Chapter 13 Dynamics - scunderwood Specialist Mathematics Complete Worked Solutions 1 Chapter 13 – Dynamics Solutions to ... Essential Specialist Mathematics Complete Worked Solutions

Essential Specialist Mathematics Complete Worked Solutions 1

Chapter 13 – Dynamics

Solutions to Exercise 13A 1

a

i. r = 7i – 2i + 5j

= 5i + 5j

ii. | r | = 52 + 52

= 50

= 5 2

≈ 7.07 N

tan ° = 55

= 45

magnitude is 7.07 N, direction is 45°

b

i. r = 2j + 7i – 11i – 6j

= – 4i – 4j

ii. | r | = 16 + 16

= 32

= 4 2 N

≈ 5.66 N

tan ° = 44

= 45

magnitude is 5.66 N, direction is 225°

c

i. r = 3i – 4i – 5j

= – i – 5j

ii. | r | = (–1)2 + (– 5)2

= 26

≈ 5.10 N

tan ° = 5

= 78.7

magnitude is 5.10 N, direction is 258.7°

d

i. r = 8i + 8j – 5i + 2j

= 3i + 10j

ii. | r | = 32 + 102

= 109

≈ 10.44 N

tan ° = 103

= 73.3°

magnitude is 10.44 N, direction is 73.3°

y

x

5

5

q

y

xq

y

xq

–5

–1

y

x

10

q

3


e

i. r = 6i – 6i + 3j – 7j

= – 4j

ii. | r | = 4 N

= 270 °

magnitude is 4 N, direction is 270 °

f

i. r = 5i + 15j + 5i – 15j

= 10i

ii. magnitude = 10 N

direction = 0°

2

R = F1 + F2 + F3

= (3i + 2j) + (6i – 4j) + (2i – j)

= 11i – 3j N

3

Using the cosine rule,

F2 = 162 + 122 – 2 16 12 cos (130°)

= 256 + 144 – 384 cos (130°)

= 646.83044…

F = 25.43 N

The magnitude of F is 25.43 N, correct to two

decimal places.

4


162 = 92 + F22 – 2 9 F2 cos (120°)

256 = 81 + F22 – 18F2 – 1

2

175 = F22 + 9F2

F22 + 9F2 – 175 = 0

Using the quadratic formula,

F 2 =

–9 ± 92

– 4 1 –175

2

=781 – 9

2

= 9.5 (F2 > 0)

| F2 | = 9.5 N, correct to two decimal

places.

5

F1 + F2 + F3 = F

F3 = F – F1 – F2

= (3i – 2j + k) – (2i – j + k) – (3i – j – k)

= – 2i + k

6

Component of F in the direction of motion

= 400 cos 15°

= 386.37033…

≈ 386 N

7

a

i.

a = 4 cos 40° i + 4 sin 40° j

b = 3i

r = 6.064i + 2.57j

y

x

4

9 16

F2

120°

400

15°

y

x

4 N

a

b40°

3 N


ii. | r | = 2257.206.6

≈ 6.59 N

tan ° = 2×576×06

= 22° 59'

magnitude ≈ 6.59 N

direction ≈ 22° 59'

b

i.

a = 6 cos 10° i + 6 sin 10° j

b = 7 cos 20° i + 7 sin 20° j

c = 8 cos 30° i + 8 sin 30° j

r = (6cos 10° + 7cos 20° + 8cos 30°) i

+ (6sin 10° + 7sin 20° + 8sin 30°) j

≈ 19.41i + 7.44j

ii. | r | = 2244.741.19

≈ 20.79 N

tan = 7×44

19×41

≈ 20° 57'



c

i.

a = 7 cos 20° i + 7 sin 20° j

b = – 6 cos 30° i + 6 sin 30° j

r = (7 cos 20° – 6 cos 30°) i

+ (7 sin 20° + 6 sin 30°) j

= 1.382i + 5.394j

ii. | r | = 49 + 36 – 84 cos 50°

≈ 5.57 N

tan ° =

7 sin 20° + 6 sin 30°7 cos 20° - 6 cos 30°

= 75° 38'



d

i.

a = 5 cos 27° i + 5 sin 27° j

b = – 5 sin 27° i – 5 cos 27° j

r = 2.19i – 2.19j

ii. | r | = 22)19.2(19.2

≈ 3.09 N

tan ° = 1

= 45°


direction ≈ 315°

e

i.

a = 10 cos 25° i + 10 sin 25° j

b = 10 cos 25° i – 10 sin 25° j

r = 18.13i

ii. | r | ≈ 18.13 N

tan ° = 0


direction ≈ 0°

y

x

8 N (c)

10°6 N (a)

7 N (b)

10°10°

x

y

ab20°30°

7 N6 N

x

ya

b

27°

27°

5 N

5 N

x

ya

b

25°

10 N

25°

10 N


f

i.

a = 8 cos 30° i + 8 sin 30° j

b = – 2 cos 45° i + 2 sin 45° j

c = –10 cos 40° i – 10 sin 40° j

r = (8 cos 30° – 2 cos 45° – 10 cos 40°) i

+ (8 sin 30° + 2 sin 45° – 10 sin 40°) j

r ≈ – 2.15i – 1.01j

ii. | r | = 2201.115.2

≈ 2.37 N

tan = 1×012×15

= 25.27° (using exact values for the i

and j components)


direction ≈ 180° + 25.27°

≈ 205.27°

≈ 205° 17’

8

a


R2 = 42 + 32 – 2 4 3 cos 140°

= 16 + 9 – 24 cos 140°

= 43.38506…

R 6.59 N

Using the sine rule,

sin(40 - q)°

3=

R

140sin

40 – = 17.02354…

= 22.97645…

The resultant force has magnitude 6.59 N and

direction 22° 59'.

c


R2 = 62 + 72 – 2 6 7 cos 50°

= 36 + 49 – 84 cos 50°

= 31.00584…

R 5.57 N

Using the sine rule,

sin(180 - (q + 30))°

7 =

R

50sin

180 – ( + 30)= 74.36726…

= 75.63273…


direction 75° 38'.

e


R2 = 102 + 102 – 2 10 10 cos 130°

= 200 – 200 cos 130°

= 328.55752…

R 18.13 N


is in the positive direction of the x axis.

30 6

20 30

7

R

10

25

130 10

R

25

25


9

a Let a = 3i + 4j

a = 15

(3i + 4j)

b = – 4i + 3j

^b = 1

5(– 4i + 3j)

c = – 2j

c = – j

r = 165

(3i + 4j) + 125

(– 4i + 3j) – 15j

= (485

– 485

) i + (645

+ 365

– 15) j

= 5j

b | r | = 5 N

direction = 90°

10

a resolved part = 12 cos 20° = 11.28 N

b resolved part = 15 cos 65° = 6.34 N

c resolved part = 8 cos 90° = 0 N

d resolved part = 11 cos 145° = – 9.01 N

11

a 8 cos 40° + 12 cos 15° ≈ 17.72 N

b 8 + 12 cos 55° ≈ 14.88 N

12

a Let a = 2i – j and b = 7i + 3j

then a = 1

5(2i – j)

b . a = (7i + 3j) . 1

5(2i – j)

= 1

5(14 – 3)

= 11 5

5

b a

a =

11 55

1

5(2i – j)

= 115

(2i – j)

The component of (7i + 3j) N in the direction

of 2i – j is 115

(2i – j) N.

b Let a = 3i + 4j and b = 2i – 3j

then a = 15

(3i + 4j)

b . a = (2i – 3j) . 15

(3i + 4j)

= 15

(6 – 12)

=

-65

b a

a =

-65

15

(3i + 4j)

=

-625

(3i + 4j)

The component of (2i – 3j) N in the direction of

3i + 4j is

-625

(3i + 4j) N.

13

a 8 + 11 cos 155° ≈ –1.97 N

b 10 + 11 cos 115° ≈ 5.35 N

c 11 + 10 cos 115° + 8 cos 155° ≈ – 0.48 N

14

15 cos 25° + 25 cos 80° + 50 cos 115°

≈ –3.20 N

15

tan = 2×56

= 22.62°

10

155° 115°

11

8

B

10 NC A

6·56

2·5

q

24 N


a 10 + 24 cos 22.62° = 10 + 24 6

6×5

≈ 32.15 N

b 24 + 10 cos 22.62° = 24 + 10 6

6×5

≈ 33.23 N

16

a

| | OX®

: 3 + 2 cos 50° ≈ 4.29 N

OX®

: 2 sin 50° ≈ 1.53 N

| r | = 2253.129.4

≈ 4.55 N

tan = 1×534×29

= 19.67° = 19° 40’

4.55 N at angle of 19° 40’ to OX®

b

| | OX®

: 10 cos 30° + 5 cos 110° + 2 cos 135°

≈ 5.54 N

OX®

: 10 sin 30° + 5 sin 110° + 2 sin 135°

≈ 11.11 N

| r | = 22)11.11()54.5(

= 14.154

≈ 12.42 N

tan ° = 11×115×54

≈ 63.52° = 63° 31’

12.42 N at angle of 63° 31’ to OX®

17

| | 10 N: 10 + 7 cos 50° ≈ 14.50 N

10 N: 5 sin 50° ≈ 5.36 N

| r | = 2236.55.14

≈ 15.46 N

18

a

| | 10 N: 10 + 8 cos 60° = 14 N

10 N: 8 sin 60° – P

8 sin 60° – P = 0

P = 8 sin 60°

= 4 3 N

≈ 6.93 N

b | r | = 142 + 02

= 14 N

19

OX®

: P + 5 sin 35° – 7 sin 35° = 0

P – 2 sin 35° = 0

P = 2 sin 35°

P ≈ 1.15 N

1·53

4·29O X

50°

7 N

10 N

60°

8 N

P N

10 N


Solutions to Exercise 13B

1

a P = 2 kg 5 m/s = 10 kg m/s

b P = 0.3 kg 0.03 m/s = 0.009 kg m/s

c P = 1000 kg 30 ´ 1000

3600 m/s

= 8333 13

kg m/s

d P = 6 kg 10 m/s = 60 kg m/s

e P = 3000 kg 50 5

18 m/s

= 41 666 23

kg m/s

2

a P = 10(i + j) kg m/s

b

i. P = 10(5i + 12j) kg m/s

ii. | P | = 102(52 + 122)

= 10 25 + 144

= 130 kg m/s

3

change in momentum = final momentum –

initial momentum

a change in momentum

= 10 3 – 10 6

= 30 – 60

= – 30 kg m/s

b change in momentum

= 10 10 – 10 6

= 100 – 60

= 40 kg m/s

c change in momentum

= 10 3 – 10 – 6

= 30 + 60

= 90 kg m/s

4

a 5 kg = 5 g N

= 5 9.8 N

= 49 N

b 3 tonnes = 3 1000 kg

= 3 1000 9.8 N

= 29 400 N

c 60 g = 0.06 kg

= 0.06 9.8 N

= 0.588 N

5

a Resultant force, F = 8 4 = 32 N

b F = ma

a = 1mF =

510

= 12

m/s2

6

a F = ma m = | F || a |

= 102×5

= 4

kg

b | F | = m | a |

= 2 3.5

= 7 N

7

R – mg = ma

96 = ma + mg

96 = m(a + g)

96

a + g = m

m = 96

1.2 + g

m 8.73 kg


8

(F – 75g)j = ma

a =

F - 75g

75

–1 =

F - 75g

75

– 75 = F – 75g

F = 75g – 75

F = 660 N

9

2.5g – mg = m

. .x

. .x = 2

2.5g = mg + 2m

2×5g

g + 2 = m

m = 2.076 kg

The reading would be 2.076 kg wt.

