Database Storage & Design Chapter 3 Using triangle OAC Method of components

12

Prob. 3.1

Using triangle OAC:

R2 = 252 + 452 2(25)(45)cos130

Prob. 3.2

1= tan-1(4/3) = 53.13

Prob. 3.3

1 = tan-1(1/2) = 26.57

= 180 26.57 30 = 123.43

= 180 = 56.57

Rx = 45 cos10 + 25 cos60 = +56.82 lb

Ry = 45 sin10 + 25 sin60 = +29.46 lb

lb 0.64

46.2982.56

22

R

13

3-2

Prob. 3.5

————————————————————

Prob. 3.6

lb 12.1530cos60

5

2

75

x

R

————————————————————

Prob. 3.7

Px = 75 cos40 = 57.45 lb

Prob. 3.8

4.39

6.165

0.136

tan

1–

————————————————————

Prob. 3.9

For sloping force

5.171;9.102

83.5

200

53

83.535 hypt triang.Slope 22

yx

y

x

FF

F

Total horiz. = 200 102.9 = 302.9 lb

14

3-3

Prob. 3.11

Force (lb) Fx (lb) F

y

(lb)

400 400 0

————————————————————

Prob. 3.12

(a)

04.2210

1

x

(b)

N 59.58

33.76cos)20)(60(22060

22

4,3

R

Prob 3.12 Contd

(c)

16.10

)41.122(59.57

14.47

92.82

1

x

————————————————————

Prob. 3.13

(a) Method of components

Force (N) Horiz. (N) Vert. (N)

18.10

41.46

33.8

tan

1

x

————————————————————

Prob. 3.14

Rx = Fx = +200 cos30 +100 cos45 400

300 cos60 = 306.1 lb

15

Prob. 3.15

R = Ry = +300 lb

Rx = Fx = 0 = 500 +F1x +240 cos30

Prob. 3.16

Rx = Fx = +1000 + 600 sin30

= +1300 lb

————————————————————

Prob. 3.17

(a) MA :

Force

(lb)

Moment

arm (ft)

Moment

(lb-ft)

20 4 80

(b) MA = 380 lb-ft (clockwise)

————————————————————

Prob. 3.18

(a) [F1] Mo = 0

lb 0.84

)4.47(4.69

22

R

[F3] Mo = 10(4) = 40 lb-ft

[F4] Mo = 0

(b) Mo = +103.9 113.1 40 = 49.2 lb-ft

4.61

716.6

304.12

tan

lb 02.14

1

x

Moment of resultant:

(a) MA

Force (kN) Arm m) Moment (kN m)

70 1.50 105.0

26.49 2.00 +52.98

14.08 1.70 23.94

16

3-5

Prob. 3.22

b =5/(tan40 ) = 5.959 ft

a = 10 5.959 = 4.041 ft

————————————————————

Prob. 3.23

Components at B:

————————————————————

Prob. 3.24

Fx = 2 cos20

= 1.879 k

Fy = 2 sin20

————————————————————

Prob. 3.26

R = 300 450 200 = 950 lb ( )

MA = 300(3) 450(8) 200(6)

Prob. 3.27

R = +5 +3 8 + 12 2 = +10 k ( )

MA = +3(3) 8(6) +12(11) 2(13)

= +67 k-ft

ft 7.610/67x

————————————————————

Prob. 3.28

R = + 40 20 20 + 80 = +80 lb ( )

MA = 20(2) 20(10) + 80(12)

= +720 lb-ft

R is located to the right of A

ft 67.53/17x

————————————————————

Prob. 3.31

Note: F1 (assumed ) is located x1 to the right of A.

+ F1 10 20 = 60 k

F1 = 30 k ( )

MA = 10(4) 20(9) 30x1

17

3-6

Prob. 3.33

————————————————————

Prob. 3.34

————————————————————

Prob. 3.35

(a) pmax = 62.4(18) = 1123 psf

————————————————————

Prob. 3.36

Resultant of the triangular load is 36 k ( ) located 3

ft right of A.

