chapter 8 exercise 8a - mathsgrinds.ie (iv) rectangle: centre of gravity is at (8, 2) area of...
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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
Centre of gravity is at (x, y)
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.
Centre of gravity is at (x, y)
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
Centre of gravity is at (0, 0)
Smaller Circle: Area = pr2
Centre of gravity is at b(2r, 0)
Remainder: Area = 16pr2 − pr2
= 15pr2
Centre of gravity is at (x, y)
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
Centre of gravity is at (20, 20)
Square: Area = 100 × 100
= 10,000
Centre of gravity is at (50, 50)
Remainder: Area = 10,000 − 1,257
= 8,743
Centre of gravity is at (x, y)
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
Centre of gravity is at (45, 40)
Smaller Section:
Area = 80 × 30
= 2,400
Centre of gravity is at (45, 95)
Remainder:
Area = 10,350 − 5,600 − 2,400
= 2,350
Centre of gravity is at (x, y)
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.
Here, then, is the diagram of the forces.
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.
Here, then, is the diagram of the forces.
The total weight of the compound body is
6W, which acts through a point which is
x mm from P.
=
W
5W6W
1524 xP P
By the Principle of Moments:
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.
W acts through a point 1 __
2 (6) = 3 mm from
the base of the hemisphere, i.e. 13 mm
from P.
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 mm from P.
=
W
3W
4W
58 x
By the Principle of Moments,
W(13) + 3W(5) = 4W(x) … divide by W
⇒ 28 = 4x
⇒ x = 7 mm
Q. 10.
P
W W
3 3
W acts through a point 1 __
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.
Here, then, is the diagram of the forces.
The total weight of the compound body
is 2W which acts through a point which is
x cm from P.
=
W W
2W
9 P P6 x
By the Principle of Moments,
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
By the Principle of Moments:
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.
Here, then, is the diagram of the
forces. The total weight of the
compound body is 3W, which acts
through a point which is x cm from P.
18.7512
3W
=
2W
W
x
By the Principle of Moments:
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
through a point 1 _ 4 h = 1
_ 4 (16) = 4 cm from
the top of the cylinder.
Q
12
Here, then, is the diagram of the forces.
The total weight of the compound body
is 5W which acts through a point which
is x cm from O.
=
W
3W5W
W
3 10 14 x
By the Principle of Moments:
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
through a point 1 _ 2 h = 1
_ 2 (4r) = 2r from
the base of the hemisphere.
Here, then, is the diagram of the
forces. The total weight of the
compound body is 5W which acts
through a point which is a distance of
x from P.
4W
5W
W
2.5r2r x=
By the Principle of Moments:
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
centroid therefore lies 2 __
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
centroid therefore lies 2 __
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)
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
3 Taking moments about B:
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
6 Taking moments about B:
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
Dividing Equation 3 by
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
Taking moments around B we get:
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.
Taking moments around B we get:
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
Dividing Equation 1 by
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.
Taking moments around B we get:
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.