10

F = ma

= 9 10–31 6 1016

= 5.4 10–14 N

11

F = ma

2i + 10j = 2a

a = i + 5j

12

F = ma

(8i + 2j + (2i – 6j) = 10a

10a = 10i – 4j

a = i – 25

j

13

a

mg – 2·5g = m

m (g – 1) = 2.5g

m =

2 × 5g

g - 1

When at rest the reading is

2 × 5g

g - 1 kg wt

≈ 2.78 kg wt.

b

T – mg = 2m

T = mg + 2m

=

(g + 2) ´ 2 × 5g

g - 1

≈ 3.35 kg wt.

14

u = 50 5

18 =

1259

m/s

v = 0

t = 10

Use v = u + at

a =

v - ut

=

0 - 1259

10

=

-2518

Now m = 25 1000

= 25 000 kg

F = ma

= 25 000

-2518

= – 34 722 29

N

F

75g

2·5g

mg

+ve

2·5g

mg

T

mg


15

R – mg = ma

R = mg + ma

= m(g + a)

= 10(9.8 + 1.5)

= 113 N

16

F = ma

= 16 0.6i

= 9.6i

and F = F1 + F2 + F3

F3 = F – F1 – F2

= 9.6i – (–10i – 15j) – (16j)

= 19.6i – j

17

u = 5, t = 3, v = 8

Using v = u + at,

a =

v - ut

=

8 - 53

= 1

F = ma

= 5 1

= 5 N

18

F = ma

a = 1m (F)

= 14

((8i + 12j) + (6i – 4j))

= 14

(14i + 8j)

= 72

i + 2j

19

F = 600 – 550 = 50

F = ma

a = Fm = 50

300 = 1

6

u = 0, t = 3

Using v = u + at

v = 0 + 16

3

= 12

The velocity after three seconds is 12

m/s.

20

R – 85g = ma

R = 85 9.8 + 85 – 2

= 85 7.8

= 663

The reaction force is 663 N.

21

a

F = (4g – Fr)i + (R – 10g) j

F = ma

(4g – Fr) i + (R – 10g)j = 10ai

i component: 4g – Fr = 10a 1

j component: R – 10g = 0

R = 10g 2

Fr = R

= 0.2 10g

= 2g 3

Substitute 3 into 1

4g – 2g = 10a

2g = 10a

R

10 g

+ve

R

85 g

+ve

4g

R

0·2R

10g


a = g

5

a ≈ 1.96 m/s2

b

v = at

= 1.96(10)

= 19.6 m/s

22

Resultant force in direction of motion

= 8000g – 4000g

= 4000g

4000g = 200 103 . .x

. .x =

4000g

200 ´ 103

= g

50

.x =

g

50t + c

but .x = 0 when t = 0

c = 0

Hence .x =

g

50t

30 km/h = 30 000 m/h = 3

25 m/s

hence when x = 3

25:

3

25 =

g

50t

t = 50 ´ 25

3g = 42.517

It takes 42.517 seconds to go from rest to 30 km/h.

23

F = ma

One man: P – FR = 250 0.15

P – FR = 37.5 1

Two men: 2P – FR = 250 0.4

2P – FR = 100 2

2 – 1 gives P = 62.5

Substitute P into 1

62.5 – FR = 37.5

FR = 25 N

Pushing force = 62.5 N, and

Resistance = 25 N

24

(F – 20 000)i = ma

a =

F - 20000200 000

0.2 =

F - 20000200 000

F = 0.2 200 000 + 20 000

F = 60 000 N

Now,

– 20 000 i = ma

a =

-20 000200 000

a = – 110

m/s2

8000g N

R

20 ´ 200g

200 ´ 103g N

P

N

FR

250g

F

200 000g

20 000 N

N

20 000


25

a = 0 since velocity is constant

10 – R = 10 0

= 10R

Now R – 10g = 0

R = 10g

= 10

10g = 1

g

= 5

49

26

a

FR = R

= 0.025 0.1g

= 0.0245 N

b

– 0.0245i = ma

a =

-0 × 02450 ×1

= – 0.245 m/s2

u = 10, t = 20

Using v = u + at

= 10 – 0.245 20

= 5.1 m/s

27

(24 – R) i + (R – 4g)j = ma i

R – 4g = 0

R = 4g

24 – 4g = 0

4g = 24

= 6g ≈ 0.612

28

a F = ma

(T – 200g) j = ma j

a = 0 m/s

T – 200g = 0

T = 200g N

T = 1960 N

b (T – 200g) j = ma j

ma = T – 200g

200 0.5 = T – 200g

T = 200g + 100 = 2060 N

29

a

For a smooth surface, = 0

(10 – 0) i + (R – 5g)j = 5a i

10 = 5a

a = 2 m/s2

b

R = 5g (20 – FR) i = ma i

20 – 0.3 5g = 5a

a =

20 - 1 × 5g

5

= 1.06 m/s2

10

R

10g

mR

R

0·1g

0·025R

T

200g

20 N

R

FR

5g

4g

24 N

R

R


Solutions to Exercise 13C 1

i : mg sin 45° i = ma i

ma = mg sin 45°

a = g sin 45°

= 9×8

2

≈ 6.93 m/s2

2

i : (– R + mg sin 45°) i = ma i

– R + mg

2 = ma 1

j : R = mg cos 45°

R = mg

2 2

Substitute 2 into 1

– mmg

2 +

mg

2 = ma

a = 2

gg

a = 2

)1( g m/s2

3

j : R = 60g cos 60°

= 30g

i : (FR – 60g sin 60°) i = ma i

ma = FR – 60g sin 60°

60 –8 = FR – 60g sin 60°

FR = 60g sin 60° – 60° 8

= 29.223 N

4

a

i component: 10 cos 30° = 5a

5 3 = 5a

a = 3 m/s2

b

vertical: R + 20 sin 30° = 5g

R = 5g – 10

horizontal: (20 cos 30° – FR) i = ma i

ma = 20 cos 30° – FR

5a = 20 3

2 – 0·3(5g – 10)

a =

10 3 - 1 × 5g + 3

5

= 1.124 m/s2

5

vertical: R + F sin 45° = 2g

R = 2g – F

2

=

4g - 2F

2

jR

mg45°

i

jmR

R

mg45°

i

j

FR

R

60g

60°

i


horizontal: (F cos 45° – R) i = 2a

( 2F2

– 12

(

4g - 2F

2)) = 2a

2a =

2 2F - 4g + 2F

4

8a = 3 2 F – 4g

F = 8a + 4g

3 2

= 8(g4 ) + 4g

3 2

= 6g

3 2

= 2g

2

= 2 g N

6

vertical:

R + 30 sin 30° – 20g = 0

R = 20 9.8 – 30 12

= 196 – 15

= 181 N

7

Resolving in the j direction:

R = mg cos

Resolving in the i direction:

P – R – mg sin = ma

a = 1m (P – mg cos – mg sin )

= Pm – g cos – g sin

8


– mg sin 30° = ma = ma

–mg

2 = ma

a = –g

2

a = –g

2 i

9

Resolving in the j direction:

R – mg cos 60° = 0

R – mg

2 = 0 = 0

R = mg

2


mg sin 60° – R = ma

mg 3

2 – 0.8

mg

2 = ma

g 3

2 –

4g

10 = a

a = 10

4g35g

= 10

)435 g( m/s2

= 4.57 m/s2

s = 5, u = 0, v = ?

Use v2 = u2 + 2as

v = 557.420

= 6.76

speed = v = 6.76 m/s


10

j : R = mg cos 20°

i : – 0.25R – mg sin 20° = ma

– 0.25(mg cos 20°) – mg sin 20° = ma

a = g(– sin 20° – 0.25 cos 20°)

= – 0.577g

≈ – 5.65 m/s2

Using v2 = u2 + 2as,

s =

2(10)

2( 5.65)

= 8.84 m

When retuning:

i : 0.25R – mg sin 20° = ma

0.25(mg cos 20°) – mg sin 20° = ma

a = g(0.25 cos 20° – sin 20°)

a = –1.05 m/s2

Using v2 = u2 + 2as

= 2 –1.05 – 8.84

v = 58.18

= 4.31 m/s

11

If tan =4

3 , then sin =

5

4

mg sin i + (R – mg cos ) j = ma

a

ma = mg sin

a = g sin

= 4g

5

≈ 7.84 m/s2

u = 0, s = x,

Using v2 = u2 + 2as

= 2 4g

5 x

= 8gx

5

v = 8gx

5

= 8gx

5

5

5

= 40 gx

25

= 40 gx

5

= 25

10gx m/s

b

Resolving vertically:

R – mg = 0

R = mg

Resolving horizontally:

0.3R = ma

0.3mg = ma

a = 0.3g

= 2.94 m/s2

So,

u = 2 10gx

5

a = –2.94 m/s2

v = 0

s = ?

using v2 = u2 + 2as

s =

v2 - u2

2a

jR

0·25R

20°

i

mg

j

R 0·25R

20°

i

mgR

mg

0·3R


s =

–4

25 10 gx

–6g

10

s = 8gx

5

10

6g

s =8x

3 m

12

a Resolve perpendicular to the plane:

F sin + R = Mg

Resolve parallel to the plane:

F cos – R = Ma

a = 1M

(F cos – (Mg – F sin ))

= FM

(cos + sin ) – g

b

Resolve perpendicular to the plane:

F sin + Mg = R


F cos – R = Ma

a = 1M

(F cos – (F sin + Mg))

= FM

(cos – sin ) – g

13

sin = 1

20

a

Resolve in the i direction

(FR – 1000g sin ) = 0

FR – 1000g

20 = 0

FR – 50g = 0

FR = 50g

= 490 N

b

(F – mg sin – FR) i + (R – mg cos ) j = ma i

F – FR – mg sin = ma

F – 1000g

20 –

1000g

20 = 1000 (a = 1)

