ch05

STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS

111

.Chapter 5 5-1 Two forces are applied to a bracket as shown in Fig. P5-1. Determine (a) The moment of force F1 about point O. (b) The moment of force F2 about point O. SOLUTION

(a) ( )1 10 10 100 lb in. = = ⋅M .........................................................................................Ans.

(b) ( )2 25 21 525 lb in. = = ⋅M ........................................................................................Ans.

5-2 Determine the moments of the 225-N force shown in Fig. P5-2 about points A, B, and C. SOLUTION

( )225 0.6 135 N m A = = ⋅M .....................................................................................Ans.

( )225 0.4 90 N m B = = ⋅M .......................................................................................Ans.

( )225 0.8 180 N m C = = ⋅M .....................................................................................Ans.

5-3 Two forces are applied at a point in the plane of a rigid steel plate as shown in Fig. P5-3. Determine the moments of

(a) The 500-lb force about points A and B. (b) The 300-lb force about points B and C. SOLUTION

(a) ( )500 30 15,000 lb in. A = = ⋅M ...............................................................................Ans.

(b) ( )500 20 10,000 lb in. B = = ⋅M ...............................................................................Ans.

(c) ( )300 30 9000 lb in. B = = ⋅M ..................................................................................Ans.

(d) ( )300 25 7500 lb in. C = = ⋅M ..................................................................................Ans.

5-4 Two forces are applied to the bridge truss shown in Fig. P5-4. Determine the moments of (a) The 3.6-kN force about points A and D. (b) The 2.7-kN force about point A. SOLUTION

(a) ( )3.6 3 10.80 kN m A = = ⋅M ....................................................................................Ans.

( )3.6 3 10.80 kN m D = = ⋅M ....................................................................................Ans.

(b) ( )2.7 3 8.1 kN m A = = ⋅M .........................................................................................Ans.

5-5 A 50-lb force is applied to the handle of a lug wrench, which is being used to tighten the nuts on the rim of an automobile tire as shown in Fig. P5-5. The diameter of the bolt circle is 5.5 in. Determine the moments of the force about the axle of the wheel (point O) and about the point of contact of the wheel with the pavement (point A).

SOLUTION

( )( )( )( )50cos 20 20cos 20

50sin 20 2.75 20sin 20O = ° °

+ ° + °

M

1047 lb in. = ⋅ ...............................................................Ans.


112

( )( )( )( )50cos 20 20cos 20

50sin 20 16.75 20sin 20A = ° °

+ ° + °

M

1286 lb in. = ⋅ ....................................................................................................Ans.

5-6 A 160-N force is applied to the handle of a door as shown in Fig. P5-6. Determine the moments of the force about the hinges A and B.

SOLUTION

( )( ) ( )( )160cos 45 0.6 160sin 45 0.8 158.4 N m A = ° + ° = ⋅M ......................Ans.

( )( ) ( )( )160sin 45 0.8 160cos 45 0.5 33.9 N m B = ° − ° = ⋅M ........................Ans.

5-7 Determine the moment of the 425-lb force shown in Fig. P5-7 about point B. SOLUTION

( )( ) ( )( )425cos35 16 425sin 35 16 9470 lb in. B = ° + ° = ⋅M ........................Ans.

5-8 A pry bar is used to extract a nail from a board as shown in Fig. P5-8. Determine the moment of the 120-N force

(a) About point A. (b) About point B. SOLUTION

(a) ( )( )( )( )120cos 20 0.75

120sin 20 0.46A = °

+ °

M

103.5 N m = ⋅ ..........................................Ans.

(b) ( )( )( )( )120cos 20 0.65

120sin 20 0.460B = °

+ °

M

92.2 N m= ⋅ .............................................Ans.

5-9 A man exerts a force P to hold a 60-ft pole in the position shown in Fig. P5-9. If the moment of the force P about point A is 4000 lb ft⋅ , determine the magnitude of the force P.

SOLUTION 60cos50 38.567 ftb = ° =

60sin 50 45.963 fth = ° =

1tan 23.72866hb

φ −= = °+

( )( ) ( )( )cos 45.963 sin 38.567 4000 lb ftA P Pφ φ= − = ⋅M

150.6 lbP = .......................................................................................................................Ans.

5-10 Two forces act on a truss as shown in Fig. P5-10. Member AC of the truss is perpendicular to members BC and CD. Determine the moment of

(a) The 4-kN force about point A. (b) The 3-kN force about point D. SOLUTION

(a) ( )4 2sin 60 6.93 kN m A = ° = ⋅M ............................................................................Ans.

(b) ( )3 4sin 60 10.39 kN m D = ° = ⋅M ..........................................................................Ans.


113

5-11 The crane and boom shown in Fig. P5-11 is lifting a 4000-lb load. Determine the moment of the load about point C.

SOLUTION

( )4000 24cos30 1 1 83,100 lb ft C = °− + = ⋅M .....................................................Ans.

5-12 Due to combustion, a compressive force P is exerted on the connecting rod of an automobile engine as shown in Fig. P5-12. The lengths of the crank throw AB and connecting rod BC are 75 mm and 225 mm, respectively. Determine the moment of the force P about the bearing at A in terms of the crank angle θ.

SOLUTION From the Law of Sines

sin sin75 225φ θ=

sinsin3θφ =

Also

2

2 sincos 1 sin 19θφ φ= − = −

Then

( )( ) ( )( )cos 75sin sin 75cosA P Pφ θ φ θ= +M

( )75 sin cos cos sinP θ φ θ φ= +

2sin sin cos75 sin 1

9 3P θ θ θθ

= − + ..........................................................Ans.

5-13 A 760-lb force acts on a bracket as shown in Fig. P5-13. Determine the moment of the force about point A (a) Using the vector approach. (b) Using the scalar approach. SOLUTION

(a) ( ) ( )10 12 760cos 40 760sin 40A = − × ° + °M i j i j

( ) ( ) ( )10 760sin 40 12 760cos 40 11,870 lb in.= ° + ° = ⋅ k k .....................Ans.

(b) ( )( ) ( )( )760cos 40 12 760sin 40 10A = ° + °M

( )11,870 lb in. 11,870 lb in.= ⋅ = ⋅k ............................................................Ans.

5-14 Determine the moment of the 675-N force shown in Fig. P5-14 about point O (a) Using the vector approach. (b) Using the scalar approach. SOLUTION

(a) ( ) ( )0.230 0.250 675cos 22 675sin 22O = + × ° − °M i j i j

( ) ( ) ( )0.230 675sin 22 0.250 675cos 22 215 N m= − ° − ° = − ⋅ k k .........Ans.

(b) ( )( ) ( )( )675cos 22 0.250 675sin 22 0.230O = ° + °M

( )215 N m 215 N m= ⋅ = − ⋅k ......................................................................Ans.


114

5-15 Two forces F1 and F2 are applied to a gusset plate as shown in Fig. P5-15. Determine the moment (a) Of force F1 about point A. (b) Of force F2 about point B. SOLUTION

(a) ( ) ( )9 4 550sin 45 550cos 45A = + − ° + °M i j × i j

( ) ( ) ( )9 550cos 45 4 550sin 45 5060 lb in.= ° + ° = ⋅ k k ............................Ans.

(b) ( ) ( )7 4 750sin 60 750cos 60B = − + × ° + °M i j i j

( ) ( ) ( )7 750cos 60 4 750sin 60 5220 lb in.= − ° − ° = − ⋅ k k ......................Ans.

