WAEC · 2009 · May/June · Further Maths · Paper 2

Question 1 ✱✱

(a)
Solve $2x^2 + x - 6 < 0$ .
Show the answer
$-2 < x < \frac32$
(b)
Express $\dfrac{5 - 2\sqrt{10}}{3\sqrt5 + \sqrt2}$ in the form $m\sqrt2 + n\sqrt5$ , where $m$ and $n$ are rational numbers.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Factorise the quadratic: $2x^2 + x - 6 = (2x - 3)(x + 2)$ .
The expression is zero at $x = \frac32$ and $x = -2$ .
These split the number line into three intervals.
A product is negative when its two brackets have opposite signs.
That happens only between the roots: for example, $x = 0$ gives $(-3)(2) = -6 < 0$ .
So the solution is $-2 < x < \frac32$ .

(b)

Multiply the top and bottom by the conjugate of the denominator, $3\sqrt5 - \sqrt2$ .
Bottom: $(3\sqrt5)^2 - (\sqrt2)^2 = 45 - 2 = 43$ .
Top: $(5 - 2\sqrt{10})(3\sqrt5 - \sqrt2) = 15\sqrt5 - 5\sqrt2 - 6\sqrt{50} + 2\sqrt{20}$ .
Simplify the surds: $6\sqrt{50} = 30\sqrt2$ and $2\sqrt{20} = 4\sqrt5$ , so the top is $19\sqrt5 - 35\sqrt2$ .
So the fraction is $-\frac{35}{43}\sqrt2 + \frac{19}{43}\sqrt5$ : $m = -\frac{35}{43}$ and $n = \frac{19}{43}$ .

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Question 2

Indices, logarithms & surds

(a)
Solve the simultaneous equations $\log_2 x - \log_2 y = 2$ , $\log_2 (x - 2y) = 3$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Use the quotient law on the first equation: $\log_2 \dfrac{x}{y} = 2$ .
Change to index form: $\dfrac{x}{y} = 2^2 = 4$ , so $x = 4y$ .
Change the second equation to index form: $x - 2y = 2^3 = 8$ .
Substitute $x = 4y$ : $4y - 2y = 8$ , so $y = 4$ .
Then $x = 4 \times 4 = 16$ .
Check: $\log_2 16 - \log_2 4 = 4 - 2 = 2$ ✓ and $\log_2 (16 - 8) = 3$ ✓.
So $x = 16$ and $y = 4$ .

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Question 3

Polynomials & quadratic roots

(a)
If the quadratic equation $(x + 1)(x + 2) = k(3x + 7)$ has equal roots, find the possible values of the constant $k$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Expand both sides: $x^2 + 3x + 2 = 3kx + 7k$ .
Bring everything to one side: $x^2 + (3 - 3k)x + (2 - 7k) = 0$ .
Equal roots need $b^2 = 4ac$ : $(3 - 3k)^2 = 4(2 - 7k)$ .
Expand: $9 - 18k + 9k^2 = 8 - 28k$ , so $9k^2 + 10k + 1 = 0$ .
Factorise: $(9k + 1)(k + 1) = 0$ .
So $k = -\frac19$ or $k = -1$ .

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Question 4

Trigonometry

(a)
Given that $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ , evaluate $\tan 15^\circ$ , leaving your answer in surd form.

Worked solution (try it first)

(a)

Put $A = 15^\circ$ : $\tan 30^\circ = \dfrac{2\tan 15^\circ}{1 - \tan^2 15^\circ}$ .
Let $t = \tan 15^\circ$ and use $\tan 30^\circ = \dfrac{1}{\sqrt3}$ : $\dfrac{1}{\sqrt3} = \dfrac{2t}{1 - t^2}$ .
Cross-multiply: $1 - t^2 = 2\sqrt3\,t$ , so $t^2 + 2\sqrt3\,t - 1 = 0$ .
Use the formula: $t = \dfrac{-2\sqrt3 \pm \sqrt{12 + 4}}{2}$
$= \dfrac{-2\sqrt3 \pm 4}{2}$
$= -\sqrt3 \pm 2$ .
$15^\circ$ is acute, so $\tan 15^\circ$ is positive.
Take the $+$ sign.
So $\tan 15^\circ = 2 - \sqrt3$ .

