Chapter 8 Exercise 8A - mathsgrinds.ie (iv) Rectangle: Centre of gravity is at (8, 2) Area of...

1

Chapter 8 Exercise 8A

Q. 1. (a) R = 4 _

› j + 5

_

› j +

_

› j = 10

_

› j N

4(0) + 5(50) + 1(100) = 10(x)

⇒ x = 35 cm

Answer: 35 cm from P

(b) R = 2 _

› j +

_

› j + 2

_

› j = 5

_

› j

2(0) + 1(2) + 2(5) = 5x

⇒ x = 2.4 m

Answer: 2.4 m from P

(c) R = _

› j − 7

_

› j +

_

› j = −5

_

› j

1(0) − 7(1) + 1(2) = −5(x)

⇒ x = 1

Answer: 1 m from P

(d) R = 3 _

› j − 9

_

› j + 3

_

› j = −3

_

› j

3(1) − 9(3) + 3(5) = −3x

⇒ x = 3

Answer: 3 m from P

(e) R = _

› j − 2

_

› j − 3

_

› j +

_

› j = −3

_

› j

1(0) − 2(2) − 3(5) + 1(6) = −3(x)

⇒ x = 4 1 __

3

Answer: 4 1 __

3 m from P

Q. 2. R = 2W + W + 3W = 6W

2W(0) + W(3) + 3W(6) = 6W(x)

⇒ x = 3 1 __

2

Answer: 3 1 __

2 M and 2 1 __

2

M from the ends.

Q. 3. Forces:

1 m

7W

3W

Resultant:

x

R = 7W + 3W = 10W

7W(0) + 3W(1) = 10W(x)

⇒ x = 3 ___

10 m

= 30 cm = Answer

Q. 4. Let the length of the plank be 1.

Let d = the distance from the resultant’s

line of action from the left-hand end.

R = W + x

∴ W(0) + x(1) = (W + x)d

⇒ d = x ______

W + x

The remainder is 1 − x ______

W + x = W

______

W + x

The ratio of these parts = x ______

W + x : W

______

W + x

= x : W

Q. 5. (i) R = 4W + W + KW = (5 + K)W

Taking moments about p.

4W(0) + W(1) + KW(2) = (5 + K)W ( 7 __

8 )

⇒ 1 + 2K = 7(5 + K)

________

8

⇒ 8 + 16K = 35 + 7K

⇒ K = 3

(ii) 4W(0) + W(1) + KW(2) = (5 + K)W ( 11 ___

10 )

⇒ 1 + 2K = 11(5 + K)

_________

10

⇒ 10 + 20K = 55 + 11K

⇒ K = 5

Exercise 8B

Q. 1. 3N at (2, 1)

2N at (4, 3) = 6N at (x, y)

1N at (10, 9) }

3(2) + 2(4) + 1(10) = 6(x)

⇒ x = 4

3(1) + 2(3) + 1(9) = 6(y)

⇒ y = 3

Answer: (4, 3)

Q. 2. 1N at (1, 1)

2N at (1, 7) = 10N at (x, y)

3N at (3, 1) } 4N at (2, 3)

Q

Q

E

Q

Q

2

1(1) + 2(1) + 3(3) + 4(2) = 10(x)

⇒ x = 2

1(1) + 2(7) + 3(1) + 4(3) = 10(y)

⇒ y = 3

Answer: (2, 3)

Q. 3. 3(4) + 2(9) = 5x

⇒ x = 6

3(1) + 2(−9) = 5y

⇒ y = 3

Answer: (6, −3)

3(4) + 2(9) + 1(x) = 6(6)

⇒ x = 6

3(1) + 2(−9) + 1(y) = 6(−2)

⇒ y = 3

Answer: (6, 3)

Q. 4. W(3) + 2W(12) = 3W(x)

⇒ x = 9

W(1) + 2W(19) = 3W(y)

⇒ y = 13

Answer: (9, 13) = g

|pg| = √__________________

(9 − 3)2 + (13 − 1)2

= √____

180

= 6 √__

5

|gq| = √____________________

(12 − 9)2 + (19 − 13)2

= √___

45 = 3 √__

5

|pg| : |gq| = 6 √__

5 : 3 √__

5

= 2 : 1

Q. 5. 2N at (1, 2)

3N at (1, 7) = 10N at (x, y)

5N at (5, −1) }

2(1) + 3(1) + 5(5) = 10(x)

⇒ x = 3

2(2) + 3(7) + 5(−1) = 10(y)

⇒ y = 2

Answer: 3 _

› i + 2

_

› j

2N at (1, 2)

3N at (1, 7) = 12N at (4, 1)

5N at (5, −1)

2N at (x, y) }

2(1) + 3(1) + 5(5) + 2(x) = 12(4)

⇒ x = 9

2(2) + 3(7) + 5(−1) + 2(y) = 12(1)

⇒ y = −4

Answer: (9, −4)

Q. 6. 1(4) + 2(1) + 3(k) = 6 ( 2 1 __

2 )

k = 3

1(1) + 2h + 3(1) = 6(4)

h = 10

Q. 7. Centroid of triangle

pqr = ( 2 + 5 + 3 _________

3 ,

1 + 3 − 1 _________

3 )

= ( 10 ___

3 , 1 )

M at (2, 1)

M at (5, 3) = 3M at (x, y)

M at (3, −1) }

M(2) + M(5) + M(3) = 3M(x)

⇒ x = 10

___

3

M(1) + M(3) + M(−1) = 3M(y)

⇒ y = 1

The centre of gravity is at ( 10 ___

3 , 1 )

which is the centroid.

M(2) + M(5) + M(3) + 2M(x) = 5M(2)

⇒ x = 0

M(1) + M(3) + M(−1) + 2M(y) = 5M(1)

⇒ y = 1

Answer: (0, 1)

Q

Q

3

Q. 8. (i) W at (6, 5)

2W at (7, −1) = 10W at (x, y)

3W at (2, 11)

4W at (6, 1) }

W(6) + 2W(7) + 3W(2) + 4W(6) = 10W(x) … divide by W

⇒ 10x = 50

⇒ x = 5

W(5) + 2W(−1) + 3W(11) + 4W(1) = 10W(y) … divide by W

⇒ 10y = 40

⇒ y = 4

⇒ The centre of gravity is at (5, 4)

(ii) W at (6, 5)

2W at (7, −1) = (6 + k)W at (11, y)

3W at (2, 11)

kW at (6, 1) }

W(6) + 2W(7) + 3W(2) + kW(6) = (6 + k)W(11) … divide by W

⇒ 66 + 11k = 26 + 6k

⇒ 5k = −40

⇒ k = −8

W(5) + 2W(−1) + 3W(11) − 8W(1) = −2W(y) … divide by W

⇒ 2y = −28

⇒ y = −14

Q. 9. 2N at (x, 5)

3N at (11, y) = 10N at (6, −5)

5N at (5, −6) }

2(x) + 3(11) + 5(5) = 10(6)

⇒ 2x + 58 = 60

⇒ x = 1

2(5) + 3(y) + 5(−6) = 10(−5)

⇒ 3y − 20 = −50

⇒ y = −10

Q. 10. 1N at (7, 8)

4N at (1, y) = (7 + x)N at (4, 2)

xN at (5, 2)

2N at (6, −3) }

1(7) + 4(1) + x(5) + 2(6) = (7 + x)(4)

⇒ 5x + 23 = 28 + 4x

⇒ x = 5

1(8) + 4(y) + 5(2) + 2(−3) = 12(2)

⇒ 4y + 12 = 24

⇒ y = 3

Q. 11. 3W at (30, 20)

2W at (20, 10) = 10W at (x, y)

5W at (50, 20) }

3W(30) + 2W(20) + 5W(50) = 10W(x)

… divide by 10W

⇒ x = 38

3W(20) + 2W(10) + 5W(20) = 10W(y)

… divide by 10W

⇒ y = 18

⇒ Centre of gravity is at (38, 18)

Q 9 2N t ( 5) 1(8) 4( ) 5(2) 2( 3) 12(2)

Q

4

Q. 12. 3N at ( − 7 __

3 , 2 )

4N at (1, 7)

5N at (3, 7) = 18N at (x, y)

6N at (4, −4) }

3 ( − 7 __

3 ) + 4(1) + 5(3) + 6(4) = 18(x)

⇒ 18x = 36

⇒ x = 2

3(2) + 4(7) + 5(7) + 6(−4) = 18(y)

⇒ 18y = 45

⇒ y = 2.5

⇒ Centre of gravity is at (2, 2.5)

Q. 13. 5 at (3, −1)

8 at (4, 2)

3 at (−1, 5) = 18 at (x, y)

2 at (2, −6) }

5(3) + 8(4) + 3(−1) + 2(2) = 18(x)

⇒ 18x = 48

⇒ x = 8

__

3

5(−1) + 8(2) + 3(5) + 2(−6) = 18(y)

⇒ 18y = 14

⇒ y = 7 __

9

⇒ Centre of gravity is at ( 8 __

3 , 7 __

9

)

Exercise 8C

Q. 1. (i) 3

2

1

1 2 3

3 at ( 1 __

2 , 1 1 __

2

) 1 at ( 1 1 __

2

, 1 __

2 ) }

= 4 at (x, y)