5-16 Two forces F1 and F2 are applied to a bracket as shown in Fig. P5-16. Determine the moment (a) Of force F1 about point O. (b) Of force F2 about point A. SOLUTION

(a) ( ) ( )0.300 0.500 5cos 45 5sin 45O = + × ° + °M i j i j

( ) ( ) ( )0.300 5sin 45 0.500 5cos 45 0.707 kN m= ° − ° = − ⋅ k k ................Ans.

(b) ( ) ( )0.265 0.375 3cos 45 3sin 45A = + × ° − °M i j i j

( ) ( ) ( )0.265 3sin 45 0.375 3cos 45 1.358 kN m= − ° − ° = − ⋅ k k .............Ans.

5-17 Determine the moment of the contact force F of the crutch shown in Fig. P5-17 about point A. The length of the crutch is 5 ft and F = 35 lb.

SOLUTION

( ) ( )5cos 70 5sin 70 35cos 45 35sin 45A = ° − ° − ° + °M i j × i j

( )( ) ( )( )5cos 70 35sin 45 5sin 70 35cos 45= ° ° − − ° − ° k

( )74.0 lb ft= − ⋅k ....................................................................................................Ans.

5-18 The moment of the force F shown in Fig. P5-18 about point A is (–2700k) N-mm, and its moment about point C is ( )7500 N mm− ⋅k . Determine the magnitude and orientation (angle θ) of the force F.

SOLUTION

( ) ( )120 75 cos sinA F Fθ θ= + × +M i j i j

( ) ( ) ( )120 sin 75 cos 2700 N mmF Fθ θ= − = − ⋅ k k

( ) ( ) ( ) ( ) ( )75 cos sin 75 cos 7500 N mmC F F Fθ θ θ= × + = − = − ⋅ M j i j k k

Therefore 75 cos 7500 N mmF θ = ⋅

120 sin 7500 2700 4800 N mmF θ = − = ⋅

sin 40tancos 100FF

θθθ

= =

21.80θ = ° .................................................................................................................Ans.

107.7 NF = ..............................................................................................................Ans.


115

5-19 A 970-lb force acts at a point in a body as shown in Fig. P5-19. Determine the moment of the force about point C.

SOLUTION

( )2 2

25 30970 620.98 745.17 lb25 30

− += = − ++

i jF i j

( ) ( )30 20 620.98 745.17C = − + − +M j k × i j

( )14,900 12,420 18,630 lb in.= − − − ⋅i j k .......................................................Ans.

5-20 A 200-N force is applied to a pipe wrench as shown in Fig. P5-20. Determine the moment of the force about point A. Express the result in Cartesian vector form.

SOLUTION

( ) ( )0.175 0.580 0.250 200A = − + + × −M i j k k

( )116.0 35.0 N m= − − ⋅i j ....................................................................................Ans.

5-21 A 650-lb force acts at a point in a body as shown in Fig. P5-21. Determine (a) The moment of the force about point A. (b) The direction angles associated with the moment vector. SOLUTION

(a) ( )2 2 2

8 16 30650 148.876 297.751 558.283 lb8 16 30+ += = + ++ +

i j kF i j k

( ) ( )15 24 148.876 297.751 558.283A = − − + +M i j × i j k

( )13,399 8374 893.2 lb in.= − + − ⋅i j k .............................................................Ans.

(b) ( ) ( ) ( )2 2 213,399 8374 893.2 15,826 lb in.AM = + + = ⋅

1 13,399cos 147.8515,826xθ

− −= = ° ......................................................................................Ans.

1 8374cos 58.0515,826yθ

−= = ° ............................................................................................Ans.

1 893.2cos 93.2415,826zθ

− −= = ° ...........................................................................................Ans.

5-22 A pipe bracket is loaded as shown in Fig. P5-22. Determine the moment of the force F about point O. SOLUTION

( )2 2 2

75 150 140875 300.40 600.80 560.75 N75 150 140

+ += = + ++ +

i j kF i j k

( ) ( )0.2 0.25 0.15 300.40 600.80 560.75O = + + + +M i j k × i j k

( )50.1 67.1 45.1 N m= − + ⋅i j k ..........................................................................Ans.


116

5-23 The magnitude of the tension T in cable CD of Fig. P5-23 is 150 lb. Determine the moment of T about point B

(a) Using a position vector from B to D. (b) Using a position vector from B to C. SOLUTION

( )2 2 2

20 22 18150 86.315 94.947 77.684 lb20 22 18

− += = − ++ +

i j kT i j k

( ) ( )28 18 86.315 94.947 77.684B = + − +M i k × i j k

( )1709 621 2660 lb in.= − − ⋅i j k ........................................................................Ans.

( ) ( )8 22 86.315 94.947 77.684B = + − +M i j × i j k

( )1709 621 2660 lb in.= − − ⋅i j k ........................................................................Ans.

5-24 A 450-N force F acts on a machine component as shown in Fig. P5-24. The direction angles of F are 70xθ = ° , 30yθ = ° , and 69zθ = ° . Determine the moment of the force about point A. Express your

answer in Cartesian vector form. SOLUTION

( ) ( )450 cos 70 cos30 cos 69 153.909 389.711 161.266 N= ° + ° + ° = + +F i j k i j k

( ) ( )0.3 0.6 0.4 153.909 389.711 161.266A = − + + +M i j k × i j k

( )253 13.18 209 N m= − + + ⋅i j k .......................................................................Ans.

5-25 The magnitude of the force F shown in Fig. P5-25 is 100 lb. Determine the moment of F about the bearing at C.

SOLUTION

( )2 2 2

10 7 6100 73.521 51.465 44.113 lb10 7 6

− + += = − + ++ +

i j kF i j k

( ) ( )9 14 73.521 51.465 44.113C = − − − + +M j k × i j k

( )324 1029 662 lb in.= + − ⋅i j k ..........................................................................Ans.

5-26 Determine the moment of the 800-N force shown in Fig. P5-26 about point D (a) Using a position vector from D to B. (b) Using a position vector from D to E. SOLUTION

(a) ( )2 2 2

400 400 600800 388.057 388.057 582.086 N400 400 600

− −= = − −+ +

i j kF i j k

( ) ( )0.4 0.8 0.6 388.057 388.057 582.086D = − + − −M i j k × i j k

( )699 466 155 N m= + + ⋅i j k .............................................................................Ans.

(b) ( ) ( )0.8 1.2 388.057 388.057 582.086D = − − −M i j × i j k

( )699 466 155 N m= + + ⋅i j k .............................................................................Ans.


117

5-27 The magnitudes of the forces F1, F2, and F3 shown in Fig. P5-27 are 550 lb, 300 lb, and 600 lb, respectively. Determine the sum of the moments of the three forces about point B.

SOLUTION

( )1 2 2 2

7 11 3550 287.763 452.198 123.327 lb7 11 3

− − += = − − ++ +

i j kF i j k

( ) ( )1 11 6 287.763 452.198 123.327B = − − − +M j k × i j k

( )1357 1727 3165 lb ft= − + + ⋅i j k

( ) ( ) ( )2 11 6 300 1800 3300 lb ftB = − = − − ⋅M j k × i j k

( ) ( ) ( )3 6 6 600 3600 lb ftB = − − = − ⋅M j k × k i

( )4960 73 135 lb ftBΣ = − − − ⋅M i j k .........................................................................Ans.

5-28 If the magnitude of the moment of the cable force T shown in Fig. P5-28 about the hinge at B is 1150 N mBM = ⋅ , determine the magnitude of the force T.