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Question 5

Probability & distributions

(a)
The probabilities of Rotey obtaining the highest mark in Mathematics, Physics and Biology tests are $0.9$ , $0.75$ and $0.8$ respectively. Calculate the probability of getting the highest marks in at least two subjects.

Worked solution (try it first)

(a)

The probabilities of not getting the highest mark are $1 - 0.9 = 0.1$ , $1 - 0.75 = 0.25$ and $1 - 0.8 = 0.2$ .
At least two means exactly two or all three.
The tests are independent, so multiply along each case.
Maths and Physics only: $0.9 \times 0.75 \times 0.2 = 0.135$ .
Maths and Biology only: $0.9 \times 0.25 \times 0.8 = 0.18$ .
Physics and Biology only: $0.1 \times 0.75 \times 0.8 = 0.06$ .
All three: $0.9 \times 0.75 \times 0.8 = 0.54$ .
Add the four cases: $0.135 + 0.18 + 0.06 + 0.54 = 0.915$ .

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Question 6

A student representative council consists of 8 girls and 6 boys. If an editorial board of 5 persons is to be formed, what is the probability that the board consists of

(a)
3 girls and 2 boys;
(b)
either all girls or all boys.

Worked solution (try it first)

(a)

The number of ways to choose any 5 of the 14 people is ${}^{14}C_5 = 2002$ .
Choose 3 of the 8 girls and 2 of the 6 boys: ${}^8C_3 \times {}^6C_2 = 56 \times 15$
$= 840$ .
Probability $= \dfrac{840}{2002} = \dfrac{60}{143}$ .

(b)

All girls: ${}^8C_5 = 56$ ways.
All boys: ${}^6C_5 = 6$ ways.
The two cases can't happen together, so add them: $56 + 6 = 62$ ways.
Probability $= \dfrac{62}{2002} = \dfrac{31}{1001}$ .

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Question 7

The coordinates of the vertices of triangle $ABC$ are $A(-2, 1)$ , $B(4, -2)$ and $C(1, 8)$ . If $D(x, y)$ is the foot of the perpendicular from $A$ to $BC$ , find:

(a)
an equation connecting $x$ and $y$ ;
(b)
the unit vector in the direction of $\overrightarrow{BC}$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Gradient of $BC$ : $\dfrac{8 - (-2)}{1 - 4} = -\dfrac{10}{3}$ .
Gradient of $AD$ , from $A(-2, 1)$ to $D(x, y)$ : $\dfrac{y - 1}{x + 2}$ .
$AD$ is perpendicular to $BC$ , so the product of the gradients is $-1$ : $\dfrac{y - 1}{x + 2} \times \left(-\dfrac{10}{3}\right) = -1$ .
Multiply out: $10(y - 1) = 3(x + 2)$ , so $10y - 10 = 3x + 6$ .
The equation is $10y - 3x - 16 = 0$ .

(b)

$\overrightarrow{BC} = (1 - 4)\mathbf i + (8 - (-2))\mathbf j$
$= -3\mathbf i + 10\mathbf j$ .
Its length is $\sqrt{(-3)^2 + 10^2} = \sqrt{109}$ .
Divide by the length: the unit vector is $\dfrac{1}{\sqrt{109}}(-3\mathbf i + 10\mathbf j)$ , about $-0.287\mathbf i + 0.958\mathbf j$ .

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Question 8

The position vector of a body, with respect to the origin, is given by $\mathbf r = 4t\,\mathbf i + (12 - 3t)\,\mathbf j$ at any time $t$ seconds.

(a)
Find the velocity of the body.
Separate values with commas, e.g. 3, −2
(b)
Calculate the magnitude of the displacement between $t = 0$ and $t = 5$ .

