force and whose direction matches the line of action.

  3. Place the first vector arbitrarily on the page.

  4. From the head (arrow tip) of the first vector, draw the second vector, maintaining its correct magnitude and direction.

  5. Continue this process for all forces. If the system is in equilibrium, the final head will meet the tail of the first vector, closing the triangle.

  6. If the triangle does not close, the resultant vector is the line joining the start point to the final head; this resultant must be added to the system to achieve equilibrium.

Suggested diagram: A vector triangle showing three coplanar forces \$\vec{F}1\$, \$\vec{F}2\$, and \$\vec{F}_3\$ arranged head‑to‑tail, forming a closed triangle.

Worked Example

Three forces act on a particle in the horizontal plane:

  • \$\vec{F}_1 = 40\ \text{N}\$ directed \$30^\circ\$ above the positive \$x\$‑axis.

  • \$\vec{F}_2 = 50\ \text{N}\$ directed \$120^\circ\$ from the positive \$x\$‑axis.

  • \$\vec{F}_3\$ is unknown in magnitude and direction.

Determine the magnitude and direction of \$\vec{F}_3\$ required for equilibrium.

Solution Steps

  1. Resolve \$\vec{F}1\$ and \$\vec{F}2\$ into components:

    \$F{1x}=40\cos30^\circ,\qquad F{1y}=40\sin30^\circ\$

    \$F{2x}=50\cos120^\circ,\qquad F{2y}=50\sin120^\circ\$

  2. Calculate the numerical components:

    \$F{1x}=34.6\ \text{N},\; F{1y}=20.0\ \text{N}\$

    \$F{2x}=-25.0\ \text{N},\; F{2y}=43.3\ \text{N}\$

  3. Find the resultant of the known forces:

    \$Rx = F{1x}+F_{2x}=9.6\ \text{N}\$

    \$Ry = F{1y}+F_{2y}=63.3\ \text{N}\$

  4. For equilibrium, \$\vec{F}_3\$ must be equal in magnitude and opposite in direction to \$\vec{R}\$:

    \$\vec{F}_3 = -\vec{R}\$

  5. Magnitude of \$\vec{F}_3\$:

    \$|\vec{F}3| = \sqrt{Rx^2+R_y^2}= \sqrt{9.6^2+63.3^2}\approx 64.0\ \text{N}\$

  6. Direction (measured from the positive \$x\$‑axis):

    \$\theta = \tan^{-1}\!\left(\frac{Ry}{Rx}\right)=\tan^{-1}\!\left(\frac{63.3}{9.6}\right)\approx 81^\circ\$

    Since \$\vec{F}_3\$ points opposite to \$\vec{R}\$, its direction is \$81^\circ+180^\circ = 261^\circ\$ from the positive \$x\$‑axis.

Summary Table

StepActionResult
1Resolve known forces into componentsObtain \$F{ix}\$ and \$F{iy}\$ for each known force
2Calculate resultant components \$Rx\$, \$Ry\$\$Rx = \sum F{ix}\$, \$Ry = \sum F{iy}\$
3Determine unknown force \$\vec{F}_3\$\$\vec{F}_3 = -\vec{R}\$ (magnitude and direction)
4Draw vector triangleClosed triangle confirms equilibrium

Key Points to Remember

  • Equilibrium requires both \$\sum Fx = 0\$ and \$\sum Fy = 0\$ for coplanar forces.

  • The vector triangle is a visual check: if the forces close, the system is in equilibrium.

  • When an unknown force is required, it is equal and opposite to the resultant of the known forces.

  • Scale drawings must keep the length of each vector proportional to its magnitude.