3 ( 1 __

2 ) + 1 ( 1 1 __

2

) = 4x

⇒ x = 3

__

4

3 ( 1 1 __

2 ) + 1 ( 1 __

2

) = 4y

⇒ y = 1 3

__

4

Answer: ( 3 __

4 , 1 1 __

4

)

tan A = 3 _ 4 ___

1 3 _ 4

= 3

__

7

= 0.4286

∴ A = 23° 12’

(ii)

1

1

2

3

4

2 3 4

4 at ( 1 __

2 , 2 )

2 at ( 2, 1 __

2 ) = 8 at (x, y)

2 at ( 2, 3 1 __

2 )

} 4 ( 1 __

2

) + 2(2) + 2(2) = 8(x)

⇒ x = 1 1 __

4

4(2) + 2 ( 1 __

2 ) + 2 ( 3 1 __

2

) = 8(y)

⇒ y = 2

Answer: ( 1 1 __

4 , 2 )

tan A = 1 1

_ 4 ___

2

= 5

__

8

= 0.625

∴ A = 32°

(iii)

1

1

2

3

4

2 3 4

4 at ( 1 __

2 , 2 )

1 at ( 1 1 __

2 , 2 1 __

2

) = 9 at (x, y)

4 at ( 2 1 __

2 , 2 )

} 4 ( 1 __

2

) + 1 ( 1 1 __

2 ) + 4 ( 2 1 __

2

) = 9(x)

⇒ x = 3

__

2

A114–

34–

114– g

A

1 14–

2

2 g

5

4(2) + 1 ( 2 1 __

2 ) + 4(2) = 9(y)

⇒ y = 37

___

18

Answer: ( 3 __

2 ,

37 ___

18 )

tan A = 3 _ 2 __

35

__ 18

= 27 ___

35

= 0.7714

∴ A = 37° 39’

(iv)

3

7

(7,3)

(7,0)(0,0)

(13,3)

Triangle piece:

Centroid is at

( 7 + 7 + 13 ___________

3 ,

3 + 3 + 0 _________

3 ) = (9, 2)

Area is 1 __

2 (6)(3) = 9 square units

21 at ( 3 1 __

2 , 1 1 __

2

) 9 at (9, 2) } = 30 at (x, y)

21 ( 3 1 __

2 ) + 9(9) = 30(x)

⇒ x = 5.15

21 ( 1 1 __

2 ) + 9(2) = 30(y)

⇒ y = 1.65

Answer: (5.15, 1.65)

tan A = 5.15

____

1.35

= 3.8148

A = 75° 19’

Q. 2.

(6,0)(3,0)

0

Larger Circle: Area = 36p, Centre = (0, 0)

Smaller Circle: Area = 9p, Centre = (3, 0)

Remainder: Area = 27p,

Centre of gravity is at (x, y)

9p at (3, 0) = 36p at (0, 0)

27p at (x, y) } ∴ 9p(3) + 27p(x) = 36p(0) ⇒ x = −1

9p(0) + 27p(y) = 36p(0) ⇒ y = 0

The distance (0, 0) to (−1, 0)

Answer: 1 cm

Q. 3. Full Square: Area = 4 × 4 = 16 m2

Small Square: Area = 1 × 1 = 1 m2

Remainder: Area = 16 − 1 = 15 m2

1 at (3.5, 3.5) = 16 at (2, 2)

15 at (x, y) } 1(3.5) + 15(x) = 16(2)

⇒ 3.5 + 15x = 32

⇒ 15x = 28.5

⇒ x = 1.9

1(3.5) + 15(y) = 16(2)

⇒ y = 1.9

⇒ Centre of gravity of the remainder is at

(1.9, 1.9)

Q. 4. (i) ( 1 + 3 + 5 _________

3 ,

1 + 5 + 0 _________

3 ) = (3, 2)

(ii) ( 1 + 9 + 11 ___________

3 ,

2 + 6 + 1 _________

3 ) = (7, 3)

(iii) Square: Centre of gravity is at

(1.5, 1.5)

Area of square = 3 × 3 = 9

Triangle: Centre of gravity is at

( 3 + 3 + 9 _________

3 ,

0 + 3 + 0 _________

3 ) = (5, 1)

Area of triangle = 1 __

2 (6)(3) = 9

9 at (1.5, 1.5) = 18 at (x, y)

9 at (5, 1) } 9(1.5) + 9(5) = 18(x)

⇒ 18x = 58.5

⇒ x = 3.25

9(1.5) + 9(1) = 18(y)

⇒ 18y = 22.5

⇒ y = 1.25

⇒ Centre of gravity of lamina is at

(3.25, 1.25)

5.15

1.651.35

g

p

A

Q

Q

3718—

3518—

32–

g

p

A

6

(iv) Rectangle: Centre of gravity is at (8, 2)

Area of rectangle = 16 × 4 = 64

Triangle: Centre of gravity is at (4, 5)

Area of triangle = 1 __

2 (16)(3) = 24

64 at (8, 2) = 88 at (x, y)

24 at (8, 5) } 64(8) + 24(8) = 88(x)

⇒ x = 8

64(2) + 24(5) = 88(y)

⇒ y = 31

___

11

⇒ Centre of gravity of lamina is at

( 8, 31

___

11 )

Q. 5.

20

16 16

20

A(0,12)

C(16,0)B

(–16,0)D

(0,4)

m

|AB|2 = |BD|2 + |AD|2

⇒ 202 = 162 + |AD|2

⇒ |AD| = 12

Triangle: Area = 1 __

2 (32)(12)

= 192

Taking D as the origin, the centre of gravity

is at the centroid of A(0, 12), B(−16, 0)

and C(16, 0) which is at

( 0 − 16 + 16 ____________

3 ,

12 + 0 + 0 ___________

3 ) = (0, 4)

Circle: Area = pr2

= 22 ___

7 ×

49 ___

4

= 77 ___

2

= 38 1 __

2

Centre of gravity is at m(0, 6)

The remainder: Area = 192 − 38 1 __

2

= 153 1 __

2


153 1 __

2 at (x, y)}

= 192 at (0, 4) 38 1 __

2

at (0, 6) } 153 1 __

2

(x) + 38 1 __

2 (0) = 192 (0)

⇒ x = 0

153 1 __

2 (y) + 38 1 __

2

(6) = 192(4)

⇒ y = 1,074

______

307

= 3.5 cm

Q. 6. Area of full disc = p(52) = 25p

Area of missing piece = p(22) = 4p

⇒ Area of remainder = 25p − 4p

= 21p

x

y

(0,0) (0,2)

21p at (x, y) = 25p at (0, 0)

4p at (2, 0) } Taking moments around the y-axis:

21p(x) + 4p(2) = 25p(0)

⇒ 21x + 8 = 0

⇒ 21x = −8

⇒ x = − 8 ___

21

⇒ Centre of gravity of the remainder is

8 ___

21 m = 38 cm from O.

Q. 7.

C

A

B

(10,3)

(10,0)(0,0)(8,0)

A: Area is 3 × 2 = 6

Centre of gravity is at ( 9, 4 1 __

2 )

B: Area = 1 __

2 (2)(3) = 3

Q

Q

7

Centre of gravity is at

( (8 + 10 + 10) _____________

3 ,

0 + 0 + 3 _________

3 ) = ( 28

___

3 , 1 )

Rectangle: Area = 10 × 6 = 60

Centre of gravity is at (5, 3)

Remainder C: Area = 60 − 6 − 3 = 51.


6 at ( 9, 4 1 __

2 )

3 at ( 28 ___

3 , 1 ) = 60 at (5, 3)

51 at (x, y) }

6(9) + 3 ( 28 ___

3 ) + 51 (x) = 60(5)

⇒ x = 218

____

51

6 ( 4 1 __

2 ) + 3(1) + 51(y) = 60(3)

⇒ y = 150

____

51

Answer: ( 218 ____

51 ,

150 ____

51 ) = (4.27, 2.94)

Q. 8. (a) 4(2) + 5(x) + 1(5) + 3(1) = 13(2)

⇒ x = 2

4(3) + 5(4) + 1(y) + 3(7) = 13(4)

⇒ y = −1

(b)

b(2r,0)

a(0,0)

Larger circle: Area = pR2

= p(4r)2

= 16pr2


Smaller Circle: Area = pr2

Centre of gravity is at b(2r, 0)

Remainder: Area = 16pr2 − pr2

= 15pr2


15pr2 at (x, y) = 16pr2 at (0, 0)

pr2 at (2r, 0) }

15pr2(x) + pr2(2r) = 16pr2 (0)

⇒ x = − 2r ___

15

15pr2(y) + pr2(0) = 16pr2 (0)

⇒ y = 0

Answer: ( − 2r ___

15 , 0 )

Q. 9.

(50,50)

(20,20)

Circle: Area = pr2

= 22 ___

7 ×

400 ____

1

= 1,257


Square: Area = 100 × 100

= 10,000


Remainder: Area = 10,000 − 1,257

= 8,743


1,257 at (20, 20) = 10,000 at (50, 50)

8,743 at (x, y) } 1,257(20) + 8,743(x) = 10,000(50)

