University of Malta

Junior College

Subject: Advanced Applied Mathematics Date: June 2014 Time: 9.00 - 12.00

End of Year Test

Worked Solutions

Question 1 (i) 3N: 3iN

2N: ji 60sin260cos2 + = ( )ji 3+ N 4N: 4cos60 i− 4sin60 j

= 2i− 2 3j( )N

Resultant ( )jiF 323)213( −+++=∑

= ( )ji 36 − N

(ii) At equilibrium: ( )∑ =+ 0RF

( )jiR 36 +−=∴ N Also, taking moments about the origin for system: Moments about origin = 0

Ca =)(60sin2

3aC = (Clockwise)

       Question 2

O   A  

B  

C    D  

   E  

x

y

4N  

3N  

2N  

System  1  a  

O   A  

B  

C    D  

   E  

x

y

4N  

3N  

2N  

System  2  a  

R  

C  

O   ⎟⎠

⎞⎜⎝

⎛ 0,83r

( )0,2r ⎟⎠

⎞⎜⎝

⎛ 0,835r

θ x

4r r

r

(i) The volume of the cylinder = 𝜋𝑟2 4𝑟 = 4𝜋𝑟! The volume of the hemisphere = !

!𝜋𝑟!

Let M be the density of the solid.

Mass of removed hemisphere: Mr332π

Mass of cylinder: MrMrr 32 4)4( ππ = By symmetry, the centre of mass lies along the common axis of the 3 sections forming the solid. The centre of mass of the cylinder = 2r The centre of mass of a hemisphere = !!

! from the centre

By taking moments, we get: 4𝜋𝑟!𝑀𝑥 = 4𝜋𝑟!𝑀 2𝑟 + !

!𝜋𝑟!𝑀 !"!

!− !

!𝜋𝑟!𝑀 !!

!

Mr

rMrrMrrMrx 3

333

483

32

835

32)2(4

π

πππ ⎟⎠

⎞⎜⎝

⎛−⎟⎠

⎞⎜⎝

⎛+=

Dividing top and bottom by Mr3π :

38

4412

358 rrrr

x =−+

=

(ii) 83

38tan == rr

θ

6.20≈θ

 

     Question 3

r

8r/3 θ

800Kg  1200Kg  T T P/v

120N 80N

(i) For maximum acceleration T = 2000N.

maF =

a800802000 =−

4.28001920 =⇒= aa ms –2

(ii) 10=v kmh–1=925 ms–1

At maximum power, T = 2000N

maF = , where F (tractive force) =

)4.2(120020001209/25

=−−P

20001202880259

++=P

89.1388895000255000

259

=×

=⇒= PP W 89.13≈ kW

Question 4

(i) Particle A: Horizontally: v = 14ms– 1, x = (14t)m Vertically: s = y u = 0 v = ? a = –10ms–2 t = t

atuv += tv 100 −=

2

21 atuts +=

250 ts −=

800Kg  

T 80N

a

1200Kg   P/v

120N

T

O(0, 0)

r

v

h 14ms–1

Pv

( )jiv t1014 −=∴ ms–1 ( )jir )5(14 2tht −+= m Particle B: Horizontally: v = 17.5ms–1, x = (17.5t)m Vertically: s = y u = 0 v = ? a = –10ms–2 t = t

atuv += tv 100 −=

2

21 atuts +=

250 ts −=

( )jiv t105.17 −=∴ ms–1 ( )jir )5(5.17 2tht −+= m

(ii) Since the j component of the position vector is the same for both particles, then they are at the same height throughout the flight, in particular when they hit the ground.

(iii) If they land 10m apart then the difference in the i component of r is 10m. i.e. 10145.17 =− tt

857.25.310

==t s

So, time of flight = 2.857s. Height of tower: When particle hit the ground the j component of r = 0 i.e. 05 2 =− th

h = 5t2 ⇒ h = 5(2.857)2 = 40.82 m    

O(0,0)

r

v

h 17.5ms–1

   Question 5 (i) Horizontally:

GW FR =

Vertically:

WR =G Taking moments about A:

)2(sin)(cos W aRaW θθ =

454.

5 WWWRRW

=⇒=

4GWF =∴

414

G

GGG ==≥⇒≤

WW

RFRF µµ

(ii) Limiting equilibrium:

Vertically:

WR 2G = Horizontally:

85)2(

165

GWWWFR ===

Taking moments about A: W cosθ(a)+Wxcosθ = RW sinθ(2a)

aWxWaW 252

85)(

5)(

5⎟⎠

⎞⎜⎝

⎛=+

axa25

=+

23

25 aaax =−=

a

a

W

RW

RG

FG A

B

θ

2

1

√5

2tan =θ

51cos =θ

52sin =θ

a

a

W

RW

RG

A

B

W

x

G165 R

   Question 6

(i) Vertical equilibrium:

mgTT += θθ coscos BA

( ) mgTTmgTT45

54

BABA =−⇒=− …(1)

Horizontally:

2ωmrmaF == TA sinθ +TB sinθ =mrω

2

TA +TB( ) 35!"#$%&=m(2.5asinθ )ω 2 =m 2.5× 35

!"#

$%&aω 2

2

BA 25

ωmaTT =+ …(2)

(1) + (2):

mgmaT45

252 2

A += ω

TA =1252maω

2 + 54mg!"#

$%&

TA =54maω

2 + 58mg =5m8 2aω 2 + g( )

B

2.5a

2.5a

TA

TB

2a

2a

rmv2

A

θ

θ

3

4 5

54cos =θ

53sin =θ

P

since from diagram r = 2.5asinθ

(2) − (1):

mgmaT45

252 2

B −= ω

TB =54maω

2 − 58mg =5m8 2aω 2 − g( )

(ii) For minimum ω, TB = 0.

agga2

2 2min

2min =⇒= ωω

∴ωmin

2 > g2a

   Question 7 Those forces marked with an asterisk are equal by symmetry.

