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.

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

Solution:

This question involves expressing vectors in Cartesian vector form. If you are unsure on how to do this, read the detailed guide on vectors in Cartesian notation.

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.

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

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

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

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)

(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.

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

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

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

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

x-component =

1.81\sin(40^0)\,=\,1.16

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)

(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.

At a given instant the position of a plane

Solution:

Final Answer:

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

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