Worked solution (try it first)

(a)

Velocity is the rate of change of position: $\mathbf v = \dfrac{d\mathbf r}{dt}$ .
Differentiate each component: $\dfrac{d}{dt}(4t) = 4$ and $\dfrac{d}{dt}(12 - 3t) = -3$ .
So $\mathbf v = 4\mathbf i - 3\mathbf j$ m s $^{-1}$ (a constant velocity).

(b)

At $t = 0$ : $\mathbf r = 0\mathbf i + 12\mathbf j$ .
At $t = 5$ : $\mathbf r = 20\mathbf i + (12 - 15)\mathbf j$
$= 20\mathbf i - 3\mathbf j$ .
Displacement $= (20\mathbf i - 3\mathbf j) - 12\mathbf j$
$= 20\mathbf i - 15\mathbf j$ .
Magnitude $= \sqrt{20^2 + 15^2} = \sqrt{625} = 25$ units.

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Question 9

Sequences, series & binomial expansion

(a)
The 3rd and 6th terms of a geometric progression (G.P.) are 2 and 54 respectively. Find the: (i) common ratio; (ii) first term; (iii) sum of the first ten terms, correct to the nearest whole number.
Separate values with commas, e.g. 3, −2
(b)
The ratio of the coefficient of $x^4$ to that of $x^3$ in the binomial expansion of $(1 + 2x)^n$ is $3 : 1$ . Find the value of $n$ .

Worked solution (try it first)

(a)(i)

The $n$ th term of a G.P. is $ar^{n-1}$ , so $ar^2 = 2$ and $ar^5 = 54$ .
Divide the second equation by the first: $r^3 = 27$ , so $r = 3$ .

(ii)

Put $r = 3$ into $ar^2 = 2$ : $9a = 2$ , so $a = \frac29$ .

(iii)

$S_{10} = \dfrac{a(r^{10} - 1)}{r - 1}$
$= \dfrac{\frac29(3^{10} - 1)}{2}$ .
$3^{10} - 1 = 59\,048$ , so $S_{10} = \dfrac{59\,048}{9} = 6560.9$ , which is $6561$ to the nearest whole number.

(b)

The coefficient of $x^r$ in $(1 + 2x)^n$ is ${}^nC_r\,2^r$ .
Set up the ratio: $\dfrac{{}^nC_4\,2^4}{{}^nC_3\,2^3} = 3$ .
$\dfrac{{}^nC_4}{{}^nC_3} = \dfrac{n - 3}{4}$ , so the ratio is $2 \times \dfrac{n - 3}{4} = \dfrac{n - 3}{2}$ .
Solve $\dfrac{n - 3}{2} = 3$ : $n = 9$ .

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Question 10

(a)
Using the same axes, sketch the curves $y = 6 - x - x^2$ and $y = 3x^2 - 2x + 3$ .
Model answer
Sketch, don't plot a table. $y = 6 - x - x^2$ opens downwards and cuts the $x$ -axis at $(-3, 0)$ and $(2, 0)$ and the $y$ -axis at $(0, 6)$ ; its maximum is at $(-\frac12, 6\frac14)$ . $y = 3x^2 - 2x + 3$ opens upwards, cuts the $y$ -axis at $(0, 3)$ , has its minimum at $(\frac13, 2\frac23)$ and never meets the $x$ -axis. The curves cross at $x = -\frac34$ and $x = 1$ .
(b)
Find the $x$ -coordinates of the points of intersection of the two curves in (a).
Separate values with commas, e.g. 3, −2
(c)
Calculate the area of the finite region bounded by the two curves in (a).

Worked solution (try it first)

(a)

For $y = 6 - x - x^2 = -(x + 3)(x - 2)$ : it opens downwards, cuts the $x$ -axis at $-3$ and $2$ , and the $y$ -axis at $6$ .
Its vertex is at $x = -\frac12$ , $y = 6\frac14$ .
For $y = 3x^2 - 2x + 3$ : it opens upwards and cuts the $y$ -axis at $3$ .
Its discriminant is $4 - 36 < 0$ , so it doesn't meet the $x$ -axis.
Its vertex is at $x = \frac13$ , $y = 2\frac23$ .