⇒ x = 54.3

Answer: 54 mm

Q. 10. 90 cm

30 cm

30 cm

115 cm

a(0,0) b(90,0)

(i) Distance = 115

____

2

= 57.5 cm

8

(ii) Whole rectangle:

Area = 90 × 115

= 10,350

Centre of gravity is at (45, 75.5)

Larger Section:

Area = 80 × 70

= 5,600


Smaller Section:

Area = 80 × 30

= 2,400


Remainder:

Area = 10,350 − 5,600 − 2,400

= 2,350


5,600 at (45, 40)

2,400 at (45, 95)

2,350 at (x, y) }

Taking moments about the x-axis:

5,600(40) + 2,400(95) + 2,350(y)

= 10,350 (57.5)

⇒ y = 60.9 cm

Q. 11. (i) Area DOPQ = 1 __

2 (base)(height)

… take [OP] as the base

⇒ Area DOPQ = 1 __

2 (12)(18)

= 108 square units

Area DOQR = 1 __

2 (base)(height)

… take [OR] as the base

⇒ Area DOQR = 1 __

2 (30)(18)

= 270 square units

(ii) Centre of Gravity of DOPQ

= ( 0 + 0 + 18 ___________

3 ,

0 + 12 + 18 ____________

3 )

= (6, 10)

Centre of Gravity of DOQR

= ( 0 + 18 + 30 ____________

3 ,

0 + 18 + 0 ___________

3 )

= (16, 6)

(iii) 108 at (6, 10) = 378 at (x, y)

270 at (16, 6) } Taking moments around the y-axis:

(108)(6) + (270)(16) = (378)(x)

⇒ 378x = 4,968

⇒ x = 13.14

Taking moments around the x-axis:

(108)(10) + (270)(6) = (378)(y)

⇒ 378y = 2,700

⇒ y = 7.14

⇒ Centre of Gravity of lamina is at

(13.14, 7.41)

Q. 12. (i) Area DOPQ = 1 __

2 |x1y2 − x2y1|

= 1 __

2 | (18)(6) − (−6)(18) |

= 108

Area DOQR = 1 __

2 (base)(height)

… take [OR] as the base

⇒ Area DOQR = 1 __

2 (36)(18)

= 324 square units

Area DOQR : Area DOQR = 108 : 324

= 1 : 3

(ii) Centre of Gravity of DOPQ

= ( 0 − 6 + 18 ___________

3 ,

0 + 6 + 18 ___________

3 )

= (4, 8)

Centre of Gravity of DOQR

= ( 0 + 18 + 36 ____________

3 ,

0 + 18 + 0 ___________

3 )

= (18, 6)

(iii) 1 at (4, 8) = 4 at (x, y)

3 at (18, 6) } Taking moments around the y-axis:

(1)(4) + (3)(18) = (4)(x)

⇒ 4x = 58

⇒ x = 14.5

Taking moments around the x-axis:

(1)(8) + (3)(6) = (4)(y)

4y = 26

⇒ y = 6.5

⇒ Centre of Gravity of lamina is at

(14.5, 6.5)

= 10,350

at (45, 57.5)

Q

9

Exercise 8D

Q. 1. (i) The centre of gravity of the cylinder

is at a height 1 __

2 h = 1 __

2

(24) = 12 cm

above the table.

(ii) The centre of gravity of the cone is at

a height 1 __

4 h = 1 __

4

(24) = 6 cm

above the table.

(iii) The centre of gravity of the solid hemisphere is at a height

3

__

8 r =

3 __

8 (24) = 9 cm above the table.

(iv) The centre of gravity of the

hemispherical shell is at a height

1 __

2 r = 1 __

2

(24) = 12 cm above the table.

Q. 2. The centre of gravity of the solid

hemisphere is at a height

3

__

8 r =

3 __

8 (16) = 6 cm above the table.

The centre of gravity of the hemispherical

shell is at a height 1 __

2 r = 1 __

2

(16) = 8 cm

above the table.

The difference in the heights of their

centres of gravity above the table is

8 − 6 = 2 cm.

Q. 3. (i) 1 __

4 h = 5

⇒ h = 20 cm

(ii) V = 1 __

3 pr2h

= 1 __

3 p(32)(20)

= 60p cm3

Q. 4. (i) 3

__

8 r = 3

⇒ r = 8 cm

(ii) V = 2 __

3 pr3

= 2 __

3 p(83)

= 1,024p

_______

3

cm3

Q. 5. (i) 1 __

2 r = 3

⇒ r = 6 cm

(ii) Curved Surface Area = 2pr2

= 2p(62)

= 72p cm2

Q. 6. W acts through a point 3

__

8 r = 3 cm from

the base of the cylinder.

3W acts through a point 1 __

2 h = 5 cm from

the base of the hemisphere.

Let P be the point at the right extreme end

of the central axis.

Here, then, is the diagram of the forces.

The total weight of the compound body is

4W which acts through a point which is

x cm from P.

PP

3W

W

4W

5=

8 x

By the Principle of Moments:

W(13) + 3W(5) = 4W(x) … divide by W

⇒ 28 = 4x

⇒ x = 7 cm from P

Q. 7.

2 7 7P

W

2W

2W acts through a point 1 __

2 h = 1 __

2 (14) = 7 cm

from P.

W acts through a point 1 __

4 (8) = 2 cm from

the base of the cone, i.e. 16 cm from P.


The total weight of the compound body

is 3W, which acts through a point which is

x cm from P.

W

2W

3W

7 P P=

9 x

By the Principle of Moments,

W(16) + 2W(7) = 3W(x) … divide by W

⇒ 30 = 3x

⇒ x = 10 cm

Q

Q

10

Q. 8.

P15159

5W

W

5W acts through a point

1 __

2 h = 1 __

2

(30) = 15 mm from P.

W acts through a point 3

__

8 (24) = 9 mm

from the base of the hemisphere,

i.e. 39 mm from P.



6W, which acts through a point which is

x mm from P.

=

W

5W6W

1524 xP P


W(39) + 5W(15) = 6W(x) … divide by W

⇒ 114 = 6x

⇒ x = 19 mm

Q. 9.

P553

3W

W

3W acts through a point

1 __

2 h = 1 __

2

(10) = 5 mm from P.


2 (6) = 3 mm from

the base of the hemisphere, i.e. 13 mm

from P.



4W, which acts through a point which is

x mm from P.

=

W

3W

4W

58 x


W(13) + 3W(5) = 4W(x) … divide by W

⇒ 28 = 4x

⇒ x = 7 mm

Q. 10.

P

W W

3 3


4 (12) = 3 cm

from the base of the hemisphere.

W acts through a point 3

__

8 (8) = 3 cm from

the base of the cone, i.e. 15 cm from P.



is 2W which acts through a point which is

x cm from P.

=

W W

2W

9 P P6 x


W(15) + W(9) = 2W(x) … divide by W

⇒ 24 = 2x

⇒ x = 12 cm

i.e. the centre of the compound body is

12 cm from P. This is at the plane where

the bases of the two solids meet.

Q. 11. (i) vcylinder = pr2h

= p(122)(40)

= 5,760p cm3

Q

Q

11

vcone = 1 __

3 pr2h

= 1 __

3 p(122)(40)

= 1 __

3 (5,760p) cm3

⇒ vcylinder = 3(vcone)

⇒ weight of cylinder = 3(weight of

cone)

(ii)

3W

W

P20 20 10

Let W be the weight of the cone.

Therefore 3W is the weight of the

cylinder. W acts through a point

1 __

4 h = 1 __

4

(40) = 10 cm from the base

of the cylinder. 3W acts through a

point 1 __

2 h = 1 __

2

(40) = 20 cm from the

base of the cone.

Here, then, is the diagram of the

forces. The total weight of the

compound body is 4W, which acts

through a point which is x cm from P.

=

3W

W

4W

20 30 x


3W(20) + W(50) = 4W(x)

… divide by W

⇒ 110 = 4x

⇒ x = 27.5 cm

Q. 12. (i) vcylinder = pr2h = p(182)(24)

= 7,776p cm3

vhemisphere = 2 __

3 pr3 = 2 __

3

p(183)

= 3,888p cm3

⇒ vcylinder = 2(vhemisphere)

⇒ weight of cylinder : weight of cone

= 2 : 1

(ii)

6.75

2W

12

W

P

Let W be the weight of the

hemisphere. Therefore 2W is the

weight of the cylinder. W acts through

a point 3 _ 8 r = 3

_ 8 (18) = 6·75 cm

from the top of the cylinder. 2W acts

through a point 1 _ 2 h = 1

_ 2 (24) = 12 cm

from the base of the hemisphere.



compound body is 3W, which acts

through a point which is x cm from P.

18.7512

3W

=

2W

W

x


2W(12) + W(30.75) = 3W(x)

… divide by W

⇒ 3x = 54.75

⇒ x = 18.25 cm

Q. 13.

3W

O

3 10 10 4

W W

Let W be the weight of the hemisphere

and of the cone. Therefore 3W is the

weight of the cylinder. The weight of

the hemisphere, W, acts through a

point 3 _ 8 r = 3

_ 8 (8) = 3 cm from the base

of the cylinder. The weight of the

cylinder, 3W, acts through a point

1 _ 2 h = 1

_ 2 (20) = 10 cm from the base of the

cylinder. The weight of the cone, W, acts


_ 4 (16) = 4 cm from

the top of the cylinder.