SOLUTION

500 mmxCD = − 1400cos30 1212.436 mmyCD = − = −

750 1400sin 30 1450 mmzCD = + =

( )2 2 2

500 1212.436 1450500 1212.436 1450

0.25574 0.62013 0.74164 N

T

T T T

− − +=+ +

= − − +

i j kT

i j k

( )0.5 1.1cos30 0.75 1.1sin 30B = − ° + + ° M i j k ×T

( )0.09966 0.70328 0.55369 N mT T T= − − ⋅i j k

( ) ( ) ( )2 2 20.09966 0.70328 0.55369

0.90062 1150 N mBM T T T

T

= + +

= = ⋅

1277 NT = .........................................................................................................................Ans.

5-29 The force F in Fig. P5-29 can be expressed in Cartesian vector form as ( )60 100 120 lb= + +F i j k . Determine the scalar component of the moment at point B about line BC.

SOLUTION

( ) ( ) ( )18 32 60 100 120 3200 240 1800 lb in.C = + + + = − − + ⋅M i k × i j k i j k

( )2 2

9 18 0.44721 0.894439 18

BC− += = − +

+i je i j

( )( ) ( )( )3200 0.44721 240 0.89443BC C BCM = = − − + −M e

1216 lb in.= ⋅ ...........................................................................................................Ans.


118

5-30 A 3000-N force is applied to the pipe shown in Fig. P5-30. Determine the scalar component of the moment of the force about the x-axis.

SOLUTION

( )2 2 2

280 300 5303000 1253.173 1342.686 2372.078 N280 300 530

− −= = − −+ +

i j kF i j k

( ) ( )0.65 0.53 1253.173 1342.686 2372.078O = + − −M j k × i j k

( )830.227 664.182 814.562 N m= − + − ⋅i j k

830 N mOx OM = = − ⋅M i .............................................................................................Ans.

5-31 A force ( )30 50 40 lb= − + −F i j k is applied to the machine component shown in Fig. P5-31. Determine the scalar component of the moment of the force about the z-axis.

SOLUTION

( ) ( )10 24 16 30 50 40O = − + − − + −M i j k × i j k

( )160 80 220 lb in.= − + + ⋅i j k

220 lb in.Oz OM = = ⋅M k .............................................................................................Ans.

5-32 The magnitude of the force F in Fig. P5-32 is 595 N. Determine the scalar component of the moment at point O about line OC.

SOLUTION

( )2 2

220 200595 440.26 400.24 N220 200

− += = − ++

i kF i k

( ) ( )0.22 0.24 440.26 400.24O = + − +M i j × i k

( )96.06 88.05 105.66 N m= − + ⋅i j k

( )2 2

220 200 0.73994 0.67267220 200

OC+= = ++

i ke i k

( )( ) ( )( )96.06 0.73994 105.66 0.67267OC O OCM = = +M e

142.2 N m= ⋅ ..........................................................................................................Ans.

5-33 A 40-lb vertical force F is applied to a lug wrench as shown in Fig. P5-33. Determine the magnitude of the component of the moment that would tighten the lug nut.

SOLUTION

( ) ( ) ( )4 8 3 40 320 160 lb in.O = + − − = − + ⋅M i j k × k i j

320 lb in.Oz OM = = ⋅M i ............................................................................................Ans.

5-34 If the magnitude of the force T shown in Fig. P5-34 is 1000-N, determine the scalar component of the moment of the force about the line CD.

SOLUTION

( )2 2

1.8 3.61000 447.21 894.43 N1.8 3.6

+= = ++

j kT j k

( ) ( )1.2 2.7 447.21 894.43C = − + +M i j × j k


119

( )2414.96 1073.32 536.65 N m= + − ⋅i j k

( )2 2

2.4 1.5 0.84800 0.530002.4 1.5

CD− −= = − −

+i je i j

( )( ) ( )( )2414.96 0.84800 1073.32 0.53000CD C CDM = = − + −M e

2620 N m= − ⋅ .........................................................................................................Ans.

5-35 A 120-lb force F is applied to a lever-shaft assembly as shown in Fig. P5-35. Determine the moment of the force about each coordinate axis.

SOLUTION

( )2 2 2

2 14 16120 11.2390 78.6732 89.9122 lb2 14 16

− + −= = − + −+ +

i j kF i j k

( ) ( )13 16 11.2390 78.6732 89.9122O = + − + −M i k × i j k

( )1258.77 989.03 1022.75 lb in.= − + + ⋅i j k

1259 lb in.OxM = − ⋅ .........................................................................................................Ans.

989 lb in.OyM = ⋅ ..............................................................................................................Ans.

1023 lb in.OzM = ⋅ ............................................................................................................Ans.

5-36 A 650-N force acts on the awning structure shown in Fig. P5-36. Determine the moment of the force about line BC. Express the result in Cartesian vector form.

SOLUTION

( ) ( ) ( )0.6 0.6 650 390.0 N mB = − − = ⋅M i k × k j

( )2 2

1.2 0.9 0.8000 0.60001.2 0.9

BC−= = −+

j ke j k

( )( )390.0 0.8000 312 N mBC B BCM = = = ⋅M e

( ) ( )312 0.8000 0.6000 250 187.2 N mBC BC BCM= = − = − ⋅M e j k j k .............Ans.

5-37 The magnitude of the force F in Fig. P5-37 is 107 lb. Determine the component of the moment of the force that rotates the door about the axis of the hinges.

SOLUTION

34 tan 20 12.375 in.Az = ° = 32 19.625 in.Az− =

( )2 2 2

36 34 19.625107 72.318 68.300 39.423 lb36 34 19.625

− − += = − − ++ +

i j kF i j k

( ) ( )36 32 72.318 68.300 39.423O = − + − − +M i k × i j k

( )2185.60 894.95 2458.80 lb in.= − + ⋅i j k

2190 lb in.OxM = ⋅ ...........................................................................................................Ans.


120

5-38 A bracket is rigidly attached to a wall at O and is subjected to a 384-N force F as shown in Fig. P5-38. Determine the component of the moment of the force

(a) That twists the bracket about the y-axis. (b) That bends the bracket about the x-axis. SOLUTION

( )2 2 2

150 250 250384 149.978 249.963 249.963 N150 250 250

− += = − ++ +

i j kF i j k

( ) ( )0.3 0.5 0.2 149.978 249.963 249.963O = + − − +M i j k × i j k

( )74.989 104.985 149.978 N m= − − ⋅i j k

(a) 105.0 N mOyM = − ⋅ .........................................................................................................Ans.

(b) 75.0 N mOxM = ⋅ ..............................................................................................................Ans.

5-39 A bar is bent in a circular arc of radius R = 6 ft and is subjected to a 660-lb force F as shown in Fig. P5-39. The force tends to twist and bend the member about the coordinate axes. Determine the twisting and bending moments and state the axes about which each occurs.

SOLUTION

( )2 2

4 4660 466.690 466.690 lb4 4−= = −+

i kF i k

( ) ( )6 6 466.690 466.690O = + −M j k × i k

( )2800.14 2800.14 2800.14 lb in.= − + − ⋅i j k

Twist: 2800 lb in.⋅ about the negative z-axis ..........................................................Ans.

Bend: 2800 lb in.⋅ about the negative x-axis ..........................................................Ans.

Bend: 2800 lb in.⋅ about the positive y-axis ...........................................................Ans.

