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ยฉ John Wiley & Sons Australia, Ltd 2009 1
WorkSHEET 8.2 Vector applications Name: _________________________ 1 Bob is located at position 3๐ โ 2๐ and Annie is
located at position โ๐ + ๐. How far apart are they?
Set; ๐ = 3๐ โ 2๐
and ๐ = โ๐ + ๐
Use; |๐| = ,๐ฅ. + ๐ฆ.
Find the vector that joins Annie to Bob โฆ this
Resultant Vector is represented as a Vector sum;
๐ = ๐ โ ๐
So, ๐ด๐ต3333 = (โ๐ + ๐) โ (3๐ โ 2๐)
= โ๐ + ๐ โ 3๐ + 2๐
โด ๐ด๐ต3333 = โ4๐ + 3๐
Now, by sub;
|๐ด๐ต| = ,โ4. + 3.
= โ25
โด |๐ด๐ต| = 5
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Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 2
2 Abby pushes Gertrude with force of 2๐ + 3๐. At the same time, Billie also pushes Gertrude with a force of 5๐ โ 2๐. If Gertrudeโs friend is sick of other students pushing Gertrude around, what force with the friend need to exert so that Gertrude didnโt move?
Set; ๐ญ๐ = โ3๐ + ๐
and ๐ญ๐ = โ3๐ โ 3๐
Because of the state of equilibrium, we can say;
๐น= + ๐น> + ๐ = 0 Where, R is the force needed to keep Gertrude
stationary By sub,
โ3๐ + ๐ + โ3๐ โ 3๐ + ๐ = 0
๐ = โ2โ3๐ + 2๐ Use;
|๐| = ,๐ฅ. + ๐ฆ.
|๐ | = @โ2โ3.+ 2. = 4
๐ = tanEF
๐ฆ๐ฅ
Here,
๐ = tanEF2
โ2โ3
= tanEFโ1โ3
= 150
The friend pushes with a force of 4N at an
angle of 150 degrees to the x axis. ** could you tell this was a question I made?
Small numbers, came out exactly! No need for calculator. Multiple things happening in the one question.
It should have ended with a conversion to a bearing too L
I wish I knew why these words indent like this when it goes to a new line? Itsโ annoying!
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Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 3
3 A body of mass 25 kg rests on a horizontal table. The coefficient of friction between the table and the object is 0.4. Determine the magnitude of the least horizontal force that must be applied to the object in order for it to move.
Draw a Diagram! Use,
๐ญ = ๐๐ and ๐ญJ = ๐๐ต Weight of the object becomes;
๐ = 25๐ Because of the state of Equilibrium we can say;
๐ต = 25๐ Now, by sub;
๐นJ = 0.4 ร 25๐
โด ๐นJ = 10๐๐๐๐ค๐ก๐๐๐
3
4 A 20 kg object is on a smooth surface inclined at 30ยฐ to the horizontal. It is stopped from sliding by a rope pulling parallel to the surface. Determine the Tension on the rope.
Draw a diagram! Use,
๐ญ = ๐๐ Clearly,
๐ = 20๐ Because of the state of Equilibrium we can say the tension on the rope is equal to the component of the weight vector, acting parallel to the slope. As such;
๐ = 20๐ sin 30
= 10๐ So, the Tension on the rope is 10g Newtons
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Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 4
5 A student brought their new cute puppy (weighing 4 kg) in for show and tell. The maths teacher put the puppy on a table and started to lift one side. It wasnโt until the table reached an angle of 60 degrees that the puppy was on the โvergeโ of sliding off. Determine the coefficient of friction between the puppy and the table.
Draw a Diagram! Use,
๐ญ = ๐๐ Weight of the object becomes โฆ ๐ = 4๐ Break up this weight vector into components parallel and perpendicular to the face of the table;
๐ต = 4๐ cos 60 = 2๐ and
๐ญโฅ = 4๐ sin 60 = 2๐โ3 Because of the state of equilibrium, the force of Friction must equal the force parallel to the table; hence
๐ญJ = ๐ญโฅ Using,
๐ญJ = ๐๐ต By sub;
๐ ร 2๐ = 2๐โ3
โด ๐ = โ3 ** Did you spot this as one of my questions? Do you see those justification statements in the working? Nice hey!
2
Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 5
6 The velocities of two planes are .
Calculate: (a) the velocity of plane 1 relative to plane 2 (b) the velocity of plane 2 relative to plane 1.
Iโll leave the text book solutions here, but I think they are โUglyโ J ?