Q

12



is 5W which acts through a point which

is x cm from O.

=

W

3W5W

W

3 10 14 x


W(−3) + 3W(10) + W(24) = 5W(x)

… divide by W

⇒ 5x = 51

⇒ x = 10.2 cm

Q. 14. (i) CSAcylinder = 2prh = 2p(r)(4r) = 8pr2

CSAhemisphere = 2pr2

⇒ CSAcylinder = 4(CSAhemisphere)

⇒ Wcylinder = 4(Whemisphere)

(ii)

4W

2r

W

P2r

Let W be the weight of the hemisphere.

Therefore 4W is the weight of the

cylinder. W acts through a point r _ 2

from the top of the cylinder. 4W acts


_ 2 (4r) = 2r from

the base of the hemisphere.



compound body is 5W which acts

through a point which is a distance of

x from P.

4W

5W

W

2.5r2r x=


4W(2r) + W(4.5r) = 5W(x)

… divide by W

⇒ 5x = 12.5r

⇒ x = 2.5r.

Exercise 8E

Q. 1. 3

1

d

a

c

24

b

___

› R =

__

› i + 2

__

› j − 3

__

› i − 4

__

› j

= −2 __

›

i − 2 __

›

j

| ___

› R | = √

_____________

(−2)2 + (−2)2

= √__

8 N

______

›

db = 3 √__

2 ( 1 ___

√

__

2 __

›

i ) − 3 √__

2 ( 1 ___

√__

2 ) __

›

j

= 3 __

›

i − 3 __

›

j

The new resultant = (−2 __

›

i − 2 __

›

j ) + (3 __

›

i − 3 __

›

j )

= __

›

i − 5 __

›

j

∴ | ___

› R | = √

__________

12 + (−5)2 = √___

26 N

Q. 2. 5

3

d

a

c

24

b

___

›

R = 3 __

›

i − 2 __

›

i + 5 __

›

i − 4 __

›

j

= 8 __

›

i − 6 __

›

j

| ___

› R | = √

__________

82 + (−6)2

= √____

100

= 10 N

Let x = the distance of its line of action

from a.

The moment of the sum = the sum of the

moments. (Taking moments about a)

−10(x) = 3(0) − 2(1) − 5(1) + 4(0)

⇒ x = 7 ___

10 m

= 70 cm

Let it intersect at a distance k from a,

therefore a distance (1 − k) from b.

E

Q

Q

13

Taking moments about the point of

intersection.

10(0) = 3(0) − 2(1−k) − 5(1) + 4(k)

⇒ k = 7 __

6 m

Answer: 1 1 __

6 m from a.

Q. 3. 7

2

d

a

c

416

b

(i) ___

› R = 2

_

› i + 4

_

› j − 7

_

› i − 16

_

› j

= −5 _

› i − 12

_

› j

(ii) ∴ | ___

› R | = √

______________

(−5)2 + (−12)2

= √____

169

= 13 N

(iii) Taking moments about a:

13(x) = 2(0) + 4(1) + 7(1) + 16(0)

⇒ x = 11 ___

13 m

Q. 4. 2

2

1 210

d

a

c

b

(i) ___

› R = 2

_

› i + 2

_

› j + 2

_

› i −

_

› j + (8

_

› i − 6

_

› j )

= 12 _

› i − 5

_

› j

∴ | ___

› R | = √

___________

122 + (−5)2

= √____

169

= 13 N

(ii) Taking moments about d:

13(x) = 2(3) + 2(4) + 2(0) + 1(0) + 10(0)

⇒ x = 14 ___

13 m

(iii) 13x = 2(3) + 2(4) + 2(0) + 1(0)

+ 10(0) + 12

⇒ x = 2 m

Q. 5.

16 NA

8 m

5 m5 m

a b

c

16

20 N

10 N

A

|ad|2 + |dc|2 = |ac|2

⇒ 42 + |dc|2 = 52

⇒ |dc| = 3 m

∴ tan A = 3

__

4

∴ sin A = 3

__

5

∴ cos A = 4 __

5

_____

› ab = 16

_

› i

_____

› bc = −10 cos A

_

› i + 10 sin A

_

› j

= −8 _

› i + 6

_

› j

____

› ca = −20 cos A

_

› i − 20 sin A

_

› j

= −16 _

› i − 12

_

› j

⇒ ___

› R = −8

_

› i − 6

_

› j

∴ | ___

› R | = √

_____________

(−8)2 + (−6)2

= √____

100

= 10 N

Taking moments about c:

Moment of the sum = the sum of the

moments

10(x) = 16(3) + 10(0) + 20(0)

⇒ x = 4.8 m

Q. 6.

60°3 N4 N

8 Na b

c

60°

(i) _____

› ab = 8

_

› i

_____

› bc = −8 cos 60°

_

› i + 8 sin 60°

_

› j

= −8 ( 1 __

2 ) _

› i + 8 ( √

__ 3 ___

2 )

_

› j

= −4 _

› i + 4 √

__ 3 _

› j

Q

Q

14

____

› ca = 4 cos 60°

__

› i + 4 sin 60

__

› j

= 2 __

› i + 2 √

__ 3 _

› j

⇒ ___

› R = 6

_

› i + 6 √

__ 3 _

› j

∴ | ___

› R | = √

___________

62 + (6 √__

3 )2

= √_________

36 + 108

= √____

144

= 12 N

(ii)

l

x l12–

x2 + ( 1 __

2 l ) = l2

⇒ x = √

__ 3 ___

2 l

Taking moments about a:

Moment of the sum = the sum of the

moments

12(x) = 8(0) + 8 ( √__

3 ___

2 l ) + 4(0)

⇒ 12x = 4 √__

3 l

⇒ x = √

__ 3 ___

3 l

= 1 ___

√__

3 l m

The moment of the forces about a

was 4 √__

3 l N m, so the moment

M = 4 √__

3 l N m.

Exercise 8F

Q. 1.

2

WW

T1 T211

1 T1 + T2 = 2W

2 W(1) + W(2) = T2(4)

⇒ T2 = 3

__

4 W

⇒ T1 = 1 1 __

4 W

Q. 2. 4 3 1

WW

T2T1

1 T1 + T2 = 2W

2 W(4) + W(7) = T2(8)

⇒ T2 = 11 ___

8 W

⇒ T1 = 5

__

8 W

Q. 3. (i)

BA

T2T11 3

242 N

1 T1 + T2 = 2

2 2(5) = T1(1) + T2(8)

⇒ T1 + T2 = 10

Solving these gives T1 = 8

__

7 N,

T2 = 6

__

7 N

(ii)T

1 3

2

4

2 N

2T

4 Nx

1 T + 2T = 6

⇒ T = 2 N

2 4(x) + 2(5) = 4(1) + 2(8)

⇒ x = 2 1 __

2 cm

Q. 4. (i)

4 m

6 m

S

mR

W

R

1 R = W

2 mR = S

Q

Q

Q

15

3 W(2) = S(6)

⇒ S = 1 __

3 W

(ii) 2 mW = 1 __

3 W

⇒ m = 1 __

3

Q. 5.

S

WR

mR

11 m

10 m

1 R = W

2 mR = S

3 W(5) = S(11) ⇒ S = 5 ___

11 W

2 mW = 5 ___

11 W ⇒ m =

5 ___

11

Q. 6. (a) (i) Friction

(ii) Moment

(b) 1 R = 245

2 0.8R = S

3 245 (a cos a) = S(2a sin a)

Equation 2

⇒ S = 0.8(245) = 196

Equation 3

⇒ 245 cos a = (196)2 sin a

⇒ 245 = 392 tan a

⇒ tan a = 245

____

392

= 5

__

8

S

a

a

α

R

25g = 245 N

0.8 R

Q. 7.

S

WR

R

A

21

Let the ladder have a length of 2l.

1 R = W

2 1 __

2 R = S

3 W(l cos A) = S(2l sin A)

⇒ S = W cos A

_______

2 sin A = W

_______

2 tan A

2 1 __

2 W = W

_______

2 tan A ⇒ tan A = 1

⇒ A = 45°.

Q. 8.

S

WR

R

A

21

S21

Let the ladder have a length of 2l.

1 R + 1 __

2 S = W ⇒ R = W − 1 __

2

S

2 1 __

2 R = S

3 W(lcos A) = S(2l sin A) + 1 __

2 S(2l cos A)

… divide by l cos A

⇒ W = 2S tan A + S

⇒ S(2 tan A + 1) = W

⇒ S = W __________

2 tan A + 1

2 1 __

2 ( W − 1 __

2

S ) = S

⇒ 1 __

2 W − 1 __

4

S = S ⇒ 2W − S = 4S

⇒ 2W = 5S

⇒ 2W = 5 W __________

2 tan A + 1

⇒ 2 tan A + 1 = 5

__

2

⇒ tan A = 3

__

4 ⇒ A = 37°

Q

Q

16

Q. 9.

R

45°

2

2

S

amR

mS

b

20g = 196

1 R + mS = 196

2 mR = S

3 Taking moments about a:

196(2 cos 45°) = S(4 sin 45°)

+ mS(4 cos 45°)

But cos 45° = sin 45° = 1 ___

√__

2

∴ 196(2) = S(4) + mS(4)

⇒ S + mS = 98

⇒ S(1 + m) = 98

⇒ S = 98 ______

1 + m

Now mR = S

⇒ R = 1 __ m S = 1 __

m ( 98 ______

1 + m ) =

98 ________

m(1 + m)

Putting these into equation 1 , we

get

98 ________

m(1 + m) + m ( 98 ______

1 + m ) = 196.