R = 36 + 5 = 41 k

MA = 36(3) 5(13) = 173 k-ft

Prob. 3.37

MA = 21(7) 15(11.5) 10(14)

= 459.5 kN m

m 99.9

46

5.459

x

————————————————————

Prob. 3.38

M = M1 + M2

= +50(6) + 30(6) = + 480 lb-ft

————————————————————

= +30(3) + 20(6) 10(5) = +160 lb-ft

Since M=Fd = +160 lb-ft and d = 2.5 in. :

F = M/d = 160/2.5 = 64 lb

————————————————————

Prob. 3.42

18

3-7

Prob. 3.42 Contd

k 9.47

)57.47()733.5(

22

R

————————————————————

Prob. 3.43

For the resultant:

Rx = Fx = +500 100 = +400 lb

Ry = Fy = 400 + 250 = 150 lb

)150(400

22

R

————————————————————

Prob. 3.44

Rx = Fx = + 200 N

Ry = Fy = 80 50 = 130 N

)130(200

22

R

————————————————————

Prob. 3.45

For the resultant:

Rx = Fx = +15 k

Ry = Fy = 50 20 = 70 k

)70(15

22

R

Rx = Fx = +100 cos20 95 cos80

110 cos70 = + 39.9 N

Ry = Fy = +95 sin80 110 sin70

100 sin20 = 44.0 N

N 5.59

)0.44(9.39

22

R

19

3-8

Prob. 3.50

For the 200-lb resultant force:

Rx= 200 cos60 = +100 lb

Ry = 200 sin60 = +173.2 lb

————————————————————

Prob. 3.51

For the resultant:

Rx = 300 cos45 = 212.1 lb

————————————————————

Prob. 3.52

—————————————————–——-

Prob. 3.53

For the resultant:

Rx = +150 cos60 = +75 lb

Ry = 150 sin60 = 129.9 lb

Also:

————————————————————

Prob. 3.54

Force

(lb)

Horiz.

(lb)

Vert.

(lb)

= 4.216 lb

Ry = +20 sin30 50 sin70 40 sin15

= 47.34 lb

lb 5.47

)34.47()216.4(

22

R

20

3-9

Prob. 3.56

Rx= +75 cos45 250 cos70 200 90 cos30

= 310.4 lb

——————————————–—————-

Prob. 3.57

For the resultant:

Rx = 100 cos20 = +93.97 lb

Ry = 100 sin20 = +34.20 lb

————————————————————

Prob. 3.58

Prob. 3.59

Hydrostatic force F = (1/2)(62.4)(18)2

=10,100 lb

(located 18/3 = 6 ft up from point A)

100,10

MA = 19,800(4) 10,100(6)

= 139,800 lb-ft

Resultant intersects the base of the dam x to the

right of point A:

Prob. 3.61

Solve the force triangle:

F 2 = 6000 2 + 50002 2(6000)(5000)cos25

F = 2570 lb

25sin

sin

21

3-10

Prob. 3.63

MB = 12(3) 14(10) + 25(4)

= 76.0 k-ft

————————————————————

Prob. 3.64

Resultant of dist. force = 10(6) = 60 N

————————————————————

Prob. 3.65

MA = 3(5) (4/5)(5)(10)

= 55.0 k-ft

————————————————————

Prob. 3.66

F1 = 0.5(900 350)(15) = 4125 lb

————————————————————

Prob. 3.67

Prob. 3.68

Ry = 53 70 = 123 kN

Rx = 21 kN

3.80

21

123

tan

kN 8.124)21()123(

1–

22

R

Prob. 3.69

(a)

21

RR

22

Prob. 3.70

3

2.

R1 = 52(0.833)(5)

3.

R1 = 104(0.833)(5)

= 433 lb

————————————————————

Prob. 3.71

(a)

ft–lb 100,19

)20(400)9)(5/3(500

)12)(5/4(500)3(1200

A

M

Prob. 3.72

R = Fy = 25(4) 12 40 = 77 k

MA = 25(4) 12(14) 40(28) = 1388 k-ft

MB = +(0.12 106)(15)+980(10)

+(8 103)sin40 (3)

= 1.825 106 N.m

1.126

N 103.126

)101.126()10128.6(

3

2323

R

Prob. 3.74

F1 + F2 = 150+100 = +250 lb

F1 = 250 F2 I.

MA = +220 = + F1(4) + F2(8) 100(12)

23

3-12

Prob. 3.75

Rx = Fx = +25 cos45 40(4/5)

= 14.32 lb

Ry = Fy = + 25 sin45 + 40(3/5) 20

= + 21.68 lb

24