5-40 The magnitude of the force F in Fig. P5-40 is 976 N. Determine (a) The component of the moment at point C parallel to line CE. (b) The component of the moment at point C perpendicular to line CE and the direction angles associated

with this moment vector. SOLUTION

(a) ( )2 2 2

350 300 160976 700.06 600.06 320.03 N350 300 160

− − += = − − ++ +

i j kF i j k

( ) ( )( )0.14 0.4 700.06 600.06 320.03

128.01 44.80 196.02 N mC = + − − +

= − + ⋅

M i j × i j k

i j k

( )2 2

210 400 0.46483 0.88540210 400

CE− += = − +

+i je i j

( )( ) ( )( )128.01 0.46483 44.80 0.88540 99.169 N mCE C CEM = = − + − = − ⋅M e

( )46.10 87.80 N mCE CE CEM= = − ⋅M e i j ................................................................Ans.

(b) ( ) ( )128.01 44.80 196.02 46.10 87.80C CE⊥ = − = − + − −M M M i j k i j

( )81.91 43.00 196.02 N m= + + ⋅i j k ................................................................Ans.


121

( ) ( ) ( )2 2 281.91 43.00 196.02 216.75 N mM⊥ = + + = ⋅

1 81.91cos 67.80216.75xθ

−= = ° ............................................................................................Ans.

1 43.00cos 78.56216.75yθ

−= = ° ...........................................................................................Ans.

1 196.02cos 25.26216.75zθ

−= = ° ............................................................................................Ans.

5-41 The magnitude of the force F in Fig. P5-41 is 781 lb. Determine (a) The component of the moment at point C parallel to line CD. (b) The component of the moment at point C perpendicular to line CD and the direction angles associated

with this moment vector. SOLUTION

(a) ( )2 2

12 10781 599.98 499.98 lb12 10

+= = ++

i kF i k

( ) ( )12 25 599.98 499.98C = − + +M i j × i k

( )12,499.60 5999.81 14,999.52 lb in.= + − ⋅i j k

( )2 2

12 10 0.76822 0.6401812 10

CD− += = − +

+i ke i k

( )( ) ( )( )12,499.60 0.76822 14,999.52 0.64018CD C CDM = = − + −M e

19,204.8 lb in.= − ⋅

( )14,753.5 12,294.6 lb in.CD CD CDM= = − ⋅M e i k .................................................Ans.

(b) ( )2253.9 5999.8 2704.9 lb in.C CD⊥ = − = − + − ⋅M M M i j k ..............................Ans.

( ) ( ) ( )2 2 22253.9 5999.8 2704.9 6956.6 lb in.M⊥ = + + = ⋅

1 2253.9cos 108.906956.6xθ

− −= = ° .......................................................................................Ans.

1 5999.8cos 30.416956.6yθ

−= = ° ...........................................................................................Ans.

1 2704.9cos 112.886956.6zθ

− −= = ° .......................................................................................Ans.

5-42 Determine the moment of the couple shown in Fig. P5-42 and the perpendicular distance between the two forces.

SOLUTION

( )( ) ( )( )760cos35 0.2 760sin 35 0.1 168.103 N m = ° + ° = ⋅M ...................Ans.

168.103 N m 760d⋅ =

0.221 m 221 mmd = = ...................................................................................................Ans.


122

5-43 A lug wrench is being used to tighten a lug nut on an automobile wheel as shown in Fig. P5-43. Two equal magnitude, parallel forces of opposite sense are applied to the wrench. If the magnitude of each force is 25 lb, determine the couple applied to a lug nut and express the result in Cartesian vector form.

SOLUTION

( )( )25 12 300 lb in.= − = − ⋅M i i ..................................................................................Ans.

5-44 To open a valve to a steam line in a power plant requires a couple of magnitude 54 N m⋅ . Determine the magnitude of each force F shown in Fig. P5-44 required to open the valve.

SOLUTION

( )54 N m 0.3M F= ⋅ =

180 NF = ..........................................................................................................................Ans.

5-45 A beam is loaded with a system of three couples as shown in Fig. P5-45. Express the resultant of the couple system in Cartesian vector form.

SOLUTION

( )( ) ( )( ) ( )( )500 30 250 14 300 30= − −M

( )2500 lb in. 2500 lb in.= ⋅ = ⋅k ..................................................................Ans.

5-46 The input and output torques (couples) from a gear box are shown in Fig. P5-46. Determine the magnitude and direction of the resultant torque T.

SOLUTION

( )750 200 125 150 cos 45 sin 45= + + + ° + °T i j k j k

( )750 306 231 N m= + + ⋅i j k

( ) ( ) ( )2 2 2750 306 231 842.36 N mT = + + = ⋅ .......................................................Ans.

1 750cos 27.08842.36xθ

−= = ° ............................................................................................Ans.

1 306cos 68.70842.36yθ

−= = ° ............................................................................................Ans.

1 231cos 74.08842.36zθ

−= = ° ............................................................................................Ans.

5-47 Three couples are applied to a rectangular block as shown in Fig. P5-47. Determine the magnitude of the resultant couple C and the direction angles associated with the resultant couple vector.

SOLUTION

( ) ( ) ( ) ( ) ( ) ( )15 10 18 8 10 20 8 15 15= − − + − + + +M j k × i i k × j i j × k

( )25 60 110 lb in.= + + ⋅i j k

( ) ( ) ( )2 2 225 60 110 127.77 lb in.M = + + = ⋅ ..........................................................Ans.

1 25cos 78.72127.77xθ

−= = ° ............................................................................................Ans.

1 60cos 61.99127.77yθ

−= = ° ............................................................................................Ans.


123

1 110cos 30.58127.77zθ

−= = ° ............................................................................................Ans.

5-48 Three couples are applied to a bent bar as shown in Fig. P5-48. Determine (a) The magnitude of the resultant couple C and the direction angles associated with the resultant couple

vector. (b) The scalar component of the resultant couple C about line OA. SOLUTION

(a) ( )80 65 95 N m= + + ⋅C i j k

( ) ( ) ( )2 2 280 65 95 140.178 N mC = + + = ⋅ ............................................................Ans.

1 80cos 55.20140.178xθ

−= = ° ..........................................................................................Ans.

1 65cos 62.37140.178yθ

−= = ° .........................................................................................Ans.

1 95cos 47.34140.178zθ

−= = ° ..........................................................................................Ans.

(b) ( )2 2 2

140 120 80 0.69653 0.59702 0.39801140 120 80

OA+ −= = + −+ +

i j ke i j k

( )( ) ( )( ) ( )( )80 0.69653 65 0.59702 95 0.39801OA OAC = = + + −C e

56.7 N m= ⋅ ..............................................................................................................Ans.

5-49 Three couples are applied to a rectangular block as shown in Fig. P5-49. Determine (a) The magnitude of the resultant couple C and the direction angles associated with the resultant couple

vector. (b) The scalar component of the resultant couple C about line OA. SOLUTION

(a) ( )90 125 75 lb in.= + + ⋅C i j k

( ) ( ) ( )2 2 290 125 75 171.318 lb in.C = + + = ⋅ .........................................................Ans.

1 90cos 58.31171.318xθ

−= = ° ..........................................................................................Ans.

1 125cos 43.14171.318yθ

−= = ° .........................................................................................Ans.

1 75cos 64.04171.318zθ

−= = ° ..........................................................................................Ans.

(b) ( )2 2 2

24 8 32 0.58835 0.19612 0.7844624 8 32

OA+ += = + ++ +

i j ke i j k

( )( ) ( )( ) ( )( )90 0.58835 125 0.19612 75 0.78446OA OAC = = + +C e

136.3 lb in.= ⋅ ............................................................................................................Ans.