… Multiply by m(1 + m)

⇒ 98 + 98m2 = 196m(1 + m).

… Divide by 98.

⇒ 1 + m2 = 2m(1 + m)

⇒ 1 + m2 = 2m + 2m2

⇒ m2 + 2m − 1 = 0

⇒ m = −2 ± √

______ 4 + 4 _____________

2 =

−2 ± 2 √__

2 __________

2

= −1 ± √__

2

Since m > 0, m = √__

2 − 1 Q.E.D.

Q. 10. (i)

W

S

5

5

R

45

R

A

1 R = W

2 4 __

5 R = S

3 Taking moments about the foot

of the ladder:

W(5 cos A) = S(10 sin A)

⇒ W cos A = 2S sin A

But S = 4 __

5 R = 4 __

5

W.

∴ W cos A = 2 ( 4 __

5 W ) sin A

⇒ cos A = 8

__

5 sin A

⇒ 1 = 8

__

5 tan A

⇒ tan A = 5

__

8

(ii)

W

2W

S

xR

6

5

45

R

A

Since tan A = 3

__

4 , cos A = 4 __

5

, sin A = 3

__

5

1 R = 2W + W = 3W

2 4 __

5 R = S

⇒ S = 4 __

5 (3W) = 12 ___

5

W

3 2W(x cos A) + W(5 cos A)

= S(10 sin A)

⇒ 2W ( 4 __

5 x ) + W(4) = S(6)

⇒ 8

__

5 xW + 4W = 6S

But S = 12 ___

5 W

∴ 8

__

5 xW + 4W = 6 ( 12 ___

5

W ) ⇒ 8x + 20 = 72

⇒ x = 6.5 m

(iii)

R

45

R

2W5

5

S

W

A

17

Assume it is just on the point of slipping

when the man reaches the top.

1 R = W + 2W = 3W

2 4 __

5 R = S

⇒ S = 4 __

5 (3W) = 12 ___

5

W

3 W(5 cos A) + 2W(10 cos A)

= S(10 sin A)

⇒ 5W = 2S tan A

But S = 12 ___

5 W,

∴ 5W = 24 ___

5 W tan A

⇒ tan A = 25

___

24

Exercise 8G

Q. 1. (i) A

B

R

8

8

F

W

4g2

S

M C

|MC| = 6 (from Pythagoras’ Theorem)

The centroid, g, is 2 cm from m, 4 cm

from C.

Assume it is on the point of slipping.

Therefore, F = mR.

(ii) 1 R = W

2 mR = S

3 Taking moments about b:

W(2) = S(8)

⇒ S = 1 __

4 W

Equation 2

⇒ mR = S

⇒ m(W) = 1 __

4 W

⇒ m = 1 __

4

The least value of is 1 __

4 .

Q. 2.

8 cm

8 cm

A

B

CD

17 cm

17 cmWR

mR

S

(i) Using Pythagoras’ Theorem

|CD|2 + 82 = 172

⇒ |CD| = √________

172 − 82

= 15 cm

[CD] is a median of the triangle. The

centroid therefore lies 2 __

3 of the way

along [CD].

2 __

3 (15) = 10

⇒ The centre of gravity is 10 cm from C.

(ii) 1 R = W

2 mR = S

3 W(5) = S(8)

⇒ S = 5

__

8 W

2 mW = 5

__

8 W

⇒ m = 5

__

8

Q. 3.

20 cm

20 cm

A

B

CD

29 cm

29 cmWR

mR

S

Using Pythagoras’ Theorem

|CD|2 + 202 = 292

⇒ |CD| = √_________

292 − 202 = 21 cm

[CD] is a median of the triangle. The


3 of the way

along [CD].

2

__

3 (21) = 14

Q

Q

18

⇒ The centre of gravity is 14 cm

from C.

1 R = W

2 mR = S

3 W(7) = S(20)

⇒ S = 7 ___

20 W

2 mW = 7 ___

20 W

⇒ m = 7 ___

20

Q. 4.

A

B

CD

7 cm

7 cm

25 cm

25 cmR

mR

WS

14S

(i) Using Pythagoras’ Theorem:

|CD|2 + 72 = 252

⇒ |CD| = √________

252 − 72

= 24 cm

[CD] is the median of the triangle. The


3 of the way

along [CD].

2 __

3 (24) = 16

⇒ The centre of gravity is 16 cm

from C.

1 R + 1 __

4 S = W

⇒ R = W − 1 __

4 S

2 mR = S

3 W(8) = S(7) + 1 __

4 S(24)

⇒ 8W = 13S

⇒ S = 8 ___

13 W

⇒ R = 11 ___

13 W

(ii) 2 m ( 11 ___ 13

W ) = 8 ___

13 W

⇒ m = 8 ___

11

Q. 5. (a) See text

(b)

W

S

R

q

p

r

mg

F

x

a

p

q

12

a

m

Let |pm| = x

x2 + ( 1 __

3 a )

2

= a2 ⇒ x = √__

3

__

2 a

∴ |pg| = 2 __

3 ( √

__

3

__

2 a ) = 1 ___

√

__ 3 a

|gm| = 1 __

3 ( √

__ 3 ___

2 a ) = 1 ____

2 √

__ 3 a

1 R = W

2 F = S

3 S ( 1 __

2 a ) = W ( 1 ____

2 √

__ 3 a )

⇒ S = 1 ___

√__

3 W

The reactions are 1 ___

√__

3 W, W

If it is on the point of slipping, then F = mR.

Equation 2

⇒ mR = S

⇒ mW = 1 ___

√__

3 W

⇒ m = 1 ___

√__

3

∴ The least value of m is 1 ___

√__

3 .

Q. 6. (a) See text

(b) T1 p(0,3)

r(0,0)q(–4,0)

T2T3

43

W W

(–2,1)43

1– )( ,

Centroid is at

( 0 + 0 + 4 _________

3 ,

0 + 3 + 0 _________

3 ) = ( − 4 __

3

, 1 )

Q

Q

19

1 T1 + T2 + T3 = 4 __

3 W + W

⇒ T1 + T2 + T3 = 7 __

3 W

2 Taking moments about the y-axis:

T2(4) + T1(0) + T3 (0) = 4 __

3 W(2) + W ( 4 __

3

) ⇒ 4T2 = 4W

⇒ T2 = W

3 Taking moments about the x-axis:

T1(3) + T2(0) + T3(0) = 4 __

3 W(1) + W(1)

⇒ T1 = 7 __

9 W

But T1 + T2 + T3 = 7 __

3 W

⇒ T3 = 5

__

9 W

Answer: 7 __

9 W, W,

5 __

9 W

Exercise 8H

Q. 1.

A

A

F

50

30

40

R

tan A = 4 __

3

⇒ cos A = 3

__

5 and sin A = 4 __

5

Component of weight acting down the

plane = 50 sin A = 50 ( 4 __

5 ) = 40 N

⇒ F = 40 N

Q. 2.

A

A

F

200

160

mR

120

R

tan A = 3

__

4

⇒ cos A = 4 __

5 and sin A =

3 __

5

Component of weight acting down the

plane = 200 sin A = 200 ( 3 __

5 ) = 120 N

Component of weight acting perpendicular

to the plane = 200 cos A = 200 ( 4 __

5 ) = 160 N

⇒ R = 160 N

⇒ Friction = mR

= 1 __

2 (160)

= 80 N

⇒ F + 80 = 120

⇒ F = 40 N

Q. 3. Here is a diagram of all the forces acting

on the particle

30°

30°

100 N

R

35

R

F

Component of weight acting down

the plane = 100 sin 30° = 100 ( 1 __

2 ) = 50 N

Component of weight acting perpendicular

to the plane = 100 cos 30° = 100 ( √__

3 ___

2 )

= 50 √__

3 N

Component of F acting along

the plane = F cos 30° = √

__ 3 ___

2 F

Component of F acting perpendicular to

the plane = F sin 30° = 1 __

2 F

⇒ R = 50 √__

3 + 1 __

2 F

But, 3

__

5 R + 50 =

√__

3 ___

2 F

⇒ 3

__

5 [ 50 √

__ 3 + 1 __

2

F ] + 50 = √

__ 3 ___

2 F

⇒ 30 √__

3 + 3 ___

10 F + 50 =

√__

3 ___

2 F

⇒ 300 √__

3 + 3F + 500 = 5 √__

3 F

⇒ F(5 √__

3 − 3) = 300 √__

3 + 500

⇒ F = 100(3 √

__ 3 + 5) _____________

5 √__

3 − 3 ~− 180 N

Q

20

Q. 4. tan a = 2

⇒ sin a = 2 ___

√__

5 cos a = 1 ___

√

__ 5

Forces

W

W

RmR

Resolved

W

√3R

W

√5

2W

√5

2W

√5

mR

1 R = 2W ____

√__

5 + W ___

√

__ 5 =

3W ____

√__

5

2 mR + W ___

√__

5 = 2W

____

√__

5

⇒ mR = W ___

√__

5

Dividing 1 by 2

⇒ m = 1 __

3

Q. 5.

R

a

a

l

WW cos a

W sin a

R tan l

The diagram shows the forces on the

particle. Since m = tan l, the friction force

is R tan l as shown.