(b)

At the intersections the $y$ -values are equal: $6 - x - x^2 = 3x^2 - 2x + 3$ .
Collect terms: $4x^2 - x - 3 = 0$ , which factorises as $(4x + 3)(x - 1) = 0$ .
So $x = -\frac34$ or $x = 1$ .

(c)

Between these points the first curve is on top, so the area is $\displaystyle\int_{-3/4}^{1} [(6 - x - x^2) - (3x^2 - 2x + 3)]\,dx = \int_{-3/4}^{1} (3 + x - 4x^2)\,dx$ .
Integrate: $\left[3x + \frac{x^2}{2} - \frac{4x^3}{3}\right]_{-3/4}^{1}$ .
At $x = 1$ : $3 + \frac12 - \frac43 = \frac{13}{6}$ .
At $x = -\frac34$ : $-\frac94 + \frac{9}{32} + \frac{9}{16} = -\frac{45}{32}$ .
Subtract: $\frac{13}{6} + \frac{45}{32} = \frac{208 + 135}{96}$
$= \frac{343}{96}$ .
The area is $3\frac{55}{96}$ square units, about $3.57$ .

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Question 11

(a)
Evaluate $\displaystyle\int_1^2 \frac{x}{\sqrt{5 - x^2}}\,dx$ .
(b)
(i) Evaluate $\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix}$ . (ii) Using your answer in (b)(i), solve the simultaneous equations $2x - 3y + z = 10$ , $y - 2z = -7$ , $x + 2y - 3z = -9$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Substitute $t^2 = 5 - x^2$ .
Then $2t\,dt = -2x\,dx$ , so $x\,dx = -t\,dt$ .
Change the limits: $x = 1$ gives $t = 2$ , and $x = 2$ gives $t = 1$ .
The integral becomes $\displaystyle\int_2^1 \frac{-t}{t}\,dt = \int_2^1 (-1)\,dt$ .
Evaluate: $[-t]_2^1 = -1 - (-2) = 1$ .

(b)(i)

Expand along the top row: $2(-3 + 4) + 3(0 + 2) + 1(0 - 1)$ , which is $2 + 6 - 1 = 7$ .

(ii)

By Cramer's rule, replace each column in turn by $(10, -7, -9)$ .
$\Delta_x = \begin{vmatrix} 10 & -3 & 1 \\ -7 & 1 & -2 \\ -9 & 2 & -3 \end{vmatrix}$
$= 14$ , $\Delta_y = -7$ and $\Delta_z = 21$ .
So $x = \frac{14}{7} = 2$ , $y = \frac{-7}{7} = -1$ and $z = \frac{21}{7} = 3$ .
Check in the third equation: $2 - 2 - 9 = -9$ ✓.

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Question 12

(a)
Use the trapezium rule with five ordinates to evaluate $\displaystyle\int_0^1 \frac{x}{1 + x^2}\,dx$ , correct to four significant figures.
(b)
If $A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$ , find the image of the point $(1, 2)$ under the linear transformation $A^2 + A + 2I$ , where $I$ is the $2 \times 2$ unit matrix.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Five ordinates means four strips, so $h = \frac{1 - 0}{4} = 0.25$ .
Ordinates of $y = \dfrac{x}{1 + x^2}$ at $x = 0, 0.25, 0.5, 0.75, 1$ : $0$ , $0.23529$ , $0.4$ , $0.48$ , $0.5$ .
Trapezium rule: $\frac{h}{2}[y_0 + y_4 + 2(y_1 + y_2 + y_3)] = 0.125[0 + 0.5 + 2(1.11529)]$ .
$= 0.125 \times 2.73059 = 0.34132$ , which is $0.3413$ to 4 significant figures.

(b)

$A^2 = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$
$= \begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix}$ .
Add $A$ and $2I$ : $A^2 + A + 2I = \begin{pmatrix} 7 + 2 + 2 & 4 + 1 \\ 12 + 3 & 7 + 2 + 2 \end{pmatrix}$
$= \begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix}$ .
Multiply by $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ : $\begin{pmatrix} 11 + 10 \\ 15 + 22 \end{pmatrix} = \begin{pmatrix} 21 \\ 37 \end{pmatrix}$ .
The image is $(21, 37)$ .