124

5-50 Replace the 3-kN force shown in Fig. P5-50 by a force at point A and a couple. SOLUTION

3 kN A = ↓F ......................................................................................................................Ans.

( )( )3000 0.15 =450 N m A = ⋅C ...............................................................................Ans.

5-51 Replace the 50-lb force shown in Fig. P5-51 by a force at point A and a couple. Express your answer in Cartesian vector form.

SOLUTION

( )50 lbA =F j .....................................................................................................................Ans.

( )( )( ) ( )50 6 300 lb in.A = − = − ⋅C k k ........................................................................Ans.

5-52 Replace the 130-N vertical force shown in Fig. P5-52 by the mechanic’s hand to the wrench by an equivalent force-couple system at the lug nut.

SOLUTION

130 N A = ↓F ....................................................................................................................Ans.

( )( )130 0.3 =39.0 N m A = ⋅C ...................................................................................Ans.

5-53 A gusset plate is riveted to a beam by three rivets as shown in Fig. P5-53. Replace the 2500-lb force by a force-couple system at the top rivet.

SOLUTION 2500 lb = →F

( )( )2500 6 15,000 lb in. = = ⋅C

5-54 Four forces are applied to a truss as shown in Fig. P5-54. Determine the magnitude and direction of the resultant of the four forces and the perpendicular distance dR from point A to the line of action of the resultant.

SOLUTION

( )3 12 kN 12.369 kN A = − =R i j 75.96° ............................................................Ans.

( )( ) ( )( ) ( )( )6 2 4 4 2 6 40 kN m 12.369A d= + + = ⋅ =C

3.23 md = .........................................................................................................................Ans.

5-55 Three forces are applied to the locked pulley shown in Fig. P5-55. Determine the magnitude and direction of the resultant of the three forces and the perpendicular distance from the axle of the pulley to the line of action of the resultant.

SOLUTION

( ) ( ) ( )( )

160cos 20 160sin 20 90

150.351 35.277 lbA = − ° + ° −

= − −

R i j j

i j

154.43 lb = 13.20° ..............................................................................................Ans.

( )( ) ( )( ) ( )( )120 1 40 1 90 2 100 lb ft 154.43A d= − − = − ⋅ = −C

0.648 ftd = .......................................................................................................................Ans.

5-56 Determine the resultant of the four forces acting on the bell crank shown in Fig. P5-56, and determine where the resultant intersects the x-axis.

SOLUTION

( ) ( )175 400cos 45 400sin 45 300 200= + ° + ° + −R i i j j j


125

( )457.84 382.84 N 596.82 N = + =i j 39.90° ...................................................Ans.

( )( ) ( )( )400 50 200 80 36,000 N mm 382.84A d= + = ⋅ =C

94.0 mm left of d O= ....................................................................................................Ans.

5-57 Three 75-lb traffic lights are suspended over a roadway as shown in Fig. P5-57. Determine the resultant of the weights of the traffic lights and locate the resultant with respect to point A.

SOLUTION

( )( )3 75 225 lb = = ↓R ..............................................................................................Ans.

( )( ) ( )( ) ( )( )75 8 75 11 75 14 2475 lb ft 225d= + + = ⋅ =C

11 ftd = ..............................................................................................................................Ans.

5-58 Four parallel forces act on a concrete slab as shown in Fig. P5-58. Determine the resultant of the forces and locate the intersection of the line of action of the resultant with the xy-plane.

SOLUTION

( ) ( )4 3 3 2 12 kN= − − − − = −R k k ...........................................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )3 1.5 4 3 3 3 1.5 3 3 4.5 2= + − + + − + + − + −C i j × k i j × k i j × k j × k

( )33 25.5 kN m= − + ⋅i j

( ) ( ) ( )12 12 12x y y x= + − = − +i j × k i j

25.5 12 2.12 mx = = ......................................................................................................Ans.

33 12 2.75 my = = ..........................................................................................................Ans.

5-59 The forces exerted on the wheels of an airplane by the runway are shown in Fig. P5-59. Determine the resultant of the three forces and locate the intersection of the line of action of the resultant with the xy-plane (the runway).

SOLUTION

( ) ( )5000 5000 2500 12,500 lb= + + =R k k ...........................................................Ans.

( ) ( ) ( ) ( ) ( )20 8 5000 20 8 5000 200,000 lb ftO = + + − = − ⋅C i j × k i j × k j

( ) ( ) ( )12,500 12,500 12,500x y y x= + = −i j × k i j

200,000 12,500 16 ftx = = ..........................................................................................Ans.

0 12,500 0 fty = = ..........................................................................................................Ans.

5-60 The magnitude of the force F acting on the casting shown in Fig. P5-60 is 2 kN. Replace the force with a force R through point A and a couple C.

SOLUTION

( )2 2 2

3 4 122000 461.538 615.385 1846.154 N3 4 12+ += = + ++ +

i j kR i j k .................Ans.

( ) ( )0.25 0.075 461.538 615.385 1846.154= + + +C j k × i j k

( )415.4 34.6 115.4 N m= + − ⋅i j k ..........................................................................Ans.


126

5-61 A force ( )50 50 200 lb+ −F = i j k acts on the wall bracket shown in Fig. P5-61. Replace the force by a force-couple system at rivet B. The eye bolt is small and the force F may be considered as acting at point C.

SOLUTION

( )50 50 200 lb= + −F i j k ............................................................................................Ans.

( ) ( )2 4 3 50 50 200= + − + −C i j k × i j k

( )650 250 100 lb in.= − + − ⋅i j k ...............................................................................Ans.

5-62 The homogeneous plate shown in Fig. P5-62 has a mass of 90 kg. The magnitude of the force T in cable BC is 800 N. Replace the weight and cable forces by an equivalent force-couple system at hinge A.

SOLUTION

( )( )( ) ( )90 9.81 882.90 N= − = −W k k

( )2 2 2

350 610 500800 324.486 565.533 463.552 N350 610 500

− − += = − − ++ +

i j kT i j k

( )324 566 419 N= + = − − −R W T i j k ...................................................................Ans.

( ) ( ) ( ) ( )0.65 0.28 324.486 565.533 463.552 0.35 882.90= + − − + + −C i j × i j k i × k

( )129.8 7.71 277 N m= + − ⋅i j k ............................................................................Ans.

5-63 Forces act at A, D, and E of the member shown in Fig. P5-63. If the equivalent force-couple system at B is ( )100 100 lb= − −R i j and a couple C, determine the magnitude of FE, the angle θ, and the required

couple C. SOLUTION

( ) ( ) ( )100 300 100 100 lbEx EyF F= − + + − = − +R i j i j

0 lbExF = 200 lbEyF =

200 lb E = ↑F ...................................................................................................................Ans.

( )( ) ( )( ) ( )( )100 10 300 14 200 20 800 lb in. = − + = ⋅C ..................................Ans.

5-64 Four forces and a couple are applied to a rectangular plate as shown in Fig. P5-64. Determine the magnitude and direction of the resultant of the force-couple system and the distance xR from point O to the intercept of the line of action of the resultant with the x-axis.

SOLUTION

( ) ( ) ( )90 60 75 50 30 25 N= − + − = +R i j i j

39.1 N = 39.81° ....................................................................................................Ans.

( )( ) ( )( ) ( )( ) ( )( )10 90 0.25 50 0.3 60 0.25 75 0.3

65.00 N m 25Rx= − − − −= − ⋅ =

C

65.00 25 2.6 mRx = − = − ..............................................................................................Ans.