Particle will slip down the plane if

W sin a > R tan l … but R = W cos a

⇔ W sin a > W cos a tan l

… divide by W cos a

⇔ tan a > tan l

⇔ a > l

Q. 6. Here is a diagram of the resolved forces

acting on the body:

7 sin A7cos A

7

14 cos A

14 sin AA

A

A

R

7 cos A + 7 = 14 sin A

⇒ cos A + 1 = 2 sin A

⇒ cos A + 1 = 2 √__________

1 − cos2 A

… square both sides

⇒ cos2 A + 2 cos A + 1 = 4(1 − cos2 A)

⇒ cos2 A + 2 cos A + 1 = 4 − 4 cos2 A

⇒ 5 cos2 A + 2 cos A − 3 = 0

⇒ (5 cos A − 3)(cos A + 1) = 0

⇒ cos A = 3

__

5 OR cos A = −1

cos A = −1 is excluded because A is an

acute angle.

Q. 7. (i) Particle about to slip down the plane

A

A9898 cos A

98 sin A

19.6

RmR

R = 98 cos A

19.6 + mR = 98 sin A

⇒ 19.6 + 98 m cos A = 98 sin A

⇒ 2 + 10 m cos A = 10 sin A

⇒ 10 sin A − 10 m cos A = 2 … 1

Particle about to slip up the plane

A

A9898 cos A

98 sin A

98

R

mR

R = 98 cos A

98 = mR + 98 sin A

⇒ 98 = 98 m cos A + 98 sin A

⇒ 10 = 10 m cos A + 10 sin A

Q

Q

21

⇒ 10 sin A + 10 m cos A = 10 … 2

Adding equations 1 and 2 we get

20 sin A = 12

⇒ sin A = 3

__

5

⇒ A = sin−1 ( 3 __ 5

) = 36.87°

(ii) From equation 2 we get

10 ( 3 __

5 ) + 10 m ( 4 __

5 ) = 10

⇒ 6 + 8m = 10

⇒ 8m = 4

⇒ m = 1 __

2

(iii) The force diagram therefore looks like this:

A

F

78.498

58.8

R F sin l

F cos l1–2R

l

R + F sin l = 78.4 1 __

2 R + 58.8 = F cos l

⇒ R = 78.4 − F sin l ⇒ R = 2F cos l − 117.6

⇒ 78.4 − F sin l = 2F cos l − 117.6

⇒ F(2 cos l + sin l) = 196

⇒ F = 196 _____________

2 cos l + sin l

= 196(2 cos l + sin l)−1

⇒ dF

___

dl = −196(2 cos l + sin l)−2 (−2 sin l + cos l)

⇒ dF

___

dl =

196(cos l − 2 sin l) __________________

(2 cos l + sin l)2

Putting dF

___

dl = 0 we get

cos l − 2 sin l = 0

⇒ 1 − 2 tan l = 0

⇒ tan l = 1 __

2

⇒ sin l = 1 ___

√__

5 and cos l = 2 ___

√

__ 5

F = 196 _____________

2 cos l + sin l

= 196 ________

4 ___

√

__ 5 + 1

___

√__

5

= 196

____

5 ___

√

__ 5

= 196

____

√__

5 N

(iii) The force diagram therefore looks like this:his:

22

Q. 8. m = tan l

The diagram shows resolved forces acting on the particle.

F cos q

F sin qR

W cos a

W sin a

mR

R + F sin q = W cos a mR + W sin a = F cos q

⇒ R = W cos a − F sin q ⇒ R tan l = F cos q − W sin a

⇒ R = F cos q − W sin a

_______________

tan l

⇒ W cos a − F sin q = F cos q − W sin a

_______________

tan l

⇒ W cos a − F sin q = (F cos q − W sin a) ( cos l

_____

sin l ) … multiply by sin l

⇒ W cos a sin l − F sin q sin l = F cos q cos l − W sin a cos l

⇒ F(cos q cos l + sin q sin l) = W(cos a sin l + sin a cos l)

⇒ F cos(q − l) = W sin(a + l)

⇒ F = W sin(a + l)

___________

cos (q − l)

(i) Force acting up along the plane ⇒ q = 0

⇒ F = W sin (a + l)

____________

cos (−l) … cos(−l) = cos l

⇒ F = W sin(a + l)

___________

cos l

(ii) Horizontal force ⇒ q = −a

⇒ F = W sin(a + l)

___________

cos(−a − l)

= W sin(a + l)

____________

cos[−(a + l)]

= W sin (a + l)

____________

cos (a + l)

= W tan (a + l)

(iii) F = W sin (a + l)

____________

cos (q − l)

Minimum force will occur when cos(q − l) is at its maximum value, i.e. cos(q − l) = 1

⇒ FMIN = W sin(a + l).

23

Q. 9. (i) The normal reaction, R, and the limiting friction, F, acting on a body which is either moving

or on the point of moving, can be added to form a resultant. The angle between this

resultant and the normal reaction is the angle of friction.

(ii)

Sa

R

W

R tan l

S tan l

Let the length of the ladder be 2l

1 R + S tan l = W

⇒ R = W − S tan l

2 R tan l = S

3 Wl sin a = S(2l cos a) + S tan l(2l sin a)

⇒ S(2 cos a + 2 tan l sin a) = W sin a

⇒ S = W sin a

___________________

2(cos a + tan l sin a)

2 (W − S tan l) tan l = S

⇒ W tan l − S tan2 l = S

⇒ S(1 + tan2 l) = W tan l

⇒ S = W tan l

_________

1 + tan2 l

⇒ W sin a

___________________

2(cos a + tan l sin a)

= W tan l

_________

1 + tan2 l

⇒ (1 + tan2 l)sin a = 2 tan l(cos a + tan l sin a)

⇒ sin a + tan2 l sin a = 2 tan l cos a + 2 tan2 l sin a

⇒ tan2 l sin a + 2 tan l cos a = sin a … divide by cos a

⇒ tan2 l tan a + 2 tan l = tan a

⇒ tan a(1 − tan2 l) = 2 tan l

⇒ tan a = 2 tan l

_________

1 − tan2 l

⇒ tan a = tan 2l

⇒ a = 2l

24

Q. 10.

R W

S

R tan l

q

a

Let the length of the ladder be 2l.

We must firstly resolve R and R tan l into horizontal and vertical components:

The horizontal component of R is R sin a.

The vertical component of R is R cos a.

The horizontal component of R tan l is R tan l cos a.

The vertical component of R tan l is R tan l sin a

1 S + R sin a = R tan l cos a

⇒ R = S ________________

tan l cos a − sin a

2 W = R cos a + R tan l sin a

⇒ W = R(tan l sin a + cos a)

3 W(l sin q) = S(2l cos q)

⇒ S = 1 __

2 W tan q … as required

2 W = ( S ________________

tan l cos a − sin a

) (tan l sin a + cos a)

⇒ W = 1 __

2 W tan q ( tan l sin a + cos a

________________ tan l cos a − sin a

) … divide top and bottom by cos a

⇒ 1 = 1 __

2 tan q ( tan l tan a + 1 ______________

tan l − tan a )

⇒ tan q(tan l tan a + 1) = 2(tan l − tan a)

⇒ tan q = 2 ( tan l − tan a

______________

1 + tan l tan a )

⇒ tan q = 2 tan(l − a) … as required.