F = 1000 + 1000g

20 2

= 1980 N

14

sin = 35

cos = 45

a


N = 0.5g cos

= 0.4g


–3N8

– 0.5g sin = 0.5 . .x

–8

3

5

2g –

g

2

5

3 = 1

2. .x

–3g

20 –

3g

10 = 1

2. .x

–9g

10 =

. .x

d ( 1

2 v2 )

dx = –

9g

10

R

Mg

F

mR q

R

Mg F

mR

q

R

1000gq

F

N

0·5gq

3N8


12

v2 = 6 – 3g

10x

12

v2 = –9gx

10 + c

When x = 0, v = 6

18 = c

12

v2 = –9gx

10 + 18

When v = 4, –10 = –9gx

10

i.e. 10 ´ 109g

= x

1009g

= x

x 1.13 m

b

When v = 0, –9gx

10 = –18

x = 20g

it goes 20g metres up the plane

i.e. when t = 0, x = 20g , v = 0

Resolve parallel to the plane: 3N8

– 0·5g sin = 0·5 . .x

8

3

5

2g –

g

2

5

3 = 1

2. .x

3g

20 –

3g

10 = 1

2. .x

–3g

10 =

. .x

. .x = –

3g

10

d ( 1

2 v2 )

dx = –

3g

10

12

v2 = –3g

10x + c

When x = 20g , v = 0

0 = –3g

10 20

g + c

c = 6

When x = 0, v2 = 12

i.e v = 12

= 2 3 m/s

3.46 m/s

15

Resolve in the j direction:

10 – 8 cos 60° = 6

Resolve in the i direction:

8 cos 30° – P

F = (8 cos 30 – P)i + 6j

| F | 2 = 48 – 8 3 P + P2 + 36

= 84 – 8 3 P + P2

For | F | = m | a |

84 – 8 3 P + P2 = 100

P2 – 8 3 P – 16 = 0

P = 8 3 ± 192 + 64

2

= 8 3 ± 16

2

= 4 3 + 8 or 4 3 – 8

= (4 3 + 8) N

P = 4 3 + 8 is the required force.

16

(F – R5

– 5g sin 30°) i + (R – 5g cos 30°) j = mai

a

R – 5g cos 30° = 0

R = 5g 3

2

and

F – R5

– 5g sin 30 = ma

N

0·5gq

3N8

O

+ve


F – g 3

2 –

5g

2 = 5 1.5

F = 5 1.5 + g 3 + 5g

2

= 40.49 N

b


F cos 60° + R = 5g cos 30°

R = 5g 3

2 – F

2

= 5g 3

2 – 1

2(15

2 +

g 3 + 5g

2)

=

9g 3 - 5g

4 –

154

= g

4(9 3 – 5) –

154


F cos 30° – 15

R – 5g cos 60° = 5 . .x

. .x = 1.22 m/s2

30°5g

F

30°

R

(152

+ g 3 + 5g

2)

32

– g

20(9 3 – 5) –

1520

– 5g

2 = 5

. .x


Solutions to Exercise 13D

1

a 10g – T = 10a 1

T – 8g = 8a 2

1 + 2 gives 2g = 18a

a = 2g

18

= g

9

Substitute into 2

T – 8g = 8g

9

T = 8g + 8g

9

= 80g

9

87.1 N

b a = g

9 m/s2 1.09 m/s2

2

a

F = 10 N

F = ma

10 = 11a

a = 1011

0.91 m/s2

b

F – S = ma

10 – S = 6 1011

S = 10 – 6011

T = S = 4.55 N

3

a

T – 1.5g = 1.5a 1

2g – T = 2a 2

1 + 2 gives 0.5g = 3.5a

a = 0×5g

3×5

= g

7

From 2 2g – T = 2(g

7)

T = 2g – 2g

7

= 12g

7

= 16.8 N

b

a = g

7 =

9×87

= 1.4 m/s2

4

a

Mg – T = M

Mg – M = T

M (g – 1) = T

M =

Tg - 1

1

10g8g

T T

R

11g

F

S

R

6g

F

1·5 kg

1·5g2g

T T

2 kg

Mg25°5g

T

T

R


T – 5g sin 25° = 5

T = 5g sin 25° + 5

= 5(g sin 25° + 1) 2

From 1 M =

5(g sin 25° + 1)

g - 1

= 2.92 kg

b

From 2 T = 5(g sin 25° + 1)

= 25.71 N

5

a

8g – T = 8a

a = g – T8

and T = 4a

a = g – 4a8

= g – a2

3a2

= g

a = 2g

3 =

9815

m/s2

b

T = 4a = 4 9815

= 26 215

N

6

a

4g – T = 4a

a = g – T4

and T – 2g sin 30° = 2a

T = g + 2a

= g + 2(g – T4

)

= g + 2g – T2

3T2

= 3g

T = 23

3g

= 2g

= 19.6 N

b

a = g – T4

= g – 19×6

4

= 4.9 m/s2

7

a

10g sin 30° – T = 10a 1

T – 5g sin 45° = 5a 2

1 + 2 gives

10g sin 30° – 5g sin 45° = 15a

5g – 5g 2

2 = 15a

10g - 5g 2

2 = 15a

a =

10g - 5g 2

30

=

2g - g 2

6

≈ 0.96 m/s2

b

T – 5g sin 45° = 5

2g - g 2

6

T = 5g sin 45° + 5

2g - g 2

6

≈ 39.4 N

T

8g

R

T4g

4g30°2g

T

T

R

T

5g30°

10g45°

R2

R1T


8

i : (T – 20)i = ma i

T – 20 = 5 0.8

T = 24 N

j : mg – T = ma

T = m(g – 0.8)

24 N = 9m

m = 2.67 kg

9

a

T – 750 = ma

T – 750 = 5000 2

T = 10 750 N

b

40 000 – T – FR = 10 000a

40 000 – 10 750 – FR = 20 000

FR = 40 000 – 10 750 – 20 000

= 9250 N

10

T = 37.5 N

xg – T = xa 1

T – 3g = 3a 2

From 1 xg – 37.5 = xa

x(g – a) = 37.5

x =

37 × 5g - a 3

From 2 a = 37×5

3 – g

a = 2.7 m/s2 4

Substitute 4 into 3

x =

37 × 5g - 2 × 7

x = 5.28 kg

11

Engine:

Truck:

a

engine:

(60 000 – 40 000g sin – T) i = ma

60 000 – 18

(40 000g) – T = 40 000a

a = 140000

(60 000 – 5000g – T)

truck:

T – 18

(8000g) = 8000a

T = 8000a + 1000g

Substitute T into a:

a = 140000

(60 000 – 6000g – 8000a)

a = 64

– 6

40g – a

5

6a5

= 3020

– 3

20g

a =

30 - 3g

24

a = 0.025 m/s2

b

T = 8000(0.025) + 1000g

T = 10 000 N

750 N

R2

5000g

T TR1

10000g

40000 N

FR

3 kgx kg

T T

q

R60000 N

T40000g

ji

engine

q

RT

8000g


12

a

5g – T = 5a 1

T = 20a 2

1 + 2 gives

5g = 25a

a = g

5 m/s2

T2 = 8a

= 8 g

5

= 5

8 g

= 15.68 N

b

T = 20a

= 20 g

5

= 4g N

= 39.2 N

c

a = g

5 m/s2

= 1.96 m/s2

13

u = 0, s = 3, t = 3

s = ut + 12

at2

s – ut = 12

at2

a =

2(s - ut )

t2

=

2(3 - 0)

9

= 69

m/s2

= 23

m/s2

0.2g – T = ma

1.96 – T = 0.2 23

T = 13775

N

and T – R = ma

13775

– 0.5g = 0.5 23

= (13775

– 0.5 23

) ÷ 0.5g

= 0.305

14

a

For A:


T – R – 4g sin 30° = ma 1


R – 4g cos 30° = 0

R = 2 3 g 2

Substituting 2 into 1 gives

T – 2 3 g – 2g = 4 3

For B:

6g – T = 6

T = 6(g – 1) 4

Substitute 4 into 3

6(g – 1) – 2 3 g – 2g = 4

4g – 10 = 2 3 g

=

4g - 10

2 3g

= 0.86

b

T = 6(g – 1) = 52.8 N

T

5g

R2

T12g

R1

8g

T2

a

mRT

0·2g

R

T0·5g

A

6g30°4g

T

TR

mR B


15

a

4.2g sin – T = ma

4.2g 0.6 – T = 4.2 2

T = 2.52g – 8.4

= 16.296 N

b

T – R2 = ma

16.296 – 3g = 3 2

3g = 10.296

= 0.35

T

3g

4·2g

R2

R

mR2

q

T


Solutions to Exercise 13E

1 F = (10 – t)2

using F = ma

10 a = (10 – t)2

a =(10 – t)

2

10

dv

dt=

(10 – t)2

10

v =1

10

(10 – t)2 dt

v = –1

30 (10 – t)

3+ c

When t = 0 , v = 0 :

c =100

3

v = –1

30 (10 – t)

3+

100

3

When t = 10 , v =100

3 m/s

i.e. the velocity after 6 seconds is100

3 m/s

Now,

x =

–1

30(10 – t)

3+

100

3

dt

x = –1

30

(10 – t)4

4 –1 +

100

3 t + d

x =1

120 (10 – t)

4+

100

3 t + d

When t = 0 , x = 0 :

d = –250

3

x =1

120 (10 – t)

4+

100

3 t –

250

3

When t = 10 , x = 250 m

Hence the distance travelled is 250 m

2

a F = 10 sin ( t)

using F = ma

5a = 10 sin ( t)

a = 2sin ( t)

v =

2sin ( t) dt

v = –2 cos ( t) + c

When t = 0 , v = 4 :

c = 6

v = –2 cos ( t) + 6

Now,

x =

(–2 cos ( t) + 6) dt

x = –2 sin ( t) + 6 t + d

When t = 0 , x = 0 :

d = 0

x = –2 sin ( t) + 6 t

b F = 10 + 5x

5a = 10 + 5x

a = 2 + x

d

dx 1

2 v

2

= 2 + x

1

2 v

2=

(2 + x) dx

v2

=

(4 + 2x) dx

v2

= 4x + x2

+ c

When v = 4 , x = 0 :

c = 16

v2

= 4x + x2

+ 16

v = 4x + x2

+ 16

When x = 4 , v = 48 = 4 3 m/s


c F = 10 cos2(t)

2

2

2

2

2

2

5 10 cos ( )

2 cos ( )

2 cos

2 cos ( )

a t

a t

d xt

dt

dxt dt

dt

Using the identity 2 21cos ( ) 1 cos

2t t

2(1 cos ( )

1sin(2 ) c

2

dxt dt

dt

dxt t

dt

When 0, 0 :dx

t vdt

0

1sin(2 )

2

c

dxt t

d t

So,

2

1sin(2 )

2

1 1cos(2 )

4 2

x t t dt

x t t d

When 0, 0 :t x

2

2

1

4

1 1 1cos(2 )