5-65 The magnitude of the resultant R of the four forces acting on the legs of the table shown in Fig. P5-65 is 80 lb. Determine the magnitude of F and the location, xR and yR, of the line of action of R.

SOLUTION

( ) ( )30 20 15 80 lbF= + + + =R j j


127

15 lbF = .............................................................................................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( )3 30 3 15 3 3 20 105 150 lb ft= + + + = − ⋅C i × k j × k i j × k i j

( ) ( ) ( )80 80 80R R R Rx y y x= + = −i j × j i j

150 80 1.875 ftRx = = ....................................................................................................Ans.

105 80 1.313 ftRy = = ....................................................................................................Ans.

5-66 Figure P5-66 shows a crankshaft-flywheel arrangement of a one-cylinder engine. A 1000-N force P is supplied by the connecting rod, and a couple C of magnitude 250 N-m is delivered to the crankshaft by the flywheel. Replace the force and couple by an equivalent force-couple system at the bearing A.

SOLUTION

( )( ) ( )( )1000sin 20 1000cos 20= ° − + ° −P j k

( )342.02 939.69 N= − −j k .......................................................................................Ans.

( ) ( ) ( )0.225 0.125 342.02 939.69 250A = − − − − + −M i j × j k i

( )132.5 211.4 77.0 N m= − − + ⋅i j k .................................................................Ans.

5-67 A farmer is using the hand winch shown in Fig. P5-67 to raise a 40-lb bucket of water from a well. When the force P and the weight of the bucket are replaced by an equivalent force-couple system at bearing D, the result is 10 60 lb= − −R j k and a couple CD Determine the force P applied to the handle of the winch and the couple CD.

SOLUTION

( ) ( )40 10 60 lbx y zP P P= + + − = − −R i j k k j k

0 lbxP = 10 lbyP = − 20 lbzP = − ..................................Ans.

( ) ( ) ( ) ( )30 2.5 40 61 20cos30 20sin 30 10 20D = + − + − ° + ° − −C i j × k i j k × j k

( )346 2420 610 lb in.= + − ⋅i j k .........................................................................Ans.

5-68 In order to remove a rusty screw from a steel plate, a worker attaches a screwdriver to the bent bar shown in Fig. P5-68. To hold the screwdriver in place, the worker applies a 30-N force at C; a 50-N force is applied to the handle of the bar in an attempt to remove the screw. Forces Fy and Fz are also applied at C. It is desirable that the screwdriver not bend about the y and z axes.

(a) Determine the forces Fy and Fz. (b) Replace the forces at B and C by an equivalent force-couple system at A. SOLUTION

(a) ( ) ( ) ( ) ( )0.45 30 0.25 0.15 50A y zF F= − + + + − −M i × i j k i j × k

( ) ( )7.5 12.5 0.45 0.45 0 0 N mz y xF F M= + − + = + + ⋅i j k i j k

0 NyF = 27.78 NzF = ............................................................Ans.

(b) ( ) ( ) ( ) ( )30 27.78 50 30 22.22 N= − + + − = − −R i k k i k ....................................Ans.

( )7.5 N m= ⋅C i ...............................................................................................................Ans.


128

5-69 Reduce the forces shown in Fig. P5-69 to a wrench and locate the intersection of the wrench with the xy-plane.

SOLUTION

( )2 2 2

3 5 4420 178.191 296.985 237.588 lb3 5 4

B− += = − ++ +

i j kF i j k

( )60 178.191 296.985 237.588 250= + − + +R i i j k j

( )238.191 46.985 237.588 lb= − +i j k ...............................................................Ans.

( ) ( ) ( ) ( ) ( ) ( )3 60 5 178.191 296.985 237.588 4 250O = + − + +M j × i j × i j k k × j

( )187.940 1070.955 lb ft= − ⋅i k

( )2 2 2

238.191 46.985 237.588 0.70120 0.13832 0.69942238.191 46.985 237.588

R− += = − ++ +

i j ke i j k

( )( ) ( )( )187.940 0.70120 1070.955 0.69942 617.264 lb ftR O RM = = + − = − ⋅M e

( )432.826 85.380 431.727 lb ftR R RM= = − + − ⋅M e i j k ....................................Ans.

O R= + ×M M r R

( )620.766 85.380 639.228 lb ftO R− = − − ⋅M M i j k

( ) ( )

( )238.191 46.985 237.588

237.588 237.588 46.985 238.191R R

R R R R

x y

y x x y

= + − +

= − − +

r × R i j × i j k

i j k

85.380 237.588 0.359 ftRx = = ..................................................................................Ans.

620.766 237.588 2.61 ftRy = = ..................................................................................Ans.

5-70 Locate the center of mass for the three particles shown in Fig. P5-70 if mA = 26 kg, mB = 21 kg, and mC = 36 kg.

SOLUTION 26 21 36 83 kgm = + + =

( ) ( ) ( )83 26 200cos 60 21 200cos30 36 200 962.693Gx = − ° + − ° + =

( ) ( )83 26 200sin 60 21 200sin 30 2403.332Gy = ° + − ° =

962.693 83 11.60 mmGx = = .......................................................................................Ans.

2403.332 83 29.0 mmGy = = ......................................................................................Ans.

5-71 Locate the center of gravity for the four particles shown in Fig. P5-71 if WA = 20 lb, WB = 25 lb, WC = 30 lb, and WD = 40 lb.

SOLUTION 20 25 30 40 115 lbW = + + + =

( ) ( ) ( ) ( )115 20 0 25 0 30 8 40 0 240Gx = + + + =

( ) ( ) ( ) ( )115 20 12 25 12 30 12 40 0 900Gy = + + + =

( ) ( ) ( ) ( )115 20 10 25 0 30 0 40 0 200Gz = + + + =


129

240 115 2.09 in.Gx = = ..................................................................................................Ans.

900 115 7.83 in.Gy = = ..................................................................................................Ans.

200 115 1.739 in.Gz = = ................................................................................................Ans.

5-72 Locate the center of mass for the four particles shown in Fig. P5-72 if mA = 16 kg, mB = 24 kg, mC = 14 kg, and mD = 36 kg.

SOLUTION 16 24 14 36 90 kgm = + + + =

( ) ( ) ( ) ( )90 16 300 24 0 14 0 36 0 4800Gx = + + + =

( ) ( ) ( ) ( )90 16 0 24 0 14 500 36 0 7000Gy = + + + =

( ) ( ) ( ) ( )90 16 0 24 0 14 0 36 400 14,400Gz = + + + =

4800 90 53.3 mmGx = = ...............................................................................................Ans.

7000 90 77.8 mmGy = = ..............................................................................................Ans.

14,400 90 160 mmGz = = ............................................................................................Ans.

5-73 Locate the center of gravity for the five particles shown in Fig. P5-73 if WA = 25 lb, WB = 35 lb, WC = 15 lb, WD = 28 lb, and WE = 16 lb.

SOLUTION 25 35 15 28 16 119 lbW = + + + + =

( ) ( ) ( ) ( ) ( )119 25 10 35 10 15 0 28 10 16 0 880Gx = + + + + =

( ) ( ) ( ) ( ) ( )119 25 0 35 11 15 16 28 16 16 0 1073Gy = + + + + =

( ) ( ) ( ) ( ) ( )119 25 0 35 0 15 0 28 11 16 11 484Gz = + + + + =

880 119 7.39 in.Gx = = ..................................................................................................Ans.

1073 119 9.02 in.Gy = = ................................................................................................Ans.

484 119 4.07 in.Gz = = ..................................................................................................Ans.

