At a given instant the position of a plane


At a given instant the position of a plane at A and a train at B are measured relative to a radar antenna at O. Determine the distance d between A and B at this instant. To solve the problem, formulate a position vector, directed from A to B, and then determine its magnitude.

At a given instant the position of a plane

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

Show me the final answer↓

We will first find out the locations of point A and point B and write them in Cartesian vector form. To do so, we will use trigonometry.

At a given instant the position of a plane

Notice how we can create a right angle triangle as highlighted in the image. Using sine and cosine functions, we can figure out the z-component of point A (the height) and F’. F’ is the force that lies on the xy plane.

z-component = 5\sin(60^0)\,=\,4.33 km

F'\,=\,5\cos(60^0)\,=\,2.5 km

(Remember, sin is opposite over hypotenuse, and cosine is adjacent over hypotenuse. In this triangle, the hypotenuse is 5)

 

Next, we will figure out the x and y components of point A.

At a given instant the position of a plane

In this new triangle we highlighted, notice how F’ is now the hypotenuse. We will use F’ and trigonometry to figure out the x and y components.

x-component = 2.5\cos(35^0)\,=\,2.05 km

y-component = 2.5\sin(35^0)\,=\,1.43 km

(remember we found that F’=2.5)

 

We can now write point A in Cartesian vector form.

A:(-2.05i-1.43j+4.33k) km

(Note the negative signs in front of the i(x-term) and the j(y-term). This is because from the diagram, we see that point A lies on the negative quadrant of the x and y axes)

 

We will now focus on point B.

At a given instant the position of a plane at

Again, we will use trigonometry to figure out the new F’ in this triangle, and the z-component (height).

z-component = 2\sin(25^0)\,=\,0.84 km

F'\,=\,2\cos(25^0)\,=\,1.81 km

 

We will now figure out the x and y-components.

At a given instant the position of a plane at

As before, note how F’ is now the hypotenuse of the triangle we just formed.

x-component = 1.81\sin(40^0)\,=\,1.16 km

y-component = 1.81\cos(40^0)\,=\,1.39 km

(remember we found that F’=1.81)

 

Let us write point B in Cartesian vector form.

B:(1.16i+1.39j-0.84k) km

(Note the negative signs in front of the k(z-term). Again, this is because from the diagram, we see that point B lies below the xy plane. In other words, it is in the negative z-axis.)

 

We can now figure out the position vector, r_{AB}.

r_{AB}\,=\,\left\{(1.16-(-2.05))i+(1.39-(-1.43))j+(-0.84-4.33)k\right\}

r_{AB}\,=\,\left\{3.21i+2.82j-5.17k\right\} km

A position vector, denoted \mathbf{r} is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was (x_A,y_A,z_A) and the coordinates of point B was(x_B,y_B,z_B), then r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k

 

The distance between the two points is equal to the magnitude of the position vector r_{AB}.

magnitude of r_{AB}\,=\,\sqrt{(3.21)^2+(2.82)^2+(-5.17)^2}

magnitude of r_{AB}\,=\,6.71 km

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was r\,=\,ai+bj+ck, then the magnitude would be, r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

 

Final Answer:

The distance between point A and B is 6.71 km.

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-95.

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