Engineering Example 3

5 Engineering Example 3

5.1 Vintage car brake pedal mechanism

Introduction

Figure 16 shows the structure and some dimensions of a vintage car brake pedal arrangement as far as the brake cable. The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. The pedal is pivoted about the point $A$ . The moments about $A$ must be equal as the pedal is stationary.

Problem in words

If the driver supplies a force of 900 $N$ , to act at point $B,$ calculate the force $(F)$ in the cable.

Mathematical statement of problem

The perpendicular distance from the line of action of the force provided by the driver to the pivot point $A$ is denoted by $x_{1}$ and the perpendicular distance from the line of action of force in the cable to the pivot point $A$ is denoted by $x_{2}$ . Use trigonometry to relate $x_{1}$ and $x_{2}$ to the given dimensions. Calculate clockwise and anticlockwise moments about the pivot and set them equal.

Figure 16 :

A diagram illustrating the force and dimensional relationships in a brake pedal mechanism.

Long description

Mathematical Analysis

The distance $x_{1}$ is found by considering the right-angled triangle shown in Figure 17 and using the definition of cosine.

Figure 17

A diagram of a right-angled triangle with labeled side lengths and angles, accompanied by a trigonometric calculation.

Long description

The distance $x_{2}$ is found by considering the right-angled triangle shown in Figure 18.

Figure 18

$A diagram of a right-angled triangle is shown, with one side labeled 75 millimeters and an angle labeled 15 degrees, next to a mathematical calculation for $x_2$.$

Long description

Equating moments about $A$ :

$900 x_{1} = F x_{2}$ so $F = 2013$ N.

Interpretation

This means that the force exerted by the cable is 2013 N in the direction of the cable. This force is more than twice that applied by the driver. In fact, whatever the force applied at the pedal the force in the cable will be more than twice that force. The pedal structure is an example of a lever system that offers a mechanical gain.

Task!

Obtain all the angles and the remaining side for the triangle shown:

A right-angled triangle labeled ABC shows the lengths of its sides as 5, 4, and c.

Long description

This is Case 3. To obtain the angle at $B$ we use $tan B = \frac{4}{5}$ so $B = {tan}^{- 1} (0.8) = 38.6 6^{\circ}$ .

Then the angle at $A$ is $18 0^{\circ} - (9 0^{\circ} - 38.6 6^{\circ}) = 51.3 4^{\circ}$ .

By Pythagoras’ theorem $c = \sqrt{4^{2} + 5^{2}} = \sqrt{41} \approx 6.40$ .

Task!

Obtain the remaining sides and angles for the triangle shown.

A right-angled triangle with the sides labeled "a" and "b," and the hypotenuse labeled "15," with one of the acute angles marked as 31 degrees and 40 minutes.

Long description

This is Case 1. Since $3 1^{\circ} 4 0^{'} = 31.6 7^{\circ}$ then $cos 31.6 7^{\circ} = \frac{a}{15}$ so $a = 15 cos 31.6 7^{\circ} = 12.77 .$

The angle at $A$ is $18 0^{\circ} - (90 + 31.6 7^{\circ}) = 58.3 3^{\circ}$ .

Finally $sin 31.6 7^{\circ} = \frac{b}{15}$ $∴ b = 15 sin 31.6 7^{\circ} = 7.85$ .

(Alternatively, of course, Pythagoras’ theorem could be used to calculate the length $b$ .)

Task!

Obtain the remaining sides and angles of the following triangle.

$A right-angled triangle labeled with sides $a$, $c$, and a vertical side of length $8$, with one acute angle marked as $34$ degrees and $20$ minutes.$

Long description

This is Case 2.

Here $tan 34.3 3^{\circ} = \frac{8}{a}$ so $a = \frac{8}{tan 34.3 3^{\circ}} = 11.7$

Also $c = \sqrt{8^{2} + 11 . 7^{2}} = 14.18$ and the angle at $A$ is $18 0^{\circ} - (9 0^{\circ} + 34.3 3^{\circ}) = 55.6 7^{\circ}$ .

Exercises

Obtain $cosec θ, sec θ$ , $cot θ$ , $θ$ in the following right-angled triangle.