5-74 Locate the center of mass for the five particles shown in Fig. P5-74 if mA = 2 kg, mB = 3 kg, mC = 4 kg, mD = 3 kg, and mE = 2 kg.

SOLUTION 2 3 4 3 2 14 kgm = + + + + =

( ) ( ) ( ) ( ) ( )14 2 300 3 150 4 300 3 300 2 0 3150Gx = + + + + =

( ) ( ) ( ) ( ) ( )14 2 240 3 400 4 400 3 0 2 200 3680Gy = + + + + =

( ) ( ) ( ) ( ) ( )14 2 0 3 0 4 270 3 270 2 270 2430Gz = + + + + =

3150 14 225 mmGx = = ................................................................................................Ans.

3680 14 263 mmGy = = ................................................................................................Ans.

2430 14 173.6 mmGz = = .............................................................................................Ans.


130

5-75 Three bodies with masses of 2, 4, and 6 slugs are located at points (2, 3, 4), (3, −4, 5), and (−3, 4, 6), respectively. Locate the mass center of the system if the distances are measured in feet.

SOLUTION 2 4 6 12 slugm = + + =

( ) ( ) ( )12 2 2 4 3 6 3 2Gx = + + − = −

( ) ( ) ( )12 2 3 4 4 6 4 14Gy = + − + =

( ) ( ) ( )12 2 4 4 5 6 6 64Gz = + + =

2 12 0.1667 ftGx = − = − ................................................................................................Ans.

14 12 1.167 ftGy = = ......................................................................................................Ans.

64 12 5.33 ftGz = = .........................................................................................................Ans.

5-76 Locate the centroid of the shaded area shown in Fig. P5-76 if b = 200 mm and h = 300 mm. SOLUTION

( )VdA hx b dx=

( )HdA b by h dy= −

2

002 2

bb

Vhx hx bhA dA dxb b

= = = =

∫ ∫

3 2

003 3

bb

y Vhx hx hbM xdA x dxb b

= = = =

∫ ∫

2 3 2 133.3 mm

2 3y

C

M hb bxA hb

= = = = ...........................................................................Ans.

2 3 2

002 3 6

hh

x Hby by by bhM y dA y b dyh h

= = − = − = ∫ ∫

2 6 100.0 mm

2 3x

CM bh hyA bh

= = = = ............................................................................Ans.

5-77 Determine the y-coordinate of the centroid of the shaded area shown in Fig. P5-77. SOLUTION

( )2 2VdA x b dx=

2HdA b by dy = −

2 3 2

002 6 6

bb

Vx x bA dA dxb b

= = = =

∫ ∫

/ 22 5/ 2 3 3 3/ 2

00

2 22 5 2 8 10 40

bb

x Hby y b b bM y dA y b by dy b = = − = − = − =

∫ ∫


131

3

2

40 36 20

xC

M b byA b

= = = ...................................................................................................Ans.

5-78 Determine the x-coordinate of the centroid of the shaded area shown in Fig. P5-78. SOLUTION

2

21VxdA b dxa

= −

2

20

2 2 2 1

0

1

sin2 4

a

V

a

xA dA b dxa

b x abx a x aa a

π−

= = −

= − + =

∫ ∫

( )2 232 220

0

13 3

aa

y Vx b baM xdA x b dx a xa a

− = = − = − = ∫ ∫

2 3 4

4 3y

C

M ba axA abπ π

= = = .................................................................................................Ans.

5-79 Locate the centroid of the shaded area shown in Fig. P5-79. SOLUTION

2

225VxdA x dx

= −

2 2

3 42

1 2 22 25 25

1 442 25 625

x C Vx xdM y dA x x dx

x xx

= = − −

= − +

502 350 2 2

00

2 833.33 in.25 75Vx xA dA x dx x

= = − = − =

∫ ∫

250

0

503 43

0

225

2 20,833.33 in.3 100

y VxM xdA x x dx

x x

= = −

= − =

∫ ∫

20,833.33 25.0 in.

833.33y

C

Mx

A= = = .................................................................................Ans.

503 4 3 4 550 2

00

3

1 4 1 442 25 625 2 3 25 3125

8333.33 in.

x xx x x x xM dM x dx

= = − + = − +

=

∫ ∫


132

8333.33 10.00 in.833.33

xC

MyA

= = = ...................................................................................Ans.


( )2 2HdA ay b dy=

2 3

2 2003 3

bb

Hay ay abA dA dyb b

= = = =

∫ ∫

2 4 2

2 2004 4

bb

x Hay ay abM y dA y dyb b

= = = =

∫ ∫

2 4 3

3 4x

CM ab byA ab

= = = ...................................................................................................Ans.


( )VdA ax x dx= −

( )2HdA y y a dy = −

( )3/ 2 2 2

00

3 2 2 6

aa

Vx x aA dA ax x dx a

= = − = − =

∫ ∫

( )5/ 2 3 3

00

5 2 3 15

aa

y Vx x aM xdA x ax x dx a

= = − = − =

∫ ∫

3

2

15 26 5

yC

M a axA a

= = = ...................................................................................................Ans.

2 3 4 3

003 4 12

aa

x Hy y y aM y dA y y dya a

= = − = − =

∫ ∫

3

2

126 2

xC

M a ayA a

= = = .....................................................................................................Ans.


4 59 9Vx xdA x dx dx = − =

929 2

33

5 5 20 mm9 18Vx xA dA dx

= = = = ∫ ∫

24 1 5 5 65

9 2 9 9 162x C Vx x x xdM y dA dx dx

= = + =


133

92 39 3

33

65 65 93.889 mm162 486x x C Vx xM dM y dA dx

= = = = =

∫ ∫ ∫

93.889 4.69 mm

20x

CMyA

= = = ....................................................................................Ans.

5-83 Determine the x-coordinate of the shaded area shown in Fig. P5-83. SOLUTION

( )3 2VdA x dx=

23 42 2

11

15 in.2 8 8Vx xA dA dx

−−

= = = =

∫ ∫

23 52 3

11

33 in.2 10 10y Vx xM xdA x dx

−−

= = = =

∫ ∫

33 10 44 1.760 in.15 8 25

yC

Mx

A= = = ≅ .............................................................................Ans.

5-84 Locate the centroid of the shaded area shown in Fig. P5-84. SOLUTION For the quarter circle

( ) ( )2 2 2x r y r r− + − =

( )22y r r x r= − − −

( )( )

( )

22

22 22

VdA r r r x r dx

r x r dx rx x dx

= − − − −

= − − = −

( )2

2 2 2 1

00

12 2 sin2 4

rr

Vx r rA dA rx x dx x r rx x rr

π− − = = − = − − + = ∫ ∫

( ) ( )

2

0

3/ 22 3 32 2 1

0

2

22 sin

3 2 4 3

r

y V

r

M x dA x rx x dx

rx x r x r r rx r rx x rr

π−

= = −

− − = − + − − + = −

∫ ∫

( ) ( )3 3

2

4 3 44 3

yC

r rM rx rA r

ππ π

−= = = − ........................................................................Ans.

Since the line x y= is an axis of symmetry,

43C Cry x rπ

= = − ...............................................................................................................Ans.


134

5-85 Locate the centroid of the volume obtained by revolving the shaded area shown in Fig. P5-85 about the x-axis.

SOLUTION

222

2 4

28

42 64

xdV r dx dx

x x dx

π π

π

= = −

= − +

2 44

0

43 53

0

42 64

4 26.808 in.6 320

x xV dV dx

x xx

π

π

= = − +

= − + =

∫ ∫

42 4 4 64 2 4

00

4 2 33.510 in.2 64 8 384yzx x x xM xdV x dx xπ π

= = − + = − + =

∫ ∫

33.510 1.250 in.26.808

yzC

Mx

V= = = ....................................................................................Ans.