(i) For the whole system, resolving horizontally: XA = 0 (one horizontal force) Vertically:

BA RWY =+

Since the rods are identical, then they are all equal and each triangle is equilateral. Hence the rod BE = 3a, and it makes an angle of 600 with the horizontal.

With respect to the above diagram BK = 3a cos 600 = 3a2

A B

C

D

E F

G

H

CAB

CBC

CCD

CDE CFD

*

CGC

*

CHB

*

TEF*

TGD

*

THC

*

TFG*

TGH

THA

YA

XA

RB

Load, W

a

B  

E  

600  K  

3a  

Taking moments about B:

23

23)( AA

WYaWaY =⇒=

The reaction at A is simply a vertical downward force of magnitude 3W/2. Also, this means that actually there is no need for a hinge. A simple support or peg from above will be enough.

(ii) Resolving vertically at A:

MN6MN33)32(23

2360sin AHAAH =⇒==== TWYT

Horizontally:

CAB = TAH cos60 = TAH

12!"#$%&= 3MN

Resolving vertically at E: CDE sin60

= 2 3⇒CDE = 4MN

Horizontally: TEF =CDE cos60

= 2MN

But, TEF = TFG = CDF = TGD = CGC = THC = CHB = 2MN Resolving vertically at D: CCD sin60 =CDF sin60 +CDE sin60⇒CCD = 2+ 4 = 6MN by using previous results Resolving vertically at G: TGH sin60 = TFG sin60 +CCG sin60⇒ TGH = 2+ 2 = 4MN by above

Resolving vertically at C: CBC sin60 =CGC sin60 +CCD sin60⇒CBC = 2+ 6 = 8MN by above

Summary: EF, FG, GD, HC: 2MN in tension. FD, GC, HB: 2MN in compression. AB: 3MN in compression. DE: 4MN in compression. CD: 6MN in compression. GH: 4MN in tension. AH: 6MN in tension. BC: 8MN in compression.

   Question 8 (i) At A:

PE = KE = EPE = 0 → Total Energy = 0 At B: PE = −mg(3a) = −3mga KE = 0

EPE = λλλ aaa

ax 2

2)2(

2

22

==

Therefore, Total Energy = mgaa 32 −λ By the principle of conservation of energy:

032 =− mgaaλ

⇒ λ = 3mg2

(ii) At B:

mga

amg

axT 3

)2(23

===λ

A A

a

3a mg

T

B

a

T = 3mg

mg

Applying F = ma vertically up: maF =

mamgmg =−3 a = 2g

(iii) Applying F = ma vertically up:

2mgmgT =−

23mgT =

Hooke’s law:

axmgmg

axT

23

23

=⇒=λ

∴ x = a Let this point be C. Then at C:

KE = 2

21mv

PE = mgaamg 2)2( −=−

EPE = λx2

2a =

3mg2

!"#

$%&a2

2a = 3mga4a

Therefore, Total Energy = 12mv2 + 3mga4 − 2mga

Applying the principle of conservation of energy

12mv

2 + 3mga4 − 2mga = 0

                                                 

T

g/2

mg

12mv

2 − 5mga4 = 0

v = 5ga2

v2 = 5ga2

   Question 9

(i) Applying F = ma horizontally down:

maF =

mavmg =− 1.0

dtdvav ==− 1.010 Substituting m = 1, g = 10 →

Hence, vdtdv 1.010 −=

(ii) ∫ ∫= =

=−

v

v

t

t

dtv

dv

0 01.010

tv tv00

1.010ln1.01

=−−

( ) 010ln1.010ln10 −=−−− tv

tv=

− 1.01010ln10

tv

1.01.010

10ln =−

tev

1.0

1.01010

=−

tev 1.0101.010 −=− ( )tev 1.01101.0 −−=

( )tev 1.01100 −−=

(iii) As ∞→t , 01.0 →− te

∴ limt→∞

v =100 ms−1

                     

v a

mg

0.1v

 Question 10 (i) S is the resultant of the reaction R and the friction µR.

λ is the angle of friction.

29.1935.0tan35.0 1 ≈=⇒= −λµ

(iii) The diagram of forces hence becomes… Using Lami’s Theorem:

71.110sin29.159sin30 F

=

35.7929.159sin71.110sin30

==

F N

Using once more Lami’s Theorem:

S makes an angle of 110.710 with the 30 N force.

 

F

30N

S

λ+50°=69.29°

110.71°

P  

Ssin900

= 30sin159.290

Hence S = 30sin159.290 = 84.83 N

S

30N

F

50° R

µR

λ 50°