2 Forces and Particle Equilibrium

Learn to represent forces as vectors, resolve them into components, and determine whether concurrent forces balance on a particle.

Representing a

A is a vector with both magnitude and direction. In SI units, is measured in newtons, written N\mathrm{N}. It may be represented by an arrow: the arrow's length represents magnitude, its orientation gives direction, and its line of action shows where the acts.

In a two-dimensional Cartesian coordinate system, a can be written as

F=Fxi+Fyj,\mathbf F = F_x\mathbf i + F_y\mathbf j,

where FxF_x and FyF_y are signed components along the xx- and yy-axes, and i\mathbf i and j\mathbf j are unit vectors. In three dimensions, the representation is

F=Fxi+Fyj+Fzk.\mathbf F = F_x\mathbf i + F_y\mathbf j + F_z\mathbf k.

The magnitude of a two-dimensional is

F=Fx2+Fy2,F = \sqrt{F_x^2+F_y^2},

and its direction angle, measured counterclockwise from the positive xx-axis, can be found using θ=atan2⁡(Fy,Fx)\theta=\operatorname{atan2}(F_y,F_x). The signs of the components identify the 's quadrant.

and resolution

are perpendicular vectors whose sum equals the original . If a of magnitude FF makes an angle θ\theta from the positive xx-axis, its components are

Fx=Fcos⁡θ,Fy=Fsin⁡θ.F_x=F\cos\theta, \qquad F_y=F\sin\theta.

The formulas depend on how the angle is measured. If the angle is measured from the vertical, sine and cosine switch roles. Assign signs according to the actual directions of the components. Component form makes vector sums straightforward: add corresponding components.

For example, a 100 N100\,\mathrm{N} acting 30∘30^\circ above the positive xx-axis has components

Fx=100cos⁡30∘=86.6 N,Fy=100sin⁡30∘=50.0 N.F_x=100\cos30^\circ=86.6\,\mathrm{N},\qquad F_y=100\sin30^\circ=50.0\,\mathrm{N}.

Thus, F=(86.6i+50.0j) N\mathbf F=(86.6\mathbf i+50.0\mathbf j)\,\mathrm{N}. The component values are not separate additional forces; together they represent the original .

Concurrent forces and diagrams

A has all its forces' lines of action passing through one common point. When an object's size and rotation are not relevant, it can be modeled as a particle, with all forces drawn as acting at that point.

A isolates the particle and shows every external acting on it, such as applied forces, cable tensions, and weight. Do not include forces the particle exerts on other objects.

The is the vector sum of the forces:

R=∑iFi.\mathbf R=\sum_i\mathbf F_i.

For example, two concurrent forces on a particle—40 N40\,\mathrm{N} to the right and 30 N30\,\mathrm{N} upward—have resultant

R=(40i+30j) N,∣R∣=402+302=50 N.\mathbf R=(40\mathbf i+30\mathbf j)\,\mathrm{N},\qquad |\mathbf R|=\sqrt{40^2+30^2}=50\,\mathrm{N}.

It points tan⁡−1(30/40)=36.9∘\tan^{-1}(30/40)=36.9^\circ above the positive xx-axis.

occurs when the is zero. In two dimensions, the component sums must each be zero:

∑Fx=0,∑Fy=0.\sum F_x=0,\qquad \sum F_y=0.

In three dimensions, the third condition is also required:

∑Fz=0.\sum F_z=0.

For the example with forces of 40 N40\,\mathrm{N} rightward and 30 N30\,\mathrm{N} upward, a third must be equal and opposite to their resultant:

F3=(−40i−30j) N.\mathbf F_3=(-40\mathbf i-30\mathbf j)\,\mathrm{N}.

This has magnitude 50 N50\,\mathrm{N} and acts down and left. The component checks are 40−40=0 N40-40=0\,\mathrm{N} horizontally and 30−30=0 N30-30=0\,\mathrm{N} vertically.

Setting up an equilibrium problem

Use this sequence to set up a particle-equilibrium problem:

  1. Isolate the particle and sketch its .

  2. Choose coordinate axes; align an axis with a known when convenient.

  3. Label each with its magnitude, direction, and component signs. For an unknown , assume a direction and keep the signs consistent. A negative solution means the actual direction is opposite to the assumed direction.

  4. Resolve angled forces into components and apply ∑Fx=0\sum F_x=0, ∑Fy=0\sum F_y=0, and, if needed, ∑Fz=0\sum F_z=0.

  5. Check that units are consistent and that the final vector directions make physical sense.

These equations apply to and concurrent forces. For an extended rigid body, forces may also cause rotation, so balance alone is not generally sufficient; moment equilibrium must also be considered.