Here are My solutions: (a) Think โฆ we need:
๐`F/`. = ๐`F/b + ๐b/`. Velocity of Plane 1 over the ground
๐`F/b = โจ200, โ40โฉ Velocity of the ground rel to Plane 2
๐b/`. = โจโ230, โ60โฉ Now;
๐`F/`. = โจ200, โ40โฉ + โจโ230, โ60โฉ and
๐`F/`. = โจโ30, โ100โฉ So, the same as the text book solution, but using the standard relative velocity formula! (b)
๐`./`F = ๐`./b + ๐b/`F
๐`./b = โจ230, 60โฉ
๐b/`F = โจโ200, 40โฉ Now;
๐`./`F = โจ230, 60โฉ + โจโ200, 40โฉ and
๐`./`F = โจ30, 100โฉ
2 60 230 and 40 200
~~2~~~1~ jivjiv +=-=
~~
2 rel 11 rel 2
~~
~~~~
212 rel 1
10030
(b)
10030
)60230()40200(
(a)
ji
vv
ji
jiji
vvv
~~
~~~
+=
-=
--=
+--=
-=
Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 6
7 A ferry is heading north-east at 20 km/h. At the same time a nearby launch is travelling south at 15 km/h. Calculate the velocity of the ferry as seen by someone on the launch.
Using the cosine rule:
Using the sine rule:
48.3ยฐ - 45ยฐ = 3.3ยฐ
*** Did you get that question correct? Do you agree with this solution? If you did or if you didnโt, then read the next box!
4
I know you hate when I talk about the Rubric, but lets check the โReasonableness of that solutionโ. The Ferry is heading North East at 20 km/h, so its โnortherlyโ component of speed is 20 cos 45 = .f
โ.=
14.14 km/h. If the launch is travelling South at 15 km/h and the Ferry has a northerly speed of 14.14 km/h โฆhow on earth is the relative velocity between the 2 just 14.2 km/h. Its not reasonable at all! I reckon their messy use of the cosine rule has muddled them up. The Cosine Rule is for Maths B. Donโt even think about using the Sine or Cosine rule in your Maths C Exams! Iโll do the question again, using the standard relative velocity formula and see how I go โฆ J
LFL rel F ~~~ vvv -=
km/h 2.14
45cos152021520 22
L rel F
=
ยดยด-+= !
~v
!
!
!
!
3.48
2.1445sin15sin
2.1445sin15sin
2.1445sin
15sin
1
=
รทรทรธ
รถรงรงรจ
รฆ ยด=
ยด=
=
-q
q
q
E of S 3.3at km/h 2.14 ThereforeL rel F
!=~v
Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 7
8 A ferry is heading north-east at 20 km/h. At the same time a nearby launch is travelling south at 15 km/h. Calculate the velocity of the ferry as seen by someone on the launch.
๐J/g = ๐J/h + ๐h/g
๐J/h = โจ20 cos 45, 20 sin 45โฉ
๐h/g = โจ0, 15โฉ
๐J/g = โจ20 cos 45, 20 sin 45โฉ + โจ0, 15โฉ
๐J/g = โจ14.14, 29.14โฉ
|๐| = โ14.14. + 29.14. = 32.39 km/h
๐ = tanEF32.3914.14 = 66.42j
convert to a bearing, thatโs 23.58j๐.
Solution done in half the work, and it is actually correct โฆ J Again check it is reasonable โฆ if you can visualise yourself on the launch heading south at 15km/h, surely the ferry heading north east is heading away from you pretty quickly, and at a more acute angle that its own heading of 045j๐.
9 A boat can travel at 16 km/h in still water. The driver of the boat wants to cross a river at right angles to the bank. If the river is flowing at 4 km/h, at what angle should the driver head in order to travel in the desired direction?
The driver should head upstream at an angle of about 76ยฐ to the shore. *** Iโm out of time โฆ surely this is not what you would give me in the exam โฆ ? Make sure you know how to set this one out Properly โฆ if you are not sure, come see me!
2
!14164sin
ยป
=
q
q
Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 8
10 Given D is the midpoint of AC and E is the midpoint of CB, Prove that DE and AB are parallel.
Clearly, ๐ด๐ต3333 = ๐ด๐ท3333 + ๐ท๐ถ3333 + ๐ถ๐ธ3333 + ๐ธ๐ต3333๐๐. 1
As D bisects AC โโด ๐ด๐ท3333 = ๐ท๐ถ3333 As E bisects CB โโด ๐ถ๐ธ3333 = ๐ธ๐ต3333 Sub into eq.1
๐ด๐ต3333 = ๐ท๐ถ3333 + ๐ท๐ถ3333 + ๐ถ๐ธ3333 + ๐ถ๐ธ3333
๐ด๐ต3333 = 2๐ท๐ถ3333 + 2๐ถ๐ธ3333
๐ด๐ต3333 = 2(๐ท๐ถ3333 +๐ถ๐ธ3333)๐๐. 2 Now further, clearly;
๐ท๐ธ3333 = ๐ท๐ถ3333 +๐ถ๐ธ3333๐๐. 3 By sub into eq.2,
๐ด๐ต3333 = 2๐ท๐ธ3333 Because BC and DE are scalars;
๐ด๐ต3333 โฅ ๐ท๐ธ3333
๐๐ธ๐ท
11 Point D lies on the line segment AB and divides it in the ratio 2 : 1. Find an expression for the position vector of the point D, in terms of the position vectors of the points A and B.
