13.06 Applications of Vectors

Given a vector with magnitude $r$r and direction $\theta$θ we can write it as an ordered pair as follows: 

$\left(r\cos\theta,r\sin\theta\right)$(rcosθ,rsinθ)

This, the addition and subtraction laws and the dot product are all useful in solving problems involving displacement, velocity and force vectors.

Displacement

Vectors can be applied in displacement problems including those involving bearings.

Worked example

example 1

A small boat sails a distance of $12$12 km on a bearing  $030^\circ$030° and then turns to bearing $250^\circ$250° and travels a further $10$10 km. How far from its starting point is the boat and what should its bearing be to return there?

Think: This problem can be solved using the sine and cosine rules in trigonometry but we will use a vector approach.

Do: A useful first step is to draw a diagram of the situation. The diagram below represents the problem as it might appear on a map.

This diagram shows the displacements as vectors in coordinate form:

We used some simple trigonometry to obtain the coordinates of the displacement vectors. The resultant vector, in red, is the sum of the two displacements. Its coordinate representation is:

$\left(12\cos60^\circ-10\cos20^\circ,12\sin60^\circ-10\sin20^\circ\right)$(12cos60°−10cos20°,12sin60°−10sin20°)

This simplifies to:

$\left(-3.40,6.97\right)$(−3.40,6.97)

This vector has length:

$\sqrt{(-3.40)^2+(6.97)^2}\approx7.76$√(−3.40)2+(6.97)2≈7.76

So, the boat has a return journey of $7.76$7.76 km.

The bearing angle of the red arrow is:

$\arctan\left(\frac{6.97}{-3.4}\right)\approx296^\circ$arctan(6.97−3.4​)≈296°

So, for the return trip, the bearing will need to be $180^\circ$180° away from this, which is $116^\circ$116°.

Practice question

Question 1

Velocity

Applications involving velocity also lend themselves to vector calculations. For example, in the area of air transport, where headwinds can play a large role in the direction of flight and fuel consumption, accurate calculation of the velocity and bearing to reach a destination is required.

Worked example

example 2

A river flows at the rate of $0.5$0.5 m/s. A swimmer starting from the river bank aims at the opposite side and maintains a speed of $1.26$1.26 m/s in a direction perpendicular to the flow of the river. What is the swimmer's velocity (magnitude and direction) relative to the river bank?

Think: The swimmer will be swept downstream. The vector representing the swimmer's motion will be the sum of vectors represting the velcoty of the river current and the swimmer.

Do: The velocity of the river can be represented as the vector $(0.5,0)$(0.5,0) and the velocity of the swimmer can be represented as $(0,1.26)$(0,1.26).

Adding these two vectors gives the resultant velocity of the swimmer:

$(0.5,1.26)$(0.5,1.26)

The magnitude of this vector is:

$\sqrt{0.5^2+1.26^2}\approx1.36$√0.52+1.262≈1.36 m/s.

The angle of the motion from the riverbank is:

$\tan^{-1}\left(\frac{1.26}{0.5}\right)\approx68.4^\circ$tan−1(1.260.5​)≈68.4°

Practice question

Question 2

Force

In physics and engineering in particular, vectors play an important role in the resolution of multiple forces into a single resultant force, the calculation of forces taking into account friction and other physical phenomena or the dividing of a force into directional components.

An important concept to understand is force equilibrium. Force equilibrium dictates that for a body at rest, with external forces applied, the sum of these forces in all directions must equal zero if the body is at rest or a constant velocity.

Worked examples

example 3

A vector has a direction angle of $27^\circ$27° and can be resolved into horizontal and vertical components such that the horizontal component has magnitude $14$14 N. What is the vertical component?

Think: We understand the direction angle to be the angle measured anticlockwise from the positive horizontal axis. Hence, we can draw the following diagram:

Do: The vector $\mathbf{v}$v, coloured green above, has magnitude related to the given information by:

$\cos27^\circ=\frac{14}{\mathbf{|v|}}$cos27°=14|v|​

Thus:

$\mathbf{|v|}=\frac{14}{\cos27^\circ}$|v|=14cos27°​

Then by Pythagoras, the vertical component has magnitude:

$\sqrt{\left(\frac{14}{\cos27^\circ}\right)^2-14^2}$√(14cos27°​)2−142

Therefore the vertical component $\approx7.1$≈7.1 N.

example 4

A $20$20 kg mass is suspended on a chain $A$A attached to a ceiling timber. This means that there is a tension force of $20\times9.8$20×9.8 N in the chain when it is hanging vertically (if we ignore the weight of the chain). 

Suppose the $20$20 kg mass is pulled horizontally by another chain $B$B until the chain attached to the ceiling is at an angle of $20^\circ$20° from the vertical. Has the tension in chain $A$A increased? 

What are the tension forces in the two chains?

Think: A diagram of the situation is drawn below.

We make use of the fact that the vector with magnitude $T_A$TA​ can be resolved into two orthogonal components. These are shown as green arrows. If the two green arrows acted, they would have the same effect as the red arrow with magnitude $T_A$TA​ alone. In effect, we replace it by the pair of components.

Do: Since $T_A>T_A\cos20^\circ$TA​>TA​cos20°, and this vertical force, magnitude $T_A\cos20^\circ$TA​cos20°,  is balancing the weight force of $196$196 N, we conclude that the tension in chain $A$A is greater than when the weight was hanging vertically.

We can form two equations:

$T_A\cos20^\circ+196=0$TA​cos20°+196=0

$T_B+T_A\sin20^\circ=0$TB​+TA​sin20°=0

From the first of these we find:

$T_A=\frac{-196}{\cos20^\circ}=-208.6$TA​=−196cos20°​=−208.6 N

We have taken the downward weight force to be positive. So, this tension force acts upward.

Then, from the second equation, we obtain:

$T_B=-(-208.6)\sin20^\circ=71.3$TB​=−(−208.6)sin20°=71.3 N

 

As an alternative strategy, we might note that in coordinate form the vector with magnitude $T_A$TA​ is $\left(T_A\sin20^\circ,T_A\cos20^\circ\right)$(TA​sin20°,TA​cos20°). The weight vector is $(0,-196)$(0,−196) and the sideways pull $T_B$TB​ is $\left(T_B,0\right)$(TB​,0).

These three vectors should add to the zero vector $(0,0)$(0,0). So, we have $\left(T_A\sin20^\circ+0+T_B,T_A\cos20^\circ-196+0\right)=(0,0)$(TA​sin20°+0+TB​,TA​cos20°−196+0)=(0,0). This vector equation leads to the same two equations as we derived above.

Practice question

Question 3

Question 4

Question 5