4 2 4

12 cos(2 ) 1

4

d

x t t

x t t

3 F =100

( t + 5)2

6a =100

( t + 5)2

a =50

3( t + 5)2

dv

dt=

50

3( t + 5)2

v =50

3

1

( t + 5)2 dt

v =50

3

( t + 5)–1

–1 + c

v = –50

3( t + 5)+ c

When t = 0 , v = 10 :

c =40

3

v = –50

3( t + 5)+

40

3

When t = 10 , v =110

9 m/s

i.e. the velocity after 10 seconds is

100

9 m/s

Now,

x =

–50

3( t + 5)+

40

3

dt

x = –50

3 log e( t + 5) +

40

3 t + d

When t = 0 , x = 0 :

d =50

3 log e(5)

x =50

3 log e

5

t + 5

+40

3 t

and when t = 10 :

x =50

3 log e

1

3

+400

3

x =

400

3–

50

3 log e(3) m


4 F = 1 – sin t

4

m = 1

using F = m a

a = 1 – sin t

4

dv

dt= 1 – sin

t

4

v =

1 – sin t

4

dt

v = t + 4cos t

4

+ c

When t = 0 , v = 0 :

c = –4

v = t + 4cos t

4

– 4

So,

x =

t + 4cos t

4

– 4

dt

x =1

2 t

2+ 16 sin

t

4

– 4 t + d

When t = 0 , x = 0 :

d = 0

x =1

2 t

2+ 16 sin

t

4

– 4 t

5 F = 1 – cos 1

2 t

a using F = ma

m = 1

a = 1 – cos 1

2 t

v =

1 – cos 1

2 t

dt

v = t – 2sin 1

2 t

+ c

When t = 0 , v = 0 :

c = 0

v = t – 2sin 1

2 t

b If v = t – 2sin 1

2 t

then

x =

t – 2sin 1

2 t

dt

x =1

2 t

2+ 4cos

1

2 t

+ d

When t = 0 , x = 0 :

d = –4

x =1

2 t

2+ 4cos

1

2 t

– 4

6 F = 12 t – 3t2

using F = m a

4a = 12 t – 3 t2

a = 3 t –3

4 t

2

v =

3 t –3

4 t

2

dt

v =3

2 t

2–

1

4 t

3+ c

When t = 0 , v = 2 :

c = 2

v =3

2 t

2–

1

4 t

3+ 2

When t = 4 , v = 10 m/s

Hence the velocity after 4 seconds is

10 m/s

7 F =t

t + 1

Using long division t

t + 1= 1 –

1

t + 1

Using F = ma

m = 1

a = 1 –1

t + 1

v =

1 –1

t + 1

dt

v = t – log e( t + 1) + c


When t = 0 , v = 0 :

c = 0

v = t – log e( t + 1)

When t = 10 , v = 10 – log e(11 ) 7.6 m/s

Hence the velocity after 10 seconds is

7.6 m/s

8 F = e

–t

2

a using F = ma

0.5 a = e

–t

2

a = 2e

–t

2

v =

2e

–t

2 dt

v = –4 e

–t

2+ c

When t = 0 , v = 0 :

c = 4

v = –4 e

–t

2+ 4

v = 4

1 – e

–t

2

m/s

b

c

distance travelled

=

0

30

(–4 e–0.5 t

+ 4) dt

=

8e–0.5 t

+ 4 t

30

0

= 8e–15

+ 120 – 8

112 m

Alternatively, using CAS to determine the

distance travelled in the first 30 seconds

we have

9 F ( t) =

14 – 2 t

100 t–2

0 t 5

t > 5

as F = ma

a =F

m=

F

10

a( t) =F ( t)

10=

1.4 – 0.2 t

10 t–2

0 t 5

t > 5

Note that the signed area under an a-t

graph gives change in velocity. This

concept can be used to determine the

speed of the body when t = 10.

a Sketch the a-t graph


t1 2 3 4 5 6 7 8 9 10 11

a

0.2

0.4

0.6

0.8

1

1.2

1.4

Change in velocity

=1

2(0.4 + 1.4 ) 5 +

5

10

10 t–2

dt

= 4.5 +

–10 t–1

10

5

= 4.5 + 1

= 5.5

Alternatively, using CAS we have

When t = 0, v = 0.

When t = 10 , v = 0 + change in velocity

v = 5.5

Speed = |v | = 5.5 m/s

b

distance travelled can be found by

determining the area under the v-t graph.

For 0 t 5:

v =

(1.4 – 0.2 t) dt

v = 1.4 t – 0.1 t2

+ c

From part a, when t = 0, v = 0:

c = 0

v = 1.4 t – 0.1 t2 for 0 t 5

For t > 5:

v =

10 t–2

dt

v = –10

t+ d

From part a, when t = 5, v = 4.5:

d = 6.5

v = –10

t+ 6.5 for t > 5

Hence,

v( t) =

1.4 t – 0.1 t2

–10

t+ 6.5

0 t 5

t > 5

for t [0, 10 ], v(t) 0

Hence the distance travelled in the first 10

seconds

=

0

5

(1.4 t – 0.1 t2) dt +

5

10

–10

t+ 6.5

dt

= 275

6– 10 log e(2) m (using CAS)


10 F = kv

using F = ma

m a = kv

a =kv

m

dv

dt=

kv

m

dt

dv=

m

kv

t =m

k

1

v dv

t =m

k log e(v) + c

When t = 0 , v = u :

c = –m

k log e(u)

t =m

k log e

v

u

kt

m= log e

v

u

e

kt

m=

v

u

v = ue

kt

m

So,

x = u

e

kt

m dt

x =um

k e

kt

m + d

When t = 0 , x = 0 :

d = –um

k

x =um

k e

kt

m–

um

k

1m

kt

ek

umx metres

11 F = – kv

using F = ma

m a = – kv

a = –kv

m

v dv

dx= –

kv

m

dv

dx= –

k

m

v = –k

m

1 dx

v = –k

m x + c

When x = 0 , v = V :

c = V

v = –k

m x + V

v = V –k

m x

12 F = b – cv

using F = ma

m a = b – cv

a =b – cv

m

dv

dt=

b – cv

m

dt

dv=

m

b – cv

t = m 1

b – cv dv

t = –m

c log e(b – cv) + d

When t = 0 , v = 0 :

d =m

c log e(b)

t =m

c log e

b

b – cv

ct

m= log e

b

b – cv

e

ct

m=

b

b – cv


b – cv =b

e

ct

m

= b e

–ct

m

cv = b – b e

–ct

m

cv = b

1 – e

–ct

m

m

ct

ec

bv 1 m/s

Terminal velocity is the limiting velocity

as t

= b

c

1 – e–

=b

c (1 – 0)

=b

c

Hence the terminal velocity is c

b m/s

13

When the body is projected upwards:

using F = ma

2

2

22 2

m g kva

m

dv m g kvv

dx m

dv m g kv

dx m v

kv2

mg positive

2

dx mv

dv mg kv

2

2

2

2

log2

e

m kvx dv

k m g kv

mx m g kv c

k

When 0, :x v u

2

2

2

log ( )2

log2

e

e

mc m g kv

k

m m g kux

k m g kv

Maximum height is reached when 0v 2

2

log2

log 1 (1)2

e

e

m mg kux

k mg

m kux

k mg

Hence the maximum height reached = 2

log 12

e

m ku

k m g

Now take the highest point as the origin.

kv2

using F = ma

2

2

Fa

m

m g kva

m

dv m g kvv

dx m

mg

positive


2

2

2

2

2

log ( )2

e

dx m v

dv m g kv

m kvx dv

k m g kv

mx m g kv c

k

At highest point 0, 0 :x v

2

log ( )2

log (2)2

e

e

mc m g

k

m m gx

k m g kv

The object returns to the point of

projection when (1) = (2).

m

2k log e

1 +ku

2

m g

=m

2k log e

m g

m g – kv2

1 +ku

2

m g=

m g

m g – kv2

m g + ku

2

m g=

m g

m g – kv2

m g

m g + ku2

=m g – kv

2

m g

(m g )

2

m g + ku2

= m g – kv2

kv2

= m g –(m g )

2

m g + ku2

kv2

=m g (m g + ku

2)

m g + ku2

–(m g )

2

m g + ku2

kv2

=m gku

2

m g + ku2

v2

=m gu

2

m g + ku2

v =m gu

2

m g + ku2

v = um g

m g + ku2

Speed = |v | = um g

m g + ku2

14 F =4

x

using F = m a

0.2 a =4

x

a =20

x

d

dx 1

2 v

2

=20

x

1

2 v

2=

20

x dx

1

2 v

2= 20 log e(x) + c

When v = 0 , x = 1 :

c = 0

1

2 v

2= 20 log e(x)

v2

= 40 log e(x)

v = ± 40 log e(x)

v = 40 log e(x)

(as the body is moving on the positive x

axis). As required.

15

vF

25

50 Usea

a =50

25 + v

dv

dt=

50

25 + v

dt

dv=

25 + v

50=

1

2+

v

50

t =

1

2+

v

50

dv

t =1

2 v +

v2

100+ c

When t = 0 , v = 0 :

c = 0

t =1

2 v +

v2

100


When t = 50 , v = 25 + 25 = 50

Hence, when t = 50 , v = 50 as required

b a =50

25 + v

v dv

dx=

50

25 + v

dv

dx=

50

25 v + v2

dx

dv=

25 v + v2

50=

1

2 v +

1

50 v

2

x =

1

2 v +

1

50 v

2

dv

x =1

4 v

2+

1

150 v

3+ c

When x = 0 , v = 0 :

c = 0

x =1

4 v

2+

1

150 v

3

When v = 50 , x = 625 +2500

3=

4375

3 m

Hence when v = 50 the distance from

O to P is 4375

3 m

1000 use 50For

2v

Ft c

v dv

dx= –

v2

1000

dv

dx= –

v

1000

dx

dv= –

1000

v

x = –1000

1

v dv

x = –1000 log e(v) + c

When v = 50 , x =4375

3:

c =4375

3+ 1000 log e(50 )

x = 1000 log e

50

v

+4375

3

When v = 25 ,

x = 1000 log e(2) +4375

3 2151.48 m

Hence when v = 25 the distance from

O to P is 2151.48 m


Solutions to Exercise 13F

1

a

PQR = 180° – 120° = 60°

RPQ = 180° – 80° = 100°

b

QPR = 180° – 100° = 80°

RQP = 180° – 120° = 60°

c

QPR = 180° – 115° = 65°

PRQ = 180° – (90° + 65°) = 25°

2

a

b

P2 = 22 + 52

P = 29 N

c

1

2tan(180 )

5

2180 tan

5

180 21.80

= 158.20 °

Hence the angle the P N force makes with the 5

N force is 158.20 °

3

a

b

P2 = 72 – 52

P = 24

P = 62 N

c

1

5cos(180 )

7

5180 cos

7

180 44.42

= 135.58 °

Hence the angle the 5 N force makes with the 7

N force is 135.58°

4

a

b


102 = P2 + P2 – 2P2 cos 140°

100 = 2P2 (1 – cos 140°)

P2 = 28.31

P = 5.32 N

R

QP

100°

60°

20°

R

Q

P80°

60°

40°

R

QP

65° 25°

5

P2

180°– q

180°– q

P

7

5

P

10Q

140°


180 – = tan–1

2

5

180 – = 21.80 °

5

a

Firstly, ° = 360° – (115° + 80°)

= 165°

Qsin 165°=

Psin 115° =

5sin 80°

Q = 5

sin 80° sin 165°

Q = 1.31 N

P = 5

sin 80° sin 115°

= 4.60 N

Hence, N 31.1 N, 60.4 QP

b

Qsin 160°

= 5

sin 130° = P

sin 70°

Q = 5

sin 130° sin 160°

= 2.23 N

P = 5

sin 130° sin 70°

= 6.13 N

Hence, N 23.2 N, 13.6 QP

6


P2 = 102 + 102 – 2(10)(10) cos 130°

P2 = 328.56

P = 18.13 N

7

Using Lami’s theorem,

Psin 120°

= 5g

sin 140°

P = 66.02 N

8

Along the force 10 N, the sum of the resolved

parts in newtons is

10 + 5 cos 120° + 5 3 cos 210° = 0

Along the force 5 N, the sum of the resolved

parts in newtons is

5 + 10 cos 120° + 5 3 cos 90° = 0

Along the 5 3 N force, the sum of the

resolved parts in newtons is

5 3 + 10 cos 210° + 5 cos 90° = 0

the particle is in equilibrium.

