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Solving Problems with Vectors

We can use vectors to solve many problems involving physical quantities such as velocity, speed, weight, work and so on.

Velocity:

The velocity of moving object is modeled by a vector whose direction is the direction of motion and whose magnitude is the speed.

Example :

A ball is thrown with an initial velocity of $70$ feet per second., at an angle of $35 °$ with the horizontal. Find the vertical and horizontal components of the velocity.

Let $v$ represent the velocity and use the given information to write $v$ in unit vector form:

$v = 70 (\cos (35 °)) i + 70 (\sin (35 °)) j$

Simplify the scalars, we get:

$v \approx 57.34 i + 40.15 j$

Since the scalars are the horizontal and vertical components of $v$ ,

Therefore, the horizontal component is $57.34$ feet per second and the vertical component is $40.15$ feet per second.

Force:

Force is also represented by vector. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces.

Example :

Two forces $F_{1}$ and $F_{2}$ with magnitudes $20$ and $30 lb$ , respectively, act on an object at a point $P$ as shown. Find the resultant forces acting at $P$ .

Math diagram

First we write $F_{1}$ and $F_{2}$ in component form:

$v \approx 57.34 i + 40.15 j$

Simplify the scalars, we get:

$\begin{array}{l} \begin{array}{l} F_{1} = (20 \cos (45 °)) i + (20 \sin (45 °)) j \\ = 20 (\frac{\sqrt{2}}{2}) i + 20 (\frac{\sqrt{2}}{2}) j \end{array} \\ = 10 \sqrt{2} i + 10 \sqrt{2} j \\ \begin{array}{l} F_{2} = (30 \cos (150 °)) i + (30 \sin (150 °)) j \\ = 30 (- \frac{\sqrt{3}}{2}) i + 30 (\frac{1}{2}) j \end{array} \\ = - 15 \sqrt{3} i + 15 j \end{array}$

So, the resultant force $F$ is

$\begin{array}{l} F = F_{1} + F_{2} \\ = (10 \sqrt{2} i + 10 \sqrt{2} j) + (- 15 \sqrt{3} i + 15 j) \\ = (10 \sqrt{2} - 15 \sqrt{3}) i + (10 \sqrt{2} + 15) j \\ \approx - 12 i + 29 j \end{array}$

Work:

The work $W$ done by a force $F$ in moving along a vector $D$ is $W = F \cdot D$ .

Example :

A force is given by the vector $F = ⟨ 2, 3 ⟩$ and moves an object from the point $(1, 3)$ to the point $(5, 9)$ . Find the work done.

First we find the Displacement.

The displacement vector is

$D = ⟨ 5 - 1, 9 - 3 ⟩ = ⟨ 4, 6 ⟩$ .

By using the formula, the work done is

$W = F \cdot D = ⟨ 2, 3 ⟩ \cdot ⟨ 4, 6 ⟩ = 26$

If the unit of force is pounds and the distance is measured in feet, then the work done is $26$ ft-lb.