Exercise 8I

Q. 1. Here is a diagram of the forces acting on

the rod PQ:

X

Y

W

T

4–5

T

3–5

T

1 Y = 3

__

5 T = W ... Equation 1

2 X + 4 __

5 T = W ... Equation 2

3 Take moments about P

W(2) = 3

__

5 T(4) ... Equation 2

Equation 3: 5W = 6T

⇒ T = 5

__

6 W

25222525

Equation 1: Y + 3

__

5 [ 5 __

6

W ] = W

⇒ 2Y + W = 2W

⇒ 2Y = W

⇒ Y = 1 __

2 W

Equation 2: X = 4 __

5 [ 5 __

6

W ] ⇒ X = 2 __

3

W

Answer: The horizontal and vertical

reactions at the hinge are

2 __

3 W and 1 __

2

W. The tension in

the string is 5

__

6 W.

Q. 2. (i) Here is a diagram of the forces acting

on the rod:

2W2W sin A

2W cos A5W

X

Y

1 Y + 2W sin A = 5W

2 X = 2W cos A

3 Take moments about P

5W(l sin A) = 2W(2l)

⇒ sin A = 4 __

5 ⇒ cos A =

3 __

5

1 Y + 8

__

5 W = 5W ⇒ Y = 17 ___

5

W

2 X = 6

__

5 W

(ii) A = sin–1 4 __

5 = 53°

Q. 3. (i) X

p

Y

a

a

q90° – q

W(W sin q)

(W cos q)

3W

q

(ii) 1 Y + W sin q = 3W

2 X + W cos q = 0

3 Taking moments about p, using

unresolved forces:

3W (a sin q) = W(2a)

⇒ sin q = 2 __

3

= 0.6667

⇒ q = 41° 49’

(iii) X = −W cos q

= −W ( √__

5 ___

3 )

= − √

__ 5 ___

3 W

Y = 3W − W sin q

= 3W − W ( 2 __

3 )

= 7W ____

3

Q. 4. Since tan A = 5 ___

12

cos A = 12 ___

13

sin A = 5 ___

13

Forces

Y

X

T

pA

W

Resolved

Y

W

XT

1213

T5

13p

1 Y + 5 ___

13 T = W

2 X = 12 ___

13 T

Q

26

3 Taking moments about p:

W(6) = 5 ___

13 T(12)

⇒ T = 13

___

10 W

∴ X = 12 ___

13 ( 13

___

10 W )

= 6

__

5 W

Y + 5 ___

13 ( 13

___

10 T ) = W

⇒ Y = 1 __

2 W

Answer: (i) 6

__

5 W, 1 __

2

W

(ii) 13

___

10 W

Q. 5. (i)

qq

B

a

a

a

a

S

CG

2W

R

A F

3W

q qR

A

S

BB

F

a

a

a c

XXY

Y

3W2W

(ii) From system ABC

1 R + S = 2W + 3W

⇒ R + S = 5W

2 F = G

3 Taking moments about a.

2W(a cos q) + 3W(3a cos q) = S(4a cos q)

⇒ 4S = 11W

⇒ S = 2 3

__

4 W

∴ R = 2 1 __

4 W, from Equation 1.

(iii) Since R < S, mR < mS

∴ slipping will occur at A first.

(iv) Let the rod AB be on the point of

slipping.

∴ F = mR

= 1 __

3 R

= 1 __

3 ( 2 1 __

4

W )

= 3

__

4 W

q

q

a

a

XY

B

A

R = 2 W14

2W

W34

1 R + Y = 2W

2 3

__

4 W = X

3 Taking moments about B:

2W(a cosq) + 3 __

4 W(2a sin q) = 2 1 __

4 (2a cos q)

⇒ 2 cos q + 3

__

2 sinq =

3 __

2 cos q

⇒ 3 sin q = 5 cosq

⇒ 3 tan q = 5

⇒ tan q = 5

__

3

⇒ q = tan−1 5

__

3

= 59°2’

Q. 6.

q qS

CA

R W 2W2

2

T T

55

3 33

B

Since cos q = 3

__

5 , sin q = 4 __

5

and tan q = 4 __

3

The system ABC

1 R + S = W + 2W

⇒ R + S = 3W

Q

27

2 T = T

3 Taking moments about A:

W ( 1 1 __

2 ) + T(2) + 2W ( 4 1 __

2

) = T(2) + S(6)

⇒ S = 1 3

__

4 W

∴ R = 1 1 __

4 W, from Equation 1

XY

B

2

44

T

A

2W

W114

The rod AB

1 1 1 __

4 W + Y = W

⇒ Y = − 1 __

4 W

2 T = X


T(2) + W ( 1 1 __

2 ) = ( 1 1 __

4

) W(3)

⇒ T = 1 1 __

8 W

Answer: T = 1 1 __

8 W,

R = 1 1 __

4 W,

S = 1 3

__

4 W

Q. 7. S

55B

5

W5

R

mR

f

f

A

2W

C

Since tan f = 4 __

3 ,

sin f = 4 __

5 ,

cos f = 3

__

5

1 R = W + 2W = 3W

2 mR = S

3 Taking moments about A:

W(5 cos f) + 2W(10 cos f + 5 cos q)

= S(10 sin f + 10 sin q)

⇒ 3W + 12W + 10W cos q

= 8S + 10S sin q

⇒ 15W + 10W cos q = 8S + 10S sin q

2W

5 5

SC

B xq

y

4 Y = 2W

5 X = S


2W(5 cos q) = S(10 sin q)

⇒ 10W cos q = 10S sin q

This means that Equation 3 reads:

15W + 10S sin q = 8S + 10S sin q

⇒ S = 15

___

8 W

Equation 2 now reads:

mR = S

⇒ m(3W) = 15

___

8 W

⇒ m = 5

__

8

Q. 8. (i) Let the length of each rod be 2l.

Here are the forces acting on the rods

AB and AC:

2W

Y2

X2

X1 X1

Y1Y3

X3

Y1

WAA

B C

Rod AB:

Taking moments around B we get:

2W(l cos b) + Y1(2l cos b)

= X1 (2l sin b) ... divide by 2l cos b

⇒ W + Y1 = X1 tan b ... Equation 1

28

Rod AC:

Taking moments around C we get:

W(l cos b) = X1(2l sin b) + Y1 (2l cos b)

... divide by l cos b

⇒ W = 2X1 tan b + 2Y1

⇒ W – 2Y1 = 2X1 tan b ... Equation 2

2W + 2Y1 = 2X1 tan b ... Equation 1

(× 2)

Add: 3W = 4X1, tan b

⇒ X1 = 3W _______

4 tan b ... substitute into

Equation 2

⇒ W – 2Y1 = 2 [ 3W ______

4 tan b ] tan b

... multiply by 4

⇒ 4W – 8Y1 = 6W

⇒ 8Y1 = –2W

⇒ Y1 = – W __

4 ... the minus sign

indicates that the actual

direction of Y1 is opposite

to the direction indicated

in the diagram.

(ii) Looking again at the two rods

separately, we have the following:

R2R1

a

a

W

4—W

4—W

4 tan b———3W

2W

4 tan b———3W

1 R1 sin a + W __

4 = 2W

⇒ R1 sin a = 7W ____

4

2 R1 cos a = 3W ______

4 tan b

Dividing Equation 1 by

Equation 2 gives:

tan a = ( 7W ____

4 ) ( 4 tan b

______

3W

) ⇒ tan a =

7 tan b

______

3

3 R2 sin a = 3W ______

4 tan b

4 R2 cos a = W + W __

4

⇒ R2 cos a = 5W

____

4


Equation 4 gives:

tan a = ( 3W ______

4 tan b ) ( 4 ____

5W

) ⇒ tan a =

3 ______

5 tan b

⇒ 7 tan b

______

3

= 3 ______

5 tan b

⇒ 35 tan2 b = 9

⇒ tan2 b = 9 ___

35

⇒ tan b = 3 ____

√___

35

Q. 9. (i) Force diagram for the system including

the friction force at C.

YX

F

R

W

W

A

B

C

Take moments about A:

2W ( l __

2 sin f ) = R(2l cos f)

… divide by l cos f

⇒ W tan f = 2R ⇒ R = 1 __

2 W tan f

Now, we look at the forces on the rod

BC in isolation:

W

R

F

Y1

X1

C

B


F(l sin f ) = R(l cos f ) + W ( 1 __

2 sin f )

… divide by l cos f

⇒ F tan f = R + 1 __

2 W tan f

… but R = 1 __

2 W tan f

⇒ F tan f = 1 __

2 W tan f + 1 __

2

W tan f

… divide by tan f

Q

29

⇒ F = 1 __

2 W + 1 __

2

W

⇒ F = W

(ii) An additional force is now applied downwards at C. The force diagram for the BC now looks like this:

Y1

X1

mR

W

R

WC

B

Because we are now at limiting friction, the friction force is mR.


W(l sin f) + W ( 1 __

2 sin f ) + R(l cos f)

= mR(l sin f) … divide by l cos f

⇒ W tan f + 1 __

2 W tan f + R = mR tan f

… R = 1 __

2 W tan f

⇒ W tan f + W tan f = m ( 1 __ 2

W tan2 f ) … divide by W tan f

⇒ 2 = m ( 1 __

2 tan f ) … multiply by 2

⇒ m tan f = 4

⇒ m = 4 _____

tan f

Q. 10. Note: The person will need to stand at

A to maximise the chance of slipping.

Because the system will then be

symmetrical, slipping will occur at B

and C simultaneously. Here are the

forces acting on the system:

W

W

WR S

F G

Taking moments around C we get:

W ( d __

2 ) + W(d) + W ( 3d

___

2 ) = R(2d)

… multiply by 2 __

d

⇒ W + 2W + 3W = 4R

⇒ 4R = 6W

⇒ R = 3

__

2 W

Now, look at the ladder [AB] in isolation.