80°

115°q°

P N

Q N

5 N

130°

70° q°

P N

Q N

5 N

50°

P N

10 N10 N

130°

10

P

10


9

a

The resultant force, P, of the two 10 N forces is

shown above.

The direction of the resultant of the two forces

occurs along the bisector of the angle between

the forces. (this is the rhombus property)

b

P + 10 cos 155° + 10 cos 155° = 0

P – 18.13 = 0

P = 18.13 N

c

18.13 N is the magnitude of P at an angle of

155° with each 10 N force.

10

Along the force P N, the sum of the resolved

parts in newtons is

P + Q cos 80° + 23 cos 135° = 0 1

Perpendicular to the force P N, the sum of the

resolved parts in newtons is

23 cos 45° + Q cos 190° = 0 2

From 2 , Q =

-23 cos 45°cos 190°

3

Substitute 3 into 1

P = – Q cos 80° – 23 cos 135°

= 23 cos 45° ´ cos 80°

cos 190° – 23 cos 135°

P = 13.40 N

Q = 16.51 N

11

Resolve in the direction of Q

Q + 10 cos 60° + 15 cos 150° = 0

Q + 5 + 15 (–3

2) = 0

Q = (15 3

2 – 5) N

≈ 7.99 N

Resolve in the direction of P

P + 15 cos 120° + 10 cos 210° = 0

P + 15 (– 12

) + 10 –3

2 = 0

P = 15 + 10 3

2

≈ 16.16 N

12

a

b


102 = 162 + 82 – 2(8)(16) cos (180° – )

cos (180° – ) = 64

55

= 180° – 30.75° = 149.25°

50°

P N

10 N10 N

155° 155°10 N

120°15 N

P N

60°Q

10 N

16 N8 N

q

180°– q

8

10

16


13

a

b


P2 = 32 + 52 – 2(3)(5) cos 80°

P2 = 28.79

P = 5.37 N

c


32 = 52 + 5.372 – 2(5)(5.37) cos (180° – )

cos (180° – ) = 0.835

= 180° – 33.41° = 146.59°

14

Note:

cos = 3

5

sin =4

5 Resolve horizontally

T cos (90° – ) = P cos (15°)

i.e. T sin = P cos (15°) 1

Resolve vertically

T cos = P cos 75° + 2g 2

From 1

T = P cos (15°)

sin q

= 5P cos (15°)

4

Substitute T into 2

5P cos (15°)

4

35

= P cos (75°) + 2g

P [ 3 cos 15°4

– cos (75°)] = 2g

P (

3 cos 15° - 4 cos 75°4

) = 2g

i.e. P =

8g

3 cos 15° - 4 cos 75°

≈ 42.09 N

T = 5P cos 15°

4

=

5 cos 15° ´ 8g

4(3 cos 15° - 4 cos 75°)

≈ 50.82 N

15

Method 1: Resolving forces

Resolve horizontally

5g = T1 cos + T2 cos

5g = T1 1213

+ T2 5

13 1

Resolve vertically

T1 cos = T2 cos

i.e. T1 5

13 = T2 12

13

i.e. T1 = 125

T2

Substitute in 1

5g = 125

1213

T2 + 5T213

65g = (144 + 25)T2

5

325169

= T2

T1 = 125

325169

g

T1 = 60g

13

≈ 45.23 N

P N

3 N5 N

q

100°

180°– q 5

P

3

80°

2g

P

Tq1·5

2

2·5

75°

105°90 – qq


Method 2: Triangle of forces

T1 = 5g sin

= 5g 1213

= 60g

13

≈ 45.23 N

16

a

Perpendicular to P N

10 cos 140° + 12 cos (220° – ) = 0

cos (220° – ) =

-10 cos 140°12

= 220° – cos–1 (0.638)

= 169.67°

b

Along the force P N

P + 10 cos 50° + 12 cos (169.67° + 50°) = 0

P = –10 cos 50° – 12 cos 219.67°

= 2.81 N

5g

T1

q T2


Solutions to Exercise 13G 1

a Limiting frictional force is Fmax = R

Fmax = 0.3 10 g = 3g N Fmax = 29.4 N

An applied force of 10 N generates a frictional

force of 10 N.

b An applied force of 30 N overcomes the

limiting friction force,

Fmax = 3g N = 29.4 N

c An applied force of 40 N overcomes the

limiting friction force,

Fmax = 3g N = 29.4 N

2

Resolving perpendicular to the plane:

R + Psin 40 ° – 1.2 g = 0 R = 1.2 g – Psin 40 °

N 2 If Pa then R = 1.2 g – 2sin 40 °

and limiting frictional force,

F max = 0.2 (1.2 g – 2sin 40 °) Fmax = 2.09 N

Resolving parallel to the plane:

2cos 40 ° – F = 0 F = 2cos 40 °

F = 1.53 N

Hence an applied force of 2 N generates a

frictional force of 1.53 N.

b If P = 3 N

then R = 1.2 g – 3sin 40 °

and limiting frictional force,

F max = 0.2 (1.2 g – 3sin 40 °) Fmax = 1.97 N


3cos 40 ° – F = 0 F = 3cos 40 °

F = 2.30 N

Hence an applied force of 3 N overcomes the

limiting frictional force.

F = 1.97 N

3

a Fmax = R Fmax = 0.3 100 g Fmax = 30 g N

Hence the greatest horizontal force that can be

applied without moving the particle is 30g N.

R

P

1.2 g

F 40°

R

F

100g


b


R + Psin 60 ° – 100 g = 0 (1)


Pcos 60 ° – F = 0 F = Pcos 60 °

R =1

2 P

0.3 R =1

2 P

R =5

3 P (2)

Substitute (2) into (1):

5

3 P + Psin 60 ° – 100 g = 0

P 5

3+

3

2

= 100 g

P = 386.94 N

Hence the magnitude of the greatest pulling

force that can be applied without moving the

particle is 386.94 N.

c


R + Psin 60 ° – 100 g = 0 (1)


– F – Pcos 60 ° = 0 F = – Pcos 60 °

0.3 R = –1

2 P

R = –5

3 P (2)

Substitute (2) into (1):

–5

3 P + Psin 60 ° – 100 g = 0

P

–5

3+

3

2

= 100 g

P = –1224.02 N

Hence the magnitude of the greatest pushing

force that can be applied without moving the

particle is 1224.02 N.

4


R – 5cos 30 ° = 0

R =5 3

2


5sin 30 ° – F = 0

F = 5sin 30 °

F = 2.5 N

100g

P

R

F 60°

100g

P

R

F60°

30°

5

RF


5


sin 20 ° – F = 0 F = sin 20 °


R – cos 20 ° = 0 R = cos 20 °

As F R

F

R

sin 20 °

cos 20 °

tan 20 °

Hence the particle will slide down the plane

when < tan 20 °

6


R – wcos = 0 R = wcos


wsin – F = 0 wsin = F wsin = 0.4 R wsin = 0.4 wcos tan = 0.4

= tan–1

(0.4 ) = 21.80 °

7

a Resolving perpendicular to the plane:

R – 3cos 25 ° = 0 R = 3cos 25 °


3sin 25 ° – F = 0 F = 3sin 25 °

3cos 25 ° = 3sin 25 °

= tan 25 °

= 0.47 (correct to two decimal places)

20°

1

RF

°

w

RF = 0.4

25°

3 kg wt

RF


b

As the particle is in limiting equilibrium,

P = F + 3sin 25 ° kg wt P = 3tan 25 ° cos 25 ° + 3sin 25 ° kg wt P = g(3tan 25 ° cos 25 ° + 3sin 25 °) N P = 9.8 2.5357 N P = 24.85 N

Hence a force of 24.85 N or more, when

applied up the plane, will cause the particle to

move.

8

On the point of moving down:


R – 24 gcos = 0 R = 24 gcos


24 gsin – F – 100 = 0 F = 24 gsin – 100 (1)

On the point of moving up:


R = 24 gcos


120 – F – 24 gsin = 0 F = 120 – 24 gsin (2)

a (1) + (2):

2F = 20 F = 10 N

b

Substitute F = 10 into (1):

24 gsin = 110

sin =55

12 g

= sin–1

55

12 g

27.88 °

c

As F = R

=F

R

=10

24 gcos = 0.0481019 . . . 0.05

25°

3 kg wt

RP

F

°

24g

R

F

100 N

°

24g

R

F

120 N


Solutions to Exercise 13H

1

F1 = 2i and F2 = – 3j

resultant force F = F1 + F2 = 2i – 3j

a

Use F = ma

2i – 3j = a = x ..

b

magnitude of acceleration

| a | = 4 + 9 = 13

c

x ..

= 2i – 3j

x . = 2ti – 3tj + c

When t = 0, x . = 0

c = 0

x . = v = 2ti – 3tj

d

speed = | x .

| = 4t2 + 9t2 = t 13

When t = 1, speed = 13

e

velocity gives direction of motion ≈ 303.69°

2

F = 4i + 6j

a

F = ma implies 2a = 4i + 6j

a = 2i + 3j

b

r ..

= 2i + 3j

r . = 2ti + 3tj + c

When t = 0, r . = 0

c = 0

r .

= 2ti + 3tj

c

r = t2i + 3 t 2

2j + c1

As r = 0 when t = 0, c1 = 0

r = t2i + 3 t 2

2j

d

x = t2 and y = 3 t 2

2

y = 3 x

2, x ≥ 0

3

r (t) = 5t2i + 2(t2 + 4)j

a

When t = 0, r (0) = 8j

b

x = 5t2, y = 2t2 + 8

5

2 xt

y = 2 x

5 + 8, x ≥ 0

c

F = ma

r . = 10ti + 4tj

r ..