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Question 13

(a)
Simplify ${}^{n+1}C_3 - {}^{n-1}C_3$ .
(b)
A fair die is thrown five times. Calculate, correct to three decimal places, the probability of obtaining (i) at most two sixes; (ii) exactly three sixes.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Use ${}^nC_3 = \dfrac{n(n - 1)(n - 2)}{6}$ : ${}^{n+1}C_3 = \dfrac{(n + 1)n(n - 1)}{6}$ and ${}^{n-1}C_3 = \dfrac{(n - 1)(n - 2)(n - 3)}{6}$ .
Take out the common factor $\dfrac{n - 1}{6}$ : $\dfrac{n - 1}{6}[(n + 1)n - (n - 2)(n - 3)]$ .
Inside the bracket: $n^2 + n - (n^2 - 5n + 6) = 6n - 6$
$= 6(n - 1)$ .
So ${}^{n+1}C_3 - {}^{n-1}C_3 = (n - 1)^2$ .

(b)

Each throw is a six with $p = \frac16$ and not a six with $q = \frac56$ .
The number of sixes in 5 throws is binomial.

(i)

$P(0) = (\frac56)^5 = \frac{3125}{7776}$ , $P(1) = 5(\frac16)(\frac56)^4$
$= \frac{3125}{7776}$ , $P(2) = 10(\frac16)^2(\frac56)^3$
$= \frac{1250}{7776}$ .
Add: $P(\le 2) = \frac{7500}{7776} = 0.965$ (3 d.p.).

(ii)

$P(3) = 10(\frac16)^3(\frac56)^2$
$= \frac{250}{7776}$
$= 0.032$ (3 d.p.).

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Question 14

Statistics & correlation

The distribution of the lives (in days) of 40 transistor batteries is shown in the table.

Battery life (days)	26–30	31–35	36–40	41–45	46–50	51–55
Frequency	4	7	13	8	6	2

(a)
Draw a histogram for the distribution.
Model answer
Draw the bars on the class boundaries 25.5, 30.5, 35.5, 40.5, 45.5, 50.5, 55.5, with no gaps, and heights equal to the frequencies 4, 7, 13, 8, 6, 2. Label both axes.

For the mode, take the tallest bar (35.5–40.5). Join its top-left corner to the top-left corner of the bar on its right, and its top-right corner to the top-right corner of the bar on its left. Read down from the crossing: about 38.2 days.
(b)
Use your graph in (a) to determine the mode for the distribution.
(c)
Using an assumed mean of 43 days, calculate the mean of the distribution.
(d)
What percentage of the batteries will live for less than 31 days or more than 45 days?

Worked solution (try it first)

(a)

The classes all have width 5, so the bar heights are the frequencies $4, 7, 13, 8, 6, 2$ over the boundaries $25.5, 30.5, \ldots, 55.5$ .

(b)

On the tallest bar, join each top corner to the top corner of the opposite neighbour.
The lines cross above the mode.
By calculation: $35.5 + \dfrac{13 - 7}{(13 - 7) + (13 - 8)} \times 5 = 35.5 + \dfrac{6}{11} \times 5$
$= 38.2$ days.

(c)

Class mid-values are $28, 33, 38, 43, 48, 53$ , so the deviations $d = x - 43$ are $-15, -10, -5, 0, 5, 10$ .
Multiply by the frequencies: $fd = -60, -70, -65, 0, 30, 20$ , and $\sum fd = -145$ .
Mean $= 43 + \dfrac{\sum fd}{\sum f}$
$= 43 + \dfrac{-145}{40}$
$= 43 - 3.625$
$= 39.375$ days.

(d)

Less than 31 days is the class 26–30: 4 batteries.
More than 45 days is 46–50 and 51–55: $6 + 2 = 8$ batteries.
That is $4 + 8 = 12$ of the 40 batteries: $\dfrac{12}{40} \times 100 = 30\%$ .