3
~~
~~~
~~
~~
32
31
)(32
AB32
AB
ba
aba
ad
ab
+=
-+=
+=
-=
ยฎ
ยฎ
Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 9
12 On the triangle ABC, point D lies on AB and divides it in the ratio 2 : 1. Point E lies on the side AC and divides it also in the ratio 2 : 1. Show that ED is parallel to BC.
From the previous question:
4
BC. toparallel is ED
BC32
)(32
31
32
31
32
DE
31
32
31
32
~~
~~~~
~~
~~~
~~~
\
=
-=
รทรธรถ
รงรจรฆ +-รท
รธรถ
รงรจรฆ +=
-=\
+=
+=
ยฎ
ยฎ
bc
abac
de
ace
abd
Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 10
13 Use vector methods to prove that the diagonals of a parallelogram bisect each other.
Set: ๐ด๐ถ3333 and ๐ต๐ท3333 are straight diagonals of parallelogram ABCD,
Hence,
๐ด๐ธ3333 โฅ ๐ธ๐ถ3333 Therefore they are scalars of each other,
๐ด๐ธ3333 = ๐๐ธ๐ถ3333 Further,
๐ต๐ธ3333 โฅ ๐ธ๐ท3333 Therefore they are scalare of each other,
๐ต๐ธ3333 = ๐๐ธ๐ท3333 Clearly,
๐ด๐ถ3333 = ๐ด๐ธ3333 + ๐ธ๐ถ3333 Now,
๐ด๐ท3333 = ๐ด๐ธ3333 + ๐ธ๐ท3333
๐ต๐ถ3333 = ๐ต๐ธ3333 + ๐ธ๐ถ3333 As,
๐ด๐ท3333 = ๐ต๐ถ3333 We can say through substitution,
๐ด๐ธ3333 + ๐ธ๐ท3333 = ๐ต๐ธ3333 + ๐ธ๐ถ3333 Rearrange,
๐ด๐ธ3333 โ ๐ธ๐ถ3333 + ๐ธ๐ท3333 โ ๐ต๐ธ3333 = 0 Via substitution (scalars)
๐๐ธ๐ถ3333 โ ๐ธ๐ถ3333 + ๐ธ๐ท3333 โ ๐๐ธ๐ท3333 = 0
Factorise
๐ธ๐ถ3333(๐ โ 1) + ๐ธ๐ท3333(1 โ ๐) = 0
Here, either BOTH ๐ธ๐ถ3333๐๐๐๐ธ๐ท3333 are Null vectors (which is a trivial solution), or BOTH
(๐ โ 1)๐๐๐(1 โ ๐) = 0
To make this happen, ๐ = 1๐๐๐๐ = 1 Therefore,
๐ด๐ธ3333 = 1๐ธ๐ถ3333 and
๐ต๐ธ3333 = 1๐ธ๐ท3333 Hence, point E is the midpoint of both diagonals.
๐๐ธ๐ท
Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2
ยฉ John Wiley & Sons Australia, Ltd 2009 11
14 Vector Proofs can be hard an there can be different methods to prove: Use vector methods to prove that the diagonals of a parallelogram bisect each other.
Setting ๐ as the intersection of the diagnoals;
We are not specifying that ๐ is the midpoint, it is simply a certain โscalarโ (use ๐&๐ as the scalars) along vector ๐w and ๐y. Hence;
๐ = ๐y + ๐๐w and
๐ = ๐w + ๐y โ ๐๐y Clearly,
๐y + ๐๐w = ๐w + ๐y โ ๐๐y and
๐๐w = ๐w โ ๐๐y
Now as,
๐w = ๐y โ๐w and
๐y = ๐y +๐w Through substitution,
๐{๐y โ๐w | = ๐w โ ๐{๐y +๐w |
๐๐y โ ๐๐w = ๐w โ ๐๐y โ ๐๐w
๐w = ๐๐y โ ๐๐w + ๐๐y + ๐๐w
๐w = (๐ + ๐)๐y + (๐ โ ๐)๐w
Clearly vector ๐y and vector ๐w are in different directions, hence the coefficient of ๐y is 0; and as๐w = ๐w , the coefficient of ๐w is 1
๐ + ๐ = 0 โ ๐๐. 1
๐ โ ๐ = 1 โ ๐๐. 2 Hence, solving simultaneously, we have,
๐ = F. and ๐ = โ F
.
Hence ๐ is the midpoint of both diagonals QED !
Which solution did you like better?