= 10i + 4j

F = 2(10i + 4j)

= 20i + 8j N

4

r (t) = 5(5 – t2)i + 5(t2 + 2) j

a

r (0) = 25i + 10j

b

x = 5(5 – t2) and y = 5(t2 + 2)

x = 25 – 5t2

5t2 = 25 – x

t2 =

25 x

5


y = 5(

25 x

5 + 2)

= 25 – x + 10

= 35 – x

Also x = 5(5 – t2) x ≤ 25

y = 35 – x for x ≤ 25

c

r .

= –10ti + 10t j

r ..

= –10i + 10j

F = m r ..

= 5(–10i + 10j)

= – 50i + 50j N

5

F1 = 2i + j and F2 = i – 2j

resultant force F = F1 + F2 = 3i – j

a

Using F = ma

3i – j = 2a

3

2i – 1

2j = a

acceleration is 3

2i – 1

2j m/s2

b

r ..

= 3

2i – 1

2j

r .

= 3

2t i – t

2j + c

The particle starts at rest and therefore c = 0

r .

= 3

2t i – t

2j

the velocity is 3

2t i – t

2j m/s

c

r = 3

4t2 i –

t 2

4j + c2

When t = 0, r = 2i – 2j

c2 = 2i – 2j

r = (3

4t2 + 2) i – (

t 2

4 + 2)j

6

a

For the acceleration

r ..

= 3

)3(927 jiji

27i + 9j = 3i + j + 3 r ..

r ..

= 1

3(24i + 8j)

r ..

= 8i + 3

8j m/s

2

b

i.

F = m r ..

=

ji

3

8810

= ji3

8080 N

ii.

| F | =

2

2

3

8080

= 80

310 N

7

r (t) = 2t2i + (t2 + 6) j

a

x = 2t2 and y = t2 + 6

y = x2

+ 6 as 2t2 ≥ 0 for all t, x ≥ 0

y = x2

+ 6 for x ≥ 0

b

r . (t) = 4ti + 2tj


c

speed = | r .

(t) | = 16t2 + 4t2

= 20t2

= 2t 5

2t 5 = 16 5

852

516 t

Hence, the speed of the particle is 16 5 m/s

after 8 seconds.

d

r ..

(t) = 4i + 2j

F = 2(4i + 2j)

= 8i + 4j N

8

F = 1

10(15i + 25j)

a

Using F = m r ..

1

2(3i + 5j) = 10 r

..

1

20(3i + 5j) = r

..

b

r .

(t) = t

20(3i + 5j) + c

r .

(0) = 3i + 5j

r .

(t) = t20

(3i + 5j) + 3i + 5j

= (3 t

20 + 3) i + ( t

4 + 5) j

c

r (t) = (3 t 2

40 + 3t) i + (

t 2

8 + 5t) j + c1

When t = 0, r (t) = 0i + 0j

c1 = 0

r (t) = (3 t 2

40 + 3t) i + (

t 2

8 + 5t) j

r (6) = ( 3 36

40 + 3 6) i + ( 6 2

8 + 5 6) j

= ji2

69

10

207

= 20.7i + 34.5j

d

x = 3 t 2

40 + 3t, y = t

2

8 + 5t

= 3

40(t2 + 40t) = 1

8(t2 + 40t)

t2 + 40t = 40 x

3

thus y = 1

8( 40 x

3)

= 5 x

3

and

t2 + 40t = 40 x

3

t2 + 40t + 400 – 400 = 40 x

3

400

2)20(t = 40 x

3

0 ,203

120040

3

120040

xxt

30 x

Hence,

y =

5x

3 , x – 30

9

y = 3x

r . = 5i + a j a

5 = 3 as velocity gives direction of motion

a = 15

r . = 5i + 15j

The speed in the direction of the y axis is 15

m/s.

The speed of the particle = | r . | = 22

155

= 250

= 5 10 m/s.


Chapter review: multiple-choice questions 1 P = 3(6 i + 8 j ) P = 18 i + 24 j

|P | = (18 )2

+ (24 )2

|P | = 900 |P| = 30

Answer is D

2 R – mg = ma R = ma + mg R = 10 4 + 10 g R = 40 + 10 g R = 138 N

Answer is E

3

R = (8)2

+ (6)2

R = 100 R = 10 N

Answer is B

4

For the 5 kg mass:

5g – T = 5a (1)

For the 3 kg mass:

T – 3g = 3a (2)

(1) + (2):

8a = 2g

a =g

4 m/s

2

Answer is B

5

*Note the positioning of θ

Resolving in the j direction: N + Tcos – mg = 0

Answer is D

6

Resolving down and parallel to the plane: mgsin = ma a = gsin

a =4g

5 m/s

2

Answer is B

8 N

6 N

R

mg

NT

mg

N

= 0


7

Using the cosine rule:

R2

= 102

+ 102

– 2(10 )(10 )cos (120 °)

R2

= 200 – 200 cos (120 °)

R2

= 300

R = 10 3 N

Answer is B

8 Resolving perpendicular to the plane: N + Tsin – W = 0 N = W – Tsin Resolving parallel to the plane: Tcos – F = 0 F = Tcos N = Tcos

=Tcos

N

=Tcos

W – Tsin

Answer is B

9

When an object moves in a circle, it is

constantly changing direction. Because of this

direction change the body is accelerating. This

acceleration is directed inwards towards the

centre of the circle. And according to Newton’s

second law of motion a body experiencing

acceleration must also be experiencing a net

force. This force is referred to as the centripetal

force.

Therefore if the external resultant force on a

body is zero then the body cannot be moving. in a circle.

Answer is B

10

Resolving perpendicular to the plane: R = 9g Resolving parallel to the plane: 54 – FR = ma 54 – R = 18 R = 36

=36

9g

=4

g 0.41

Answer is C

10

10

R

120°

54 NF R

9g

R


Chapter review: short answer questions

1 mass of man = 75 kg

mass of the lift = 500 kg

acceleration of the lift = 2 m/s2

acceleration due to gravity = g m/s2

a The force R exerted by the floor on the man is

given by

R – 75 g = 75 2

or R = 75(g + 2) N

= 885N

b total mass of the lift and the man = 575 kg

T – Mg = Ma

i.e. T – 575 g = 575 2

T = 575(g + 2) N

= 6785 N

2 For the 3 kg mass,

T – 3g = 3a (Newton’s 2nd law) 1

For the 5 kg mass,

5g – T = 5a 2 a Adding 1 and 2 gives 2g = 8a

a =

1

4 g m/s2

b 5 1 gives 5T – 15g = 15a

3 2 gives 15g – 3T = 15a

Subtracting yields 8T – 30g = 0

T = 30

8 g

= 15

4 g (g = acceleration due to gravity)

3 m = mass of the skier

Resolving in i direction: R – mg sin = ma 1

Resolving in j direction: R = mg cos 2

Substituting 2 into 1 gives

mg cos – mg sin = ma

a = g(sin – cos )

R

mg

T

3 kg 5 kg

T T

mg

i

j R R


4 By Newton’s second law,

a 100 – R = ma

i.e. 100 – 0.4(10g) = 10a

a = (10 – 0.4g) m/s2

b If another block of mass 10 kg is placed on top of the first one, then

m = 20 kg

100 – (20g) = 20a

a = (5 – 0.4g) m/s2

5 m = 5 kg, F = 20

(t + 1)2 N at t seconds

a F = ma

at t seconds, a = 20

(t + 1)2 1

5

= 4

(t + 1)2 m/s2

b velocity v = a

1

1dt

= 4

(t + 1)2

1

1dt

= 4(t + 1)–2

1

1dt

= 4(t + 1)1

1 + c1

= 4

t + 1 + c1

Since the body starts from rest, v (0) = 0

4

t + 1 + c1 = 0

or c1 = 4

v = 4

t + 1 + 4

= 4t

t + 1 m/s

c displacement d = v

1

1dt

= 4

t + 1 + 4

1

1dt

= – 4 loge (t + 1) + 4t + c2, t + 1 0

At t = 0, d = 0,

– 4 loge (0 + 1) + 4(0) + c2 = 0

c2 = 0

d = (4t – 4 loge (t + 1)) m

6 mass of the car m = 1000 kg

initial velocity u = 60 km/h

mg

R

R 100 N


= 60

3.6 m/s

final velocity v = 24

3.6 m/s

t = 5 s

a = v u

t

= 24

3.6 60

3.6

5

= 24 60

3.6 5

= – 2 m/s2

the retarding force using F = ma is

1000 – 2 = – 2000 N

i.e., the retarding force acting in a direction opposing motion is 2000 N.

7 Since the body is in limiting equilibrium, all the

forces balance each other. Resolving mg in the

directions of i and j, we see that mg sin acts

along the inclined plane and mg cos acts

perpendicular to it. for equilibrium, R = mg cos

and R = mg sin ( = coefficient of friction)

= mg sin

R

= mg sin

mg cos

= tan

Then the inclination of the plane is increased to

the body will begin to slide down.

Let the acceleration be a.

mg sin – tan mg cos = ma

(since = tan and R = mg cos )

mg sin – sin

cos mg cos = ma

a = g sin cos g cos sin

cos

= g

cos sin( – )

8 When the car is cruising down the slope, the velocity

is constant.

Let resistance = R,

a = 0

a 1000 g sin – R = ma = 0

R = 1000 g 1

20

= 50g N

= 490 N

R

mg

R

i j

R

mg

R

N

1000g

R


b maximum acceleration = 1 m/s2

Let the force be F when going up.

F – mg sin – R = ma

F = 1000g 1

20 + 50g + 1000 1

= (1000 + 100g) N

= 1980 N

9 The forces acting on the parcel as the conveyor belt

moves upwards are as shown.

a At the instant it is about to slip down,

a = g

4

R – mg sin 30° = ma R = mg cos 30°

mg 3

2 – mg

1

2 = m

g

4

3

2 =

1

4 +

1

2

= 3

4

2

3

= 3

2

b When the belt is stopped, the parcel is moving at 7 m/s.

Now frictional force R acts down the plane and the parcel slides up some distance

before coming to stop.

mg sin 30° + R = ma

i.e. ma = mg

2 +

3

2 mg

3

2

a = g

1

2 +

3

4

= 5

4 g

Now using v2 – u2 = 2as, find the distance s.

0 – 49 = 2

5g

4 s

s = 49 2

5g

= 98

5g m

= 2 m

10 The maximum possible tension is 400 kg wt.

a When the particle is hauled upwards,

T – 320g = 320a

i.e. 400g – 320g = 320a

80g = 320a

a = 80g

320

= g

4 m/s2

maximum acceleration= g

4 m/s2.

mg 30°

R R

mg

T


b If a particle of mass 480 kg is to be lowered by the same rope, the maximum tension is

again 400 kg wt.

Now 480g – T = 480a

or 480g – 480a = T

480g – 480a ≤ 400g

i.e. 80g ≤ 480a

a ≥ 80g

480

i.e. a ≥ g

6 m/s2

11 Given that F = 3 + 6x

F = ma implies 3 + 6x = 3x..

