On occasion objects move within a medium that is itself in motion with respect to an observer. For example, an airplane usually encounters a wind as it flies. The speed of the airplane with respect to a stationary observer situated on the ground will not be the same as the speedometer reading for the plane. In order to calculate the true speed of the plane, it is essential to include the effect due to the wind. In some cases, planes encounter headwinds which will diminish the speed of the plane relative to a stationary observer. I other cases, planes may encounter tail winds which will increase the speed relative to a stationary observer.

Question 1: Let us consider the example where an airplane is traveling with a speed of 880 km/hr with respect to the air. Suppose the airplane encounters a tailwind of velocity 30 km/hr. What is the resultant velocity of the airplane relative to an observer on the ground?

This problem can be addressed using the concept of vector addition. We will represent the velocity of the plane relative to the air as the vector V₁ and that of the tail wind as the vector V₂. The velocity of the plane relative to a stationary observer on the ground can be found by performing the vector addition of V₁ and V₂. This addition is accomplished by placing the tail of vector V2 to the head of vector V1. The result, 910 km/m, is shown in Figure 1 (a).

Figure 1: Combining velocities.

Let us now consider a different situation.

Question 2: Consider the same airplane as in Question 1. However, assume that the airplane is encountering a head wind of 30 km/hr. What is the speed of the plane relative to a stationary observer on the ground?

In this case, the head wind serves to slow the airplane. The situation is reflected in Figure 1 (b). We note that the vector V2 is directed opposite to that of vector V1 due to the fact that the head wind opposes the motion of the airplane. The resulting speed relative to an observer on the ground is computed as 850 km/hr.

Now, we will consider a situation in which we must use our knowledge of vectors and trigonometry to find the correct result.

Question 3: Consider the same airplane as in Question 1. However, assume that the airplane is encountering a cross wind of 30 km/hr. What is the speed of the plane relative to a stationary observer on the ground?

This situation is depicted in Figure 2.

Figure 2: Adding orthogonal velocities.

In this case, the speed of the airplane is represented by the length of the sum of vectors, V₁ + V₂. Using the x-y coordinate system shown in the figure, we note that

and

we may write an expression for this sum in vector form as

To determine the speed of the airplane relative to a stationary observer we must find the magnitude of the vector sum. We can do so by applying the Pythagorean Theorem

Thus, we determine speed of the airplane relative to a stationary observer on the ground to be 881 km/hr.

We note that the heading of the plane has been redirected in light of the cross wind. Let us find the new heading of the plane. We use the tangent function as follows

Applying the inverse tangent leaves us with

So we conclude the heading of the airplane will be directed 1.953⁰ from its initial heading.

Let us now consider another example that will allow us to apply 2-D vectors. Suppose that a force of 750 pounds is required to pull a boat and trailer up a ramp that is inclined at an angle of 20⁰ from the horizon. This situation is depicted in Figure 3 (a). Under the assumption of no friction, what is the combined weight of the boat and trailer?

Figure 3: Depiction of a boat on a landing ramp along with a trigonometric description.

Let us now use the figure to interpret the vectors shown in Figure 3. The vector BA→BA→ size 12{ widevec { ital "BA"} } {} represents the combined weight of the boat and the trailer. This is the quantity that we need to find. The vector BC→BC→ size 12{ widevec { ital "BC"} } {} represents the force against the ramp. The vector AC→AC→ size 12{ widevec { ital "AC"} } {} which is parallel to the ramp represents the force applied to the boat and trailer. From the problem description we know that it has a magnitude of 750 pounds. We use the fact that this is a right triangle to simplify our efforts.

We can apply the definition of the sine function to obtain

Substitution leads to the equation

We can now solve for the weight of the boat and trailer

A 200-pound weight is suspended from a ceiling. The weight is supported by two cables. One cable makes a 20⁰ angle away from the vertical and the other a 30⁰ angle as shown in Figure 4. Find the tension in each of the support cables.

Figure 4: Mass suspended from a beam with two cables.

Statics is the field of engineering that is used to solve problems of this sort. Because the object does not move, it is said to be static. Another way to look at this, is that the object is at equilibrium. At equilibrium, the forces acting on an object must balance. Otherwise, the object would indeed move. To better analyze the situation, engineers often make use of what is known as a free body diagram. Figure 5 shows the free body diagram related to our problem.

Figure 5: Free body diagram of mass suspended from a beam with two cables.

The free body diagram shows the three forces that act on the object. The Figure also shows the x-y coordinate system employed in this solution. The tension in cable 1 can be resolved into its x and y components and written in vector notation as

Similarly, T₂ can be written as

The weight (W) is expressed as

The sum of the forces acting on the object must be 0. Thus

or

This problem can be simplified by writing an equation for solely the x-component

We can do the same for the y-component

We begin by substituting in for the trigonometric function values to yield the following set of equations

and

First, we find

This expression can then be substituted into the other equation

This leads to the solution

Next, we solve for the other variable

So we conclude that the tension in the cables are 130.5 lbs and 89.3 lbs respectively.