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Question 15

Statistics & correlation

The table shows the corresponding values of two variables $x$ and $y$ .

$x$	33	31	28	25	23	22	19	17	16	14
$y$	4	6	4	10	12	10	14	15	18	22

(a)
Plot a scatter diagram to represent the data.
Model answer
Plot the ten points $(33, 4), (31, 6), (28, 4), (25, 10), (23, 12), (22, 10), (19, 14), (17, 15), (16, 18), (14, 22)$ with $x$ across and $y$ up. Don't join them. The points fall from left to right.
(b)
Calculate $\bar x$ , the mean of $x$ , and $\bar y$ , the mean of $y$ .
Separate values with commas, e.g. 3, −2
(c)
Draw the line of best fit to pass through $(\bar x, \bar y)$ .
Model answer
Plot the mean point $(22.8, 11.5)$ and draw a straight line through it that follows the trend, with about as many points above it as below. A good line passes near $(14, 19.2)$ and $(33, 2.6)$ .
(d)
From your graph in (c), determine the: (i) relationship between $x$ and $y$ ; (ii) value of $y$ when $x$ is 24.

Worked solution (try it first)

(a)

Plot each $(x, y)$ pair as a point.
The pattern is a downward trend.

(b)

$\sum x = 228$ , so $\bar x = \dfrac{228}{10} = 22.8$ .
$\sum y = 115$ , so $\bar y = \dfrac{115}{10} = 11.5$ .

(c)

Mark $(22.8, 11.5)$ and draw a straight line through it with the points spread evenly on either side.

(d)(i)

Read two points on your line, for example $(14, 19.2)$ and $(33, 2.6)$ .
The gradient is $\dfrac{2.6 - 19.2}{33 - 14} \approx -0.88$ .
Using $(22.8, 11.5)$ : $c = 11.5 + 0.88 \times 22.8 \approx 31.5$ .
So $y \approx -0.88x + 31.5$ : as $x$ increases, $y$ decreases.

(ii)

Read up from $x = 24$ to the line and across: $y \approx 10.5$ .

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Question 16

Vectors

(a)
The position vectors of points $L$ and $M$ are $(5\mathbf i + 6\mathbf j)$ and $(13\mathbf i + 4\mathbf j)$ respectively. If point $K$ lies on $LM$ such that $LK : KM = 2 : 3$ , find the position vector of $K$ .
Separate values with commas, e.g. 3, −2
(b)
Three poles are situated at points $A$ , $B$ and $C$ on the same horizontal plane such that $\overrightarrow{AB} = (8\text{ km}, 060^\circ)$ and $\overrightarrow{BC} = (12\text{ km}, 130^\circ)$ . Calculate: (i) $|AC|$ , correct to three decimal places; (ii) the bearing of $C$ from $A$ , correct to the nearest degree.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

By the ratio theorem, $\mathbf k = \dfrac{3\mathbf l + 2\mathbf m}{2 + 3}$ (each end is weighted by the part of the ratio next to the other end).
$3(5\mathbf i + 6\mathbf j) + 2(13\mathbf i + 4\mathbf j) = 41\mathbf i + 26\mathbf j$ .
Divide by 5: $\mathbf k = 8.2\mathbf i + 5.2\mathbf j$ .

(b)(i)

At $B$ , the back bearing to $A$ is $060^\circ + 180^\circ = 240^\circ$ .
The angle between $BA$ and $BC$ is $240^\circ - 130^\circ = 110^\circ$ .
Cosine rule: $|AC|^2 = 8^2 + 12^2 - 2(8)(12)\cos 110^\circ$
$= 208 + 65.666$
$= 273.666$ .
So $|AC| = 16.543$ km (3 d.p.).

(ii)

Sine rule for $\theta = \angle BAC$ : $\dfrac{\sin\theta}{12} = \dfrac{\sin 110^\circ}{16.543}$ , so $\sin\theta = 0.6816$ and $\theta = 42.97^\circ$ .
The bearing of $C$ from $A$ is $060^\circ + 42.97^\circ = 102.97^\circ$ , which is $103^\circ$ to the nearest degree.