1 + 2x = x..

1 + 2x = d(

1

2 v2 )

dx since a =

d( 1

2 v2 )

dx

Integrating both sides:

x + x2 + c = 1

2 v2

When x = 0, v = 2

c = 2

1

2 v2 = x + x2 + 2

v = 2(x + x2 + 2)

When x = 2, v = 2(2 + 4 + 2)

= 16

= 4 m/s

12 m = 3 kg, F = 3i + 6j N, v (0) = i + 2j

a F = ma implies a = 3i + 6j

3

= i + 2j

b i a = i + 2j

v (t) = t i + 2t j + c1

Now v (0) = i + 2j,

i + 2j = c1

v (t) = t i + 2t j + i + 2j

= (t + 1)i + (2t + 2)j

= (t + 1)(i + 2j)

ii Speed = | v |

= (t + 1) 1 + 4

= 5 (t + 1)


c v (t) = (t + 1)(i + 2j)

r (t) =

t2

2 + t (i + 2j) + c2

Since the particle is initially at the origin, r (0) = 0,

c2 = 0

i.e. r (t) =

t2

2 + t (i + 2j)

d The equation of the straight line in which the particle moves is given by

r = k(i + 2j), k ≥ 0

or y = 2x, x 0

13 Using s = ut + 1

2 at2, t = 20 and s = 500,

500 = 20u + 200a

25 = u + 10a 1

Using s = ut + 1

2 at2, t = 50 and s = 1000,

1000 = 50u + 1250a

20 = u + 25a 2

1 2 gives 5 = 15a

a = 1

3 (which shows deceleration)

Substituting in 1 , 25 = u 10

3

u = 85

3

Using v = u + at, and a = 1

3 , u =

85

3 and t = 50,

v = 85

3

1

3 50 =

35

3

Using v2 = u2 + 2as, and a = 1

3 , u =

35

3 and v = 0,

0 =

35

3

2

2 1

3 s

s = 1225

9

3

2 = 204

1

6

The train will travel a further 204 1

6 kilometres before coming to rest.

14 a = v u

t

= 15 0

60

= 1

4 m/s2

F = ma

= 9000 1

4

= 2250 N


15 The train is travelling with uniform velocity on level ground. As it begins the ascent, the

initial velocity is u = 20 m/s.

F = R, where F = force of engine, R = resistance.

Also, on the incline, N = mg cos , since N and mg cos balance each other.

F – R – mg sin = ma

3

50 mg = ma

or a = 3

50 g

Now v2 = u2 + 2as

gives 0 = (20)2 – 2 3

50 g s

s = 50 202

6g

= 10 000

3g m

16 Let the mass of the body in the lift be m kg and the force

of the lift on the body be F.

Then F – mg = mf

(since the lift moves upwards)

i.e. F = m (g + f) N

17 Initial velocity of the bullet = 200 m/s

At s = 10 cm (= 0.1 m), its velocity is 0

Using v2 – u2 = 2as

gives 0 – 2002 = 2 a 0.1

a = 2002

2 0.1

= –200 000

= –2 105 m/s

If the board is 5 cm thick, then s = 0.05 m

and v2 – 2002 = 2 –2 105 0.05

v2 = 2002 – 2 2 105 0.05

= 20 000

v = 20 000 since v 0

= 100 2 m/s

18 Let the mass of the body in the lift be m kg.

When the lift accelerates upwards,

T – mg = ma

10g – mg = ma 1

mg

N

R F R

F

N0

mg

F

mg

T

mg


When the lift accelerates downwards,

mg – T = m (2a)

mg – 7g = 2ma 2

a 2 – 2 1 gives

– 27g + 3mg = 0

m = 27g

3g

= 9 kg

weight of the particle is 9 kg wt.

b 1 + 2 gives 3g = 3ma

a = 3g

3m

= g

9 m/s2

19 By Newton’s second law,

for particle A, m1g – T = m1a 1

and for particle B, T – m2g = m2a 2

a Adding 1 and 2 gives

m1g – m2g = m1a + m2a

a = (m1 m2)g

m1 + m2 m/s2

b From 1 and 2 , m1g T

m1 =

T m2g

m2

m1m2g – m2T = m1T – m1m2g

T(m1 + m2) = 2m1m2g

T = 2 m1 m2 g

m1 + m2 N

20 The forces involved are represented in the sketch.

Friction can be neglected as the surface is smooth.

We have T = m2a 1

and m1g – T = m1a 2


m1g = (m1 + m2)a

a = m1 g

m1 + m2 m/s2

b Therefore A is pulled along the table towards the pulley with acceleration a.

Now T = m2a

= m1 m2 g

m1 + m2 N

T T

A B

m2g

T

T

A

B

m1g

m2g

R

m1g


21 Given that m1 > m2, then the tension will cause A to

be pulled up the plane.

Both A and B will move with the same acceleration,

say a.

For A, T – m2g sin = m2a 1

For B, m1g – T = m1a 2


m1g – m2g sin = (m1 + m2)a

a = g(m1 m2 sin )

m1 + m2

particle A will move up the plane with an acceleration

a = g(m1 m2 sin )

m1 + m2 m/s2

b From 1 and 2 we have

T m2g sin

m2 =

m1g T

m1

m1T – m1m2g sin = m1m2g – m2T

T(m1 + m2) = m1m2g + m1m2g sin

T = m1 m2 g (1 + sin )

m1 + m2 N

22 Resolving the weight mg along the plane and

perpendicular to the plane, we get:

R = mg cos

and by Newton’s second law:

mg sin – R = ma

ma = mg sin – mg cos

a = g(sin – cos )

23 Friction can be neglected as the table is smooth.

For B, 6g – T = 6a 1

For A, T = 10a 2


6g = 16a

a = 3

8 g m/s2

b From 2 we get T = 30

8 g

= 15

4 g N

A

T

T R

B

m1g m2g

mg

R

R


c Two forces acting on the pulley are shown in the

diagram. Since the forces are equal in magnitude, they

are represented by the sides of an isosceles triangle.

Further they are at right angles.

By the triangle law, the resultant force is

T2 + T2 = 2T

in the direction of 45° to the horizontal, i.e., 15 2

4 g N.

d Using s = ut + 1

2 at2, 1 = 0 +

1

2

3

8 g t2

t2 = 16

3g

t = 4

3g since t 0

0.74 s

e From part d, it takes B, and also A, 4

3 g seconds to travel 1 metre. Now as B has the

reached the floor, there is no longer any tension in the string, so A will travel the next 1

metre with constant velocity (there are now no horizontal forces acting on A as the

horizontal table is smooth).

For the constant velocity, use 2 2

2v u as where 3

0, and 18

u a g s

2 30

4

3

2

v g

gv

Now using 3

12

2

3

s vt

gt

tg

So the time it takes A to reach the edge is

4 2 6

3 3 3g g g

24 Friction can be neglected for the smooth surface.

For particle A, forces along the surface are 10g sin

and T in opposite directions.

Let a m/s2 be the acceleration of A down the plane.

10g sin – T = 10a 1

For B, T – 3g = 3a 2

a If = 60°, 1 + 2 give

45°

T

T

T

45° 2T

T

A

3g 10g

T

T R

B


10 sin 60° – 3g = 13a

a = 10g sin 60 3g

13

= g(5 3 3)

13 m/s2

b T – 3g = 3a

T = 3g + 3 g(5 3 3)

13

= 3g(5 3 + 10)

13 N

25 Given that

mA = 5 kg, mB = 3 kg, = 0.2

The forces acting on each body are shown in

the diagram.

For A, T – R = 5a

i.e. T – 0.2 5g = 5a

T g = 5a 1

For B, 3g – T = 3a 2

a 1 + 2 gives 3g – g = 8a

a = 2g

8

= g

4 m/s2

b B is hanging 1 metre from the ground.

Using v2 – u2 = 2as, we can find v when s = 1.

v2 – 0 = 2 g

4 1

v2 = g

2

or v = g

2 m/s

The velocity of A as B reaches the ground is g

2 m/s towards the pulley.

c Once B reaches the ground, T = 0. However, A continues to move on with

R N acting opposite to the direction of motion.

Now for A, u = g

2 , v = 0 and a =

R

m

= 0.2 5g

5

= – 0.2g

Using v2 = u2 + 2as, 0 = g

2 + 2 – 0.2g s

0.4s = 1

2

A R

T

3g

R

T 5g

B

1 m


or s = 1

0.8

= 5

4 m

26 If the two forces each equal to P N act at a point, inclined at 120°, we can use the triangle law

to find the resultant.

It can be seen that the forces make an angle of 60° in

the triangle. But the two forces being equal make the

remaining two angles of the triangle equal 60°.

Thus, we have an equilateral triangles and hence the

resultant R must also have a magnitude of P N.

Alternative method

As before, we can resolve along the direction of one

force and a perpendicular direction.

This gives

P + P cos 120° = P(1 + cos 120°)

and P cos 30°

R = P2(1 + cos 120)2 + P2 cos2 30°

= P 1

= P

27 The body is in limiting equilibrium. Thus the

resolved part of its weight along the plane is just

balanced by the friction R.

i.e. 10g sin 30° = R

= 10g cos 30°

= 10g sin 30

10g cos 30

= tan 30°

If the plane is now raised to an incline of 60°, the

body will start sliding down the plane. Let F be the

force required to prevent sliding.

Then 10g sin 60° = R + F

= 10g cos 60° + F

F = 10g sin 60° – tan 30° . 10g cos 60°

= 10g[sin 60° – tan 30° cos 60°]

= 10g

3

2 –

1

3

1

2

= 10g

3 1

2 3

= 10g

3 N

R

P N

60°

P N

R P

P cos 120° + P

P cos 30°

120°

30° 10g

R R

60°

10g

R

R + F


28 The particle rests in equilibrium.

Let T be the tension in the string, now inclined at

60° to the vertical. Hence, the angles between the

three forces are 90°, 150° and 120° as shown in the

diagram.

By Lami’s theorem:

P

sin 120 =

T

sin 90 =

5g

sin 150

P = 5g

sin 150 sin 120°

= 5g

3

2

1

2

= 5g 3 N

29 The final position of the particle in equilibrium is as

shown in the diagram. From the right triangle,

cos = 2

2.5

= 4

5

= cos1

4

5

The three angles between the forces are 90°, 180 and 90 + .

P

sin (90 + ) =

T

sin 90 =

2g

sin (180 )

P

cos = T =

2g

sin

P = 2g

sin cos

= 10g

3

4

5 since sin =

3

5

= 8g

3 N

and T = 2g

sin

= 10g

3 N

150° O

120°

5g

P N T

2g

P

2·5 m

2 m

O


30 The forces may be represented by a triangle of

vectors.

T1 = 5g cos

= 5g 5

13

= 25g

13 N

T2 = 5g sin

= 5g 12

13

= 60g

13 N

The tension in the two strings is 25g

13 N and

60g

13 N.

T1

T2

12

5

13

5g


Chapter review: extended-response questions

1 a 42 – T – 4g = 0

T = 42 – 4g

2.8 N

b 42 – 4g = 4x..

x..