Long description
Write down $sin θ, cos θ, tan θ, cosec θ$ for each of the following triangles:

Long description
If $θ$ is an acute angle such that $sin θ = 2 ∕ 7$ obtain, without use of a calculator, $cos θ$ and $tan θ$ .
Use your calculator to obtain the acute angles θ satisfying
1. $sin θ = 0.5260$ ,
2. $tan θ = 2.4$ ,
3. $cos θ = 0.2$
Solve the right-angled triangle shown:

Long description
A surveyor measures the angle of elevation between the top of a mountain and ground level at two different points. The results are shown in the following figure. Use trigonometry to obtain the distance $z$ (which cannot be measured) and then obtain the height $h$ of the mountain.
$A diagram showing two right triangles formed by a vertical line segment of height $h$ and an angled hypotenuse, with horizontal bases labeled $0.5$ kilometers and $z$ kilometers.$
Long description
As shown below two tracking stations $S_{1}$ and $S_{2}$ sight a weather balloon ( $W B$ ) between them
at elevation angles $α$ and $β$ respectively.

Long description
Show that the height $h$ of the balloon is given by $h = \frac{c}{cot α + cot β}$
A vehicle entered in a ‘soap box derby’ rolls down a hill as shown in the figure. Find the total
distance $(d_{1} + d_{2})$ that the soap box travels.

Long description

$h = \sqrt{1 5^{2} + 8^{2}} = 17, cosec θ = \frac{1}{sin θ} = \frac{17}{8} sec θ = \frac{1}{cos θ} = \frac{17}{15} cot θ = \frac{1}{tan θ} = \frac{15}{8}$
$θ = {sin}^{- 1} \frac{8}{17} (for example) ∴ θ = 28.0 7^{\circ}$
1. $sin θ = \frac{2}{5} cos θ = \frac{\sqrt{21}}{5} tan θ = \frac{2 \sqrt{21}}{21}$ $cosec θ = \frac{5}{2}$
2. $sin θ = \frac{y}{\sqrt{x^{2} + y^{2}}} cos θ = \frac{x}{\sqrt{x^{2} + y^{2}}} tan θ = \frac{y}{x}$ $cosec θ = \frac{\sqrt{x^{2} + y^{2}}}{y}$
Referring to the following diagram
$A right-angled triangle labeled ABC shows an angle theta at vertex A, the side opposite the right angle (the hypotenuse) labeled 7, and the side adjacent to angle theta labeled $\ell$ with a length of 2.$
Long description
Hence $cos θ = \frac{3 \sqrt{5}}{7} tan θ = \frac{2}{3 \sqrt{5}} = \frac{2 \sqrt{5}}{15}$
1. $θ = {sin}^{- 1} 0.5260 = 31.7 3^{\circ}$
2. $θ = {tan}^{- 1} 2.4 = 67.3 8^{\circ}$
3. $θ = {cos}^{- 1} 0.2 = 78.4 6^{\circ}$
$β = 90 - α = 32 . 5^{\circ}$ , $b = \frac{10}{tan 57 . 5^{\circ}} ≃ 6.37$ $c = \frac{10}{sin 57 . 5^{\circ}} ≃ 11.86$
$tan 3 7^{\circ} = \frac{h}{z + 0.5} tan 4 1^{\circ} = \frac{h}{z}$ from which
$h = (z + 0.5) tan 3 7^{\circ} = z tan 4 1^{\circ}$ , so $z tan 3 7^{\circ} - z tan 4 1^{\circ} = - 0.5 tan 3 7^{\circ}$

$∴ z = \frac{- 0.5 tan 3 7^{\circ}}{tan 3 7^{\circ} - tan 4 1^{\circ}} ≃ 3.2556 km$ , so $h = z tan 4 1^{\circ} = 3.2556 tan 4 1^{\circ} ≃ 2.83$ km
Since the required answer is in terms of $cot α$ and $cot β$ we proceed as follows:
Using $x$ to denote the distance $S_{1} P$ $cot α = \frac{1}{tan α} = \frac{x}{h} cot β = \frac{1}{tan β} = \frac{c - x}{h}$

Adding: $cot α + cot β = \frac{x}{h} + \frac{c - x}{h} = \frac{c}{h}$ $∴ h = \frac{c}{cot α + cot β}$ as required.
From the smaller right-angled triangle $d_{1} = \frac{200}{sin 2 8^{\circ}} = 426.0 m$ . The base of this triangle then has length $ℓ = 426 cos 2 8^{\circ} = 376.1 m$
From the larger right-angled triangle the straight-line distance from START to FINISH is $\frac{200}{sin 1 5^{\circ}} = 772.7$ m. Then, using Pythagoras’ theorem $(d_{2} + ℓ) = \sqrt{772 . 7^{2} - 20 0^{2}} = 746.4 m$ from which $d_{2} = 370.3$ m $∴ d_{1} + d_{2} = 796.3$ m

5 Engineering Example 3

5.1 Vintage car brake pedal mechanism

Task!

Answer

Task!

Answer

Task!

Answer

Exercises

Answer