Here are the forces when it is just on the

point of slipping:

Lengths

WR

mR

2—W

YX

l

d

l 2 – d2

Taking moments around A we get:

R(d) = W ( d __

2

) + mR ( √_______

l2 − d2 )

… R = 3

__

2

W

⇒ 3

__

2 Wd =

1

__

2 Wd + m ( 3

__

2 W ) √

_______

l2 − d2

… multiply by 2

__

W

⇒ 3d = d + 3m √_______

l2 − d2

⇒ 3m √_______

l2 − d2 = 2d

⇒ m = 2d

_________

3 √_______

l2 − d2

⇒ In order to avoid slipping

m ≥ 2 d

_________

3 √_______

l2 − d2

Q. 11. (i) Rod [AB] Ring C

Y

X

4W

T sin a

T cos a W

T cos a

T sin a

mR

R

Taking moments about A we get:

4W(l cosa) = T cos a(2l sin a) +

T sin a(2l cos a)

… divide by 2l cos a

⇒ 2W = T sin a + T sin a

⇒ 2W = 2T sin a

⇒ W = T sin a

30

From the diagram of forces on the ring

C we can see that

mR = T cos a … Equation 1

and R = T sin a + W

… but W = T sin a

⇒ R = 2T sin a … Equation 2


Equation 2 we get:

m = 1 _______

2 tan a

⇒ 2m tan a = 1

⇒ tan a = 1 ___

2m … (i)

(ii) W = T sin a

⇒ T = W/sin a … tan a = 1 ___

2m

⇒ sin a = 1 _________

√________

1 + 4m2

⇒ T = W √________

1 + 4m2

Q. 12. (i) Rod

80 N

R

X

Y

Hemisphere

40 N

SR

mS

Lengths

45°

45°

90° 1

3

1

(ii) Taking moments around the hinge

we get:

80 ( 1.5 ___

√__

2 ) = R(1)

R = 120

____

√__

2 = 60 √

__ 2 N

(iii) S = R ___

√__

2 + 40

mS = R ___

√__

2

⇒ m ( R ____

√__

2 + 40 ) = R ___

√

__ 2

… but R = 60 √__

2

⇒ m(60 + 40) = 60

⇒ 100m = 60

⇒ m = 0.6

Q. 13. Assume ladders are on the point of

slipping. We will then be at the smallest

value of m that will prevent slipping from

occurring. Here is the force diagram for

the system:

W W

q q

R R

mR mR

Note: Because both ladders have the same

weight, both reaction forces are the same

and slipping occurs simultaneously at A

and C.

2R = 2W

⇒ R = W

Now, we examine the forces on the ladder

[AB] in isolation:

R

mR

W

X

q

31

X is the supporting force from the other

ladder. There is no Y component as the

vectors in the Y direction already sum to

zero.


R(l sin q) = mR(l cos q) + W ( l __

2 sin q )

… multiply by 2 ______

l cos q and let W = R

⇒ 2R tan q = 2mR + R tan q

… divide by R

⇒ 2 tan q = 2m + tan q

⇒ 2m = tan q

⇒ m = 1 __

2 tan q

This is the minimum value of q that will

prevent slipping from occurring.

⇒ m ≥ 1

__

2

tan q

Q. 14. (i) Let x equal the distance from P to

where the block touches the rod.

sin a = 0.8y

____ x … tan a = 4 __

3

⇒ sin a = 4 __

5

⇒ 4 __

5 =

0.8y ____ x

… multiply both sides by 5x

⇒ 4x = 4y

⇒ x = y

(ii) Rod

Ra

a

Y

X

2g

Block

a

6g

S

mS

R

(iii) Firstly, examine the diagram of the rod.

Taking moments around the hinge we

get:

2g(3 cos a) = R(1) … cos a = 3

__

5

⇒ R = 18

___

5 g

Now, examine the diagram of the

block:

1 S = R cos a + 6g

⇒ S = ( 18 ___

5 g ) ( 3 __

5

) + 6g

⇒ S = 204

____

25 g

2 mS = R sin a

⇒ m ( 204 ____

25 g ) = ( 18

___

5 g ) ( 4 __

5

) … multiply by

25 ___ g

⇒ 204m = 72

⇒ m = 72 ____

204 =

6 ___

17

(iv) Looking at the diagram of the rod:

X = R sin a

= ( 18 ___

5 g ) ( 4 __

5

) = 28.224

Y + R cos a = 2g

⇒ Y = 2g − R cos a

= 2g − ( 18 ___

5 g ) ( 3 __

5

)

= −1.568

… i.e. 1.568 N downwards

⇒ Reaction at hinge

= 28.224 _

› i − 1.568

_

› j N

Magnitude

= √__________________

(28.224)2 + (1.568)2

= 28 N

32

Exercise 8J

Q. 1. Forces:

Wr

r

R

A 3r

Resolved Forces:

T14

T√154

WR

(1) √

___ 15 ____

4 T = W

⇒ T = 4W ____

√___

15

(2) R = 1 __

4 T

= 1 __

4 ( 4W

____

√___

15 )

= W ____

√___

15

Q. 2. Forces:

TT

30 N

Resolved Forces:

30 T

12–T

12–

T32– T

32–

(1) √

__ 3 ___

2 T +

√__

3 ___

2 T = 30

⇒ T = 30

___

√__

3

= 10 √__

3 N

Q. 3. (i) “.…… are concurrent”

(ii)

XY

a

A

P

b

240

1.5

k

2.5

|ak|2 + |kb|2 = |ab|2

⇒ |ak|2 + 2.25 = 6.25

⇒ |ak| = 2

(1) Y = 240

(2) P + X = 0 ⇒ X = −P

(3) Taking moments about a:

240(0.75) = P(2)

⇒ P = 90 N

___

› X = −90

__

› i , ___

› Y = 240

__

› j

Resultant = −90 _

› i + 240

_

› j = |

___

› R |

= √______________

(−90)2 + (240)2

= 256 N

Q. 4. (i) Forces:

R

2

F

W

3 5

A

Since sin A = 3

__

5 , cos A = 4 __

5

Resolved Forces:

F

W

R45

R35

(ii) In accordance with Theorem 8.7

Q

Q

sin A = r ______

3r + r = 1 __

4

∴ cos A = √

___ 15 ____

4

33

(iii) (1) 3

__

5 R = W ⇒ R =

5 __

3 W

(2) F = 4 __

5 R = 4 __

5

( 5 __

3 W ) = 4 __

3

W

Q. 5. (a) Forces:

100

R A

150

150

3g

T

cos A = 150

____

250 =

3 __

5

∴ sin A = 4 __

5

Resolved Forces:

3g

R

T35

T45

(1) 4 __

5 T = 3g ⇒ T =

15 ___

4 g

Q. 6. Step 1. To find centre of gravity:

Forces:

mg

l12

mg12

l12

Resultant:

mg

x

112

The sum of the moment = the moment of

the sum

1 __

2 mg ( 1 __

2

l ) + mg(l ) = 1 1 __

2 mg(x)

⇒ x = 5

__

6 l

32–mg

76– l

56–R

gA

A

y

O

xT2

T1

The resultant weight, 3

__

2 mg acts through g,

which must be below in accordance with

Theorem 8.7

Taking moments about g:

T1(x) = T2(y)

But x = 5

__

6 l sin A, y = 7 __

6

l sin A

∴ T1 ( 5 __

6 l sin A ) = T2 ( 7 __

6 l sin A )

⇒ T1 __

T2

= 7 __

5

Q. 7.

Ar5–

A

R

35–r

45– r

W

Pq

sin A = 4 __

5

∴ cos A = 3

__

5

Q

34

Taking moments about the top of the

obstacle:

r

OBSTACLE

P

r sin (A + q)A + q

W ( 3 __

5 r ) = P(r sin (A + q))

⇒ P = 3W __________

5 sin (A + q)

(i) In this case q = 0

∴ P = 3W ______

5 sin A =

3W ____

4

(ii) In this case we want a minimum value

for 3W ___________

5 sin (A + q) . This value is attained

when sin (A + q) = 1, and is 3W

____

5

Q. 8.

S5

53

12

B

W

mAg

B

4

c

b

a

Let m be the midpoint [ab].

|cm| = 4 (by Pythagoras), |gm| = 1 and

|gc| = √___

17

sin A = 4 ____

√___

17 and sin B = 4 __

5

Taking moments about a:

W (2 sin A) = S(6 sin B)

⇒ W ( 8 ____

√___

17 ) = S ( 24 ___

5

) ⇒ S =

5W _____

3 √___

17

Taking moments about b:

W(4 sin A) = S(6 sin B)

⇒ W ( 16 ____

√___

17 ) = S ( 24 ___

5

) ⇒ R =

10W _____

3 √___

17

Q. 9.

rr

R

S

y

x

W

B

BA

A

Since tan A = 1 __

2 , sin A = 1 ___

√

__ 5 and cos A = 2 ___

√

__ 5

Since tan B = 3

__

4 , sin B =

3 __

5 and cos B = 4 __

5

Also, sin (A + B) = sin A cos B + cos A sin B

= ( 1 ___

√__

5 ) ( 4 __

5

) + ( 2 ___

√__

5 ) ( 3 __

5

) =

10 ____

5 √__

5

= 2 ___

√__

5

Taking moments about x:

W(r sin A) = S(r sin (A + B))

⇒ W ( 1 ___

√__

5 r ) = S ( 2 ___

√

__ 5 r )

⇒ S = 1 __

2 W Q.E.D.

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Chapter 8 Exercise 8A - mathsgrinds.ie (iv) Rectangle: Centre of gravity is at (8, 2) Area of...

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