= 42 4g

4

= 0.7 m/s2

T 4g

42 N

c i For constant acceleration, s = ut + 1

2 at2

5 = 0 t + 1

2

21 2g

2 t2

20

21 2g = t2

t = 20

21 2g since t 0

3.78

It takes approximately 3.78 seconds to reach the surface.

ii The velocity of the buoy as it reaches the surface is given by

v = u + at

= 0 + 21 2g

2

20

21 2g

= 5(21 2g) m/s

2.65

The velocity of the buoy is approximately 2.65 m/s when it reaches the surface.

d When the buoy leaves the water it is acting under only gravitational force, therefore use

v2 = u2 + 2as

Now u = 5(21 2g), a = –g, v = 0,

0 = 5(21 2g) – 2gs

s = 5(21 2g)

2g

0.357

The buoy reaches a height of approximately 0.357 metres.

2 a i For the 2.8 kg mass

T2 – 2.8g = 2.8a 1

For the 2.2 kg mass

T1 + 2.2g – T2 = 2.2a 2

For the 3 kg mass

3g – T1 = 3a 3

T1

2.8g

2.2g

3g

T2 T2

T1


Substitute from 3 in 2 for T1 ,

3g – 3a + 2.2g – T2 = 2.2a

5.2g – T2 = 5.2a 4

Add 1 and 4 ,

T2 – 2.8g + 5.2g – T2 = 8a

2.4g = 8a

0.3g = a

ii Substitute in 3 ,

3g – T1 = 3 0.3g

3g – 0.9g = T1

T1 = 2.1g

b x..

= 0.3g

x. = 0.3gt + c

When t = 0, x. = 0,

c = 0

x. = 0.3gt

When t = 1.5, x. = 0.3 1.5g

= 0.45g

2.8g – T = 2.8a

and T – 2.2g = 2.2a

0.6g = 5a

a = 0.12g

(in the opposite direction to the initial velocity)

2.2g

T T

2.8g Using v2 = u2 + 2as, 0 = u2 – 2 0.12g s

with u = 0.45g s = (0.45g)2

0.24g

= 8.26875

The 2.2 kg mass falls a further distance of 8.27 metres, correct to two decimal places.

3 Let F be the tractive force of the engine.

a i Resistance force for the engine = 50

1000 60 000

= 3000 N

Resistance force for the truck = 30

1000 12 000

= 360 N

Resolving parallel to the plane with the whole

system:

F – 72 000g sin – 3000 – 360 = 0

F = 72 000g 1

200 + 3000 + 360

= 360g + 3360 6888 N

F R2

T

360

R1

3000

T

60 000g

12 000g


ii Let T N be the tension in the coupling.

Resolving parallel to the plane for the engine:

F – T – 60 000g 1

200 – 3000 = 0

T = (360g + 3360) – 60 000g

200 – 3000

= 60g + 360

948 N

b i F – 72 000g sin – 3000 – 360 = 72 000 0.1

F = 360g + 10 560

14 088 N

ii T = (360g + 10 560) – 60 000 0.1 – 60 000g

200 – 3000

= 60g + 1560

2148 N

4 a i 0.4g – T = 0.4x..

1

T – 0.4 0.2g = 0.2x..

2

1 + 2 gives

0.4g – 0.08g = 0.6x..

x.. =

0.32g

0.6

= 8g

15

5.23 m/s2

R

T

0.4g

R

T 0.2g

ii 0.4g – 0.4 8g

15 = T

T = 14g

75

1.83 N

b Consider the system before it strikes the floor.

v2 = u2 + 2as

v2 = 2 1.5 8g

15

v2 = 8g

5

For the particle on the table, after the 0.4 kg particle hits the floor,

T – R = mx..

T = 0, = 0.4, R = 0.2g,

–0.2g 0.4 = 0.2x..

–0.4g = x..


Use v2 = u2 + 2as

When v = 0, 0 = 8g

5 – 2 0.4g s

s = 8g

5 0.8g

= 2

The particle goes a further two metres after the 0.4 kg particle hits the floor.

5 a i x..

= –(a + bv2)

v dv

dx = –(a + bv2)

dv

dx =

(a + bv2 )

v

ii From i, dx

dv =

v

a + bv2

Let w = a + bv2, dw

dv = 2bv and

dx

dv =

v

w

x = v

w

1

1dv

= 1

2b 1

w

1

1dw

= 1

2b loge (a + bv2 ) + c, a + bv2 0

When x = 0, v = u,

c = 1

2b loge (a + bu2 )

x = 1

2b (loge (a + bu2 ) – loge (a + bv2 ))

x = 1

2b loge

a + bu2

a + bv2

The train comes to rest when v = 0,

i.e. x = 1

2b loge

a + bu2

a

= 1

2b loge

1 +

bu2

a

b i dv

dt = –(a + bv2 )

dt

dv =

1

a + bv2

= 1

b( a

b + v2 )

= 1

ab

a

b

a

b + v2

t = 1

ab tan–1

b

a v + c


When t = 0, v = u,

c = 1

ab tan–1

b

a u

t = 1

ab tan–1

b

a u –

1

ab tan–1

b

a v

When v = 0, t = 1

ab tan–1

b

a u

ii If b = 0.005, a= 2 and u = 25,

then t = 1

0.01 tan–1

0.005 25

2

= 10 tan–1 (1.25) 8.96

It takes 8.96 seconds for the train to stop.

6 a mg – 0.02mv2 = mx..

x..

= g – 0.02v2

x..

= g – v2

50 =

50g v2

50

v dv

dx =

50g v2

50

dv

dx =

50g v2

50v

dx

dv =

50v

50g v2

Let w = 50g – v2

dw

dv = –2v

x = –251

w

1

1dw

= –25 loge (50g – v2 ) + c, 50g – v2 0

When x = 0, v = 0,

c = 25 loge (50g)

and x = 25 loge

50g

50g v2

b e

x

25 =

50g

50g v2

v = 50g

1 e

x

25

c v

0 x

v = 50g


7 a T – 200g sin 60° – 200g cos 60° = 0

200g – 100g 3 – 100g = 0

= 200 100 3

100

= 2 – 3

0.2679

60°

Mg

200g

T T

50 m

b If the value of M is 200, the crate is on the point of moving up the plane. Consider the

crate being on the point of moving down the plane.

Then Mg sin 60° – T – (2 – 3 ) R = 0

Mg 3

2 – 200g – (2 – 3 ) Mg

1

2 = 0

Mg

2 ( 3 – (2 – 3 )) = 200g

Mg

2 (2 3 – 2) = 200g

M = 100( 3 + 1) kg

The crate will remain stationary for 200 ≤ M ≤ 100( 3 + 1)

c i Let M = 150. The mass will move up the plane.

200g – T = 200x.. 1

T – 150gcos 30° – 150g sin 30° = 150x..

200g – 75g 3 – 75g = 350x..

200g – 75g 3 – 75g(2 – 3 ) = 350x..

200g – 150g = 350x..

200g 60°

150g

T T

50g = 350x..

g

7 = x

..

The acceleration is g

7 m/s2.

ii 200g – T = 200x.. (from 1 )

T = 200g – 200 g

7

= 1200g

7

The tension is 1200g

7 N.

iii When the rope breaks, the 200 kg mass has a speed of g

7 2 =

2g

7 m/s and has

travelled 1

2 at2 =

2g

7 metres in the two seconds.


Using v2 = u2 + 2as, with u = 2g

7, a = g and s = 50 –

2g

7,

v2 =

2g

7

2

+ 2 g

50 –

2g

7

v2 = 4

49 g2 + 100g –

4

7 g2

v 30.54, since v 0

The speed of the 200 kg weight when it hits the ground is 30.54 m/s.

8 a v = 125(1 – e–0.1t )

dv

dt = 12.5e–0.1t

b i P – 20v = 250 12.5e–0.1t

P = 3125e–0.1t + 20v

= 2500(1 – e–0.1t ) + 3125e–0.1t

= 2500 + 625e–0.1t

= 625 (4 + e–0.1t)

ii P = 20v + 3125e–0.1t

= 20v + 3125

1 –

v

125

= 20v + 3125 – 25v

= 3125 – 5v

= 5(625 – v)

iii When v = 20, P = 3025 N

iv When t = 30, P = 2500 + 625e3

2531.117 N

c

0

3125

2500

t

P

9 a Resolving perpendicular to the plane:

T sin + R = Mg cos

R = (Mg cos – T sin

b Resolving parallel to the plane:

Mg

R T

T cos – Mg sin – 0.1R = 0

T cos – Mg sin – 1

10 (Mg cos – T sin ) = 0

T = 0.1 Mg cos + Mg sin

cos + 0.1 sin N


c i T = Mg (0.1

3

5 + 4

5 )

cos + 0.1 sin

= Mg

5 4.3

cos + 0.1 sin

= 43 Mg

50(cos + 0.1 sin )

= 8.6 g

cos + 0.1 sin N

ii Let y = cos + 0.1 sin

dy

d = –sin + 0.1 cos

dy

d = 0 implies tan =

1

10

= 0.09967c

or = 5.71° = 5°43´

This maximises cos + 0.1 sin , and minimises 8.6g

cos + 0.1 sin .

iii Minimum T = 86 101

101 g N, since sin =

1

101 and cos =

10

101 .

d Resolving the forces produces the same expression to be maximised as before, i.e.,

clearly no effect. So θ = 5°43´

10 If sin = 5

13 , then cos =

12

13 .

a i N = 50g cos

= 50g 12

13

= 600g

13 N

50g

N

ii Friction force = N

= 0.1 N

= 60g

13 N

b 50g sin – 60g

13 = 50x

..

250g

13 –

60g

13 = 50x

..

19g

65 = x

..

The acceleration down the plane is 19g

65 m/s2.


c i Constant acceleration

Use v2 = u2 + 2as

= 2 19g

65 10

= 76g

13

v 7.569 m/s

The speed of the particle at the bottom of the plane is 7.569 m/s.

ii v = at

76g

13 =

19g

65 t

t 2.64

The time taken to reach the bottom is 2.6 seconds.

d i 300 – 250t + 50g sin – 5g cos = 50x..

300 – 250t + 250g

13 –

60g

13 = 50x

..

x.. 8.86 – 5t

ii Let dv

dt = 8.86 – 5t

50g

F

v = 8.86t – 5t2

2 + c

When t = 0, v = 0,

c = 0

v = 8.86t – 5t2

2

i.e. dx

dt = 8.86t –

5t2

2

x = 8.86t2

2 –

5t3

6 + c

When t = 0, x = 0,

c = 0

x = 4.43t2 – 5

6 t3

When x = 10, 10 = 4.43t2 – 5

6 t3

Using a CAS calculator to find t, the particle reaches the bottom of the slope when

t = 1.865.

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