add and subtract coplanar vectors

Scalars and \cdot ectors

In physics a quantity is either a scalar or a vector. Scalars have only magnitude (size), whereas vectors have both magnitude and direction.

Key Definitions

• Scalar: A quantity described by a single number and its unit (e.g., mass \$m\$, temperature \$T\$, energy \$E\$).
• Vector: A quantity described by a magnitude and a direction (e.g., displacement \$\vec{s}\$, velocity \$\vec{v}\$, force \$\vec{F}\$).
• Coplanar vectors: Vectors that lie in the same plane.

Representing \cdot ectors

Vectors are commonly represented by arrows. The length of the arrow is proportional to the magnitude, and the arrow points in the direction of the vector.

Adding Coplanar \cdot ectors

1. Graphical (Tip‑to‑Tail) Method

1. Place the tail of the second vector at the tip of the first vector.
2. The resultant vector \$\vec{R}\$ is drawn from the tail of the first vector to the tip of the second vector.

2. Parallelogram Method

1. Place the tails of the two vectors at the same point.
2. Complete the parallelogram formed by the two vectors.
3. The diagonal of the parallelogram starting from the common tail is the resultant \$\vec{R}\$.

3. Component Method

When vectors are expressed in components, addition is performed component‑wise:

\$\vec{A} = Ax \hat{i} + Ay \hat{j}, \qquad \vec{B} = Bx \hat{i} + By \hat{j}\$

\$\vec{R} = \vec{A} + \vec{B} = (Ax + Bx)\hat{i} + (Ay + By)\hat{j}\$

The magnitude and direction of \$\vec{R}\$ are then obtained from:

\$R = \sqrt{(Ax + Bx)^2 + (Ay + By)^2}\$

\$\thetaR = \tan^{-1}\!\left(\frac{Ay + By}{Ax + B_x}\right)\$

Subtracting Coplanar \cdot ectors

Subtracting \$\vec{B}\$ from \$\vec{A}\$ is equivalent to adding \$\vec{A}\$ to the negative of \$\vec{B}\$:

\$\vec{A} - \vec{B} = \vec{A} + (-\vec{B})\$

Graphically, reverse the direction of \$\vec{B}\$ (keep its magnitude) and then use the addition methods above.

Component Method for Subtraction

\$\vec{A} - \vec{B} = (Ax - Bx)\hat{i} + (Ay - By)\hat{j}\$

Magnitude and direction are obtained in the same way as for addition.

Worked Example

Problem: Two forces act in the horizontal plane. \$\vec{F}1 = 30\;\text{N}\$ at \$30^\circ\$ north of east, and \$\vec{F}2 = 40\;\text{N}\$ at \$120^\circ\$ measured anticlockwise from east. Find the resultant force \$\vec{F}R = \vec{F}1 + \vec{F}_2\$.

1. Resolve each force into \$x\$ (east) and \$y\$ (north) components.

   \$F{1x}=30\cos30^\circ,\quad F{1y}=30\sin30^\circ\$

   \$F{2x}=40\cos120^\circ,\quad F{2y}=40\sin120^\circ\$

2. Calculate numerical components (use a calculator):

   \$F{1x}=30(0.866)=25.98\;\text{N},\qquad F{1y}=30(0.5)=15.0\;\text{N}\$

   \$F{2x}=40(-0.5)=-20.0\;\text{N},\qquad F{2y}=40(0.866)=34.64\;\text{N}\$

3. Add the components:

   \$F{Rx}=F{1x}+F_{2x}=25.98-20.0=5.98\;\text{N}\$

   \$F{Ry}=F{1y}+F_{2y}=15.0+34.64=49.64\;\text{N}\$

4. Find magnitude and direction:

   \$F_R=\sqrt{(5.98)^2+(49.64)^2}=50.0\;\text{N}\$

   \$\theta_R=\tan^{-1}\!\left(\frac{49.64}{5.98}\right)=83.1^\circ\;\text{north of east}\$

Summary Table

Key Points to Remember

• Scalars have no direction; vectors do.
• Coplanar vectors can be added or subtracted using graphical or component methods.
• When using components, always keep consistent sign conventions (e.g., east = +\$x\$, north = +\$y\$).
• Check the quadrant of the resultant when evaluating the inverse tangent.