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Question 17 ✱

Statics: forces, equilibrium & moments

Coplanar forces of 4 N, 8 N, 6 N, 4 N and 5 N act at a point as shown in the diagram (not drawn to scale). If the 6 N force acts in the direction $090^\circ$ , calculate the:

(a)
magnitude of the resultant force;
(b)
direction of the resultant force.

Worked solution (try it first)

(a)

Take east (the 6 N direction) as the $x$ -axis and north as the $y$ -axis.
Measured anticlockwise from east, the forces act at: 8 N at $55^\circ$ , 6 N at $0^\circ$ , 4 N at $-25^\circ$ , 4 N at $155^\circ$ ( $100^\circ$ beyond the 8 N) and 5 N at $205^\circ$ ( $130^\circ$ beyond the lower 4 N).
East components: $8\cos 55^\circ + 6 + 4\cos 25^\circ - 4\cos 25^\circ - 5\cos 25^\circ = 4.5886 + 6 - 4.5315$
$= 6.0571$ .
North components: $8\sin 55^\circ + 0 - 4\sin 25^\circ + 4\sin 25^\circ - 5\sin 25^\circ = 6.5532 - 2.1131$
$= 4.4401$ .
Magnitude: $|\mathbf F| = \sqrt{6.0571^2 + 4.4401^2}$
$= \sqrt{56.403}$
$= 7.51$ N (2 d.p.).

(b)

The resultant makes an angle $\tan^{-1}\dfrac{4.4401}{6.0571} = 36.2^\circ$ with east, towards north.
As a bearing: $090^\circ - 36.2^\circ = 053.8^\circ$ , which is $054^\circ$ to the nearest degree.

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Question 18 ✱

(a)
The position vectors of points $A$ , $B$ and $C$ are $\mathbf i + 5\mathbf j$ , $3\mathbf i + 9\mathbf j$ and $-\mathbf i + \mathbf j$ respectively. (i) Show that the points $A$ , $B$ and $C$ are collinear. (ii) Determine the ratio $|AB| : |BC|$ .
Show the answer
(ii) $|AB| : |BC| = 1 : 2$
(b)
A uniform beam $XY$ of mass 10 kg and length 24 m is hung horizontally from a cross bar by two vertical inextensible strings, one attached to $X$ and the other at a point $M$ , 4 m away from $Y$ . A mass of 50 kg is suspended at a point $N$ which is 8 m from $X$ . If the system remains in equilibrium, calculate the tensions in the strings. [Take $g = 10\text{ m s}^{-2}$ ]
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

$\overrightarrow{AB} = (3 - 1)\mathbf i + (9 - 5)\mathbf j$
$= 2\mathbf i + 4\mathbf j$ .
$\overrightarrow{AC} = (-1 - 1)\mathbf i + (1 - 5)\mathbf j$
$= -2\mathbf i - 4\mathbf j$
$= -\overrightarrow{AB}$ .
$\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel and share the point $A$ , so $A$ , $B$ and $C$ are collinear.

(ii)

$|AB| = \sqrt{2^2 + 4^2}$
$= \sqrt{20}$
$= 2\sqrt5$ .
$\overrightarrow{BC} = -4\mathbf i - 8\mathbf j$ , so $|BC| = \sqrt{16 + 64}$
$= \sqrt{80}$
$= 4\sqrt5$ .
So $|AB| : |BC| = 2\sqrt5 : 4\sqrt5 = 1 : 2$ .

(b)

Weights: beam $10 \times 10 = 100$ N at its midpoint, 12 m from $X$ .
Load $50 \times 10 = 500$ N at $N$ , 8 m from $X$ .
The string at $M$ is $24 - 4 = 20$ m from $X$ .
Take moments about $X$ (this removes $T_1$ ): $T_2 \times 20 = 500 \times 8 + 100 \times 12$
$= 5200$ .
So $T_2 = 260$ N.
Resolve vertically: $T_1 + T_2 = 500 + 100$ , so $T_1 = 600 - 260 = 340$ N.
The tension at $X$ is 340 N and the tension at $M$ is 260 N.

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