Since the x-axis is an axis of symmetry,

0C Cy z= = .........................................................................................................................Ans.

5-86 Locate the centroid of the volume obtained by revolving the shaded area shown in Fig. P5-86 about the x-axis.

SOLUTION

( )2 50dV r dx x dxπ π= =

2 2

0050 25 25b b

V dV x dx x bπ π π = = = = ∫ ∫

( )3 3

00

50 50503 3

bb

yzx bM xdV x x dx π ππ

= = = =

∫ ∫

3

2

50 3 225 3

yzC

M b bxV b

ππ

= = =

( )2 100

66.7 mm3

= ≅ .......................................Ans.

Since the x-axis is an axis of symmetry,

0C Cy z= = .........................................................................................................................Ans.

5-87 Locate the centroid of the curved homogeneous slender rod shown in Fig. P5-87. SOLUTION

2 6x y= 3dx dy y=

( ) ( )2 22 2 211 3 1 93

dL dx dy dx dy dy y dy y dy= + = + = + = +


135

( ) 1212 2 2 2

0 0

1 19 9 9ln 9 27.8807 in.3 6

L dL y dy y y y y = = + = + + + + = ∫ ∫

( ) ( )

212 2

0

12232 2 2 2

0

1 96 3

1 9 99 9 ln 9 304.962 in.18 4 8 8

yyM xdL y dy

y yy y y y

= = +

= + − + − + + =

∫ ∫

304.962 10.94 in.27.8807

yC

Mx

L= = = ...................................................................................Ans.

( )12

12 32 2 2

00

1 19 9 207.278 in.3 3 3xyM y dA y dy y = = + = + = ∫ ∫

207.278 7.43 in.27.8807

xC

MyL

= = = .....................................................................................Ans.

5-88 Locate the centroid of the curved homogeneous slender rod shown in Fig. P5-88 if b = 50 mm. SOLUTION

2 2y x b= dy dx x b= 50 mmb =

( ) ( )2 22 2 2 211 1 50 5050

dL dx dy dy dx dx x dx x dx= + = + = + = +

( )

( ) ( )

5050 2 2 2 2 2 2 2

0 0

2 2 2

1 150 50 50 ln 5050 100

1 2 50 50 ln 50 2 50 50 ln 50 57.3897 mm100

L dL x dx x x x x = = + = + + + +

= + + − =

∫ ∫

( )50

50 32 2 2 2 2

00

1 150 50 1523.689 mm50 50 3yxM xdL x dx x = = + = + = ∫ ∫

1523.689 26.5 mm57.3897

yC

Mx

L= = = ................................................................................Ans.

( ) ( )

250 2 2

0

502 432 2 2 2 2 2

02

1 50100 50

1 50 5050 50 ln 505000 4 8 8

525.198 mm

xxM y dL x dx

x x x x x x

= = +

= + − + − + +

=

∫ ∫

525.198 9.15 mm57.3897

xC

MyL

= = = ..................................................................................Ans.

5-89 Locate the centroid of the volume of the portion of a right circular cone shown in Fig. P5-89. SOLUTION

( )ˆ rr h zh

= −


136

( )2 2

22

ˆ4 4r rdV dz h z dz

hπ π= = −

( ) ( )32 2 22

2 200

4 4 3 12

hh h zr r r hV dV h z dzh h

π π π −= = − = − =

∫ ∫

( )2

220

2 2 2 3 4 2 2

20

4

24 2 3 4 48

h

xy

h

rM z dV z h z dzh

r h z hz z r hh

π

π π

= = −

= − + =

∫ ∫

2 2

2

4812 4

xyC

M r h hzV r h

ππ

= = = .............................................................................................Ans.

( )3

33

ˆ43 3yzr rdM dV h z dz

hπ= = −

( ) ( )43 3 33

3 300

3 3 4 12

hh

yz yz

h zr r r hM dM h z dzh h

−= = − = − =

∫ ∫

3

2

1212

yzC

M r h rxV r hπ π

= = = ...............................................................................................Ans.

Since the plane x y= is a plane of symmetry,

C Cy x r π= = ....................................................................................................................Ans.

5-90 Locate the mass center of the hemisphere shown in Fig. P5-90 if the density ρ at any point P is proportional to the distance from the xy-plane to the point P.

SOLUTION

( )22 2r y r z= + −

( )22 2 22y r r z rz z= − − = −

( ) ( )2 2 32dm dV kz y dz k rz z dzρ π π= = = −

( )3 4 4

2 3

00

2 523 4 12

rr rz z krm dm k rz z dz k ππ π

= = − = − =

∫ ∫

( )4 5 5

2 3

00

2 324 5 10

rr

xyrz z krM z dm z k rz z dz k ππ π

= = − = − =

∫ ∫

5

4

3 10 185 12 25

xyG

M kr rzm kr

ππ

= = = .........................................................................................Ans.

By symmetry,

0G Gx y= = ........................................................................................................................Ans.


137

5-91 Locate the mass center of the right circular cone shown in Fig. P5-91 if the density ρ at any point P is proportional to the distance from the xy-plane to the point P.

SOLUTION

r̂ rz h=

( )2 2

22

ˆ r zdm dV kz r dz kz dzh

ρ π π

= = =

2 2 4 2 2

32 20

04 4

hh kr kr z kr hm dm z dzh h

π π π = = = =

∫ ∫

2 2 5 2 3

42 20

05 5

hh

xykr kr z kr hM z dm z dzh h

π π π = = = =

∫ ∫

2 3

2 2

5 44 5

xyG

M kr h hzm kr h

ππ

= = = ...........................................................................................Ans.

By symmetry,

0G Gx y= = ........................................................................................................................Ans.

5-92 Locate the centroid of the volume of the tetrahedron shown in Fig. P5-92. SOLUTION By similar triangles

( )ˆ aa c zc

= −

( )ˆ bb c zc

= −

( )22

1 ˆˆ2 2

abdV abdz c z dzc

= = −

( )2 220

32 2

20

22

2 3 6

c

c

abV dV c cz z dzc

ab z abcc z czc

= = − +

= − + =

∫ ∫

( )2 2 3 4 2

2 2 32 20

0

222 2 2 3 4 24

cc

xyab ab c z cz z abcM z dV c z cz z dzc c

= = − + = − + =

∫ ∫

2 24

6 4xy

C

M abc czV abc

= = = ...............................................................................................Ans.

Similarly

4Cax = .................................................................................................................................Ans.

4Cby = .................................................................................................................................Ans.


143

( )( )2 0.316 3.5343 1.1168 lbW = − = −

( )( )( ) 3

3

9 9 0.520.2500 in.

2V = = ( )( )3 0.100 20.2500 2.0250 lbW = =

348.5243 in.totV = 10.9597 lbtotW =

( )

1 1

4 93.8197 in.

3C Cx yπ

= = =

(a) ( )( ) ( )( ) ( )( )3.8197 31.8086 3.5 3.5343 0.25 20.2500

2.35 in.48.5243Cx

+ − += = ......Ans.

( )( ) ( )( ) ( )( )3.8197 31.8086 3 3.5343 3 20.2500

3.54 in.48.5243Cy

+ − += = ................Ans.

( )( ) ( )( ) ( )( )0.25 31.8086 0.25 3.5343 3.5 20.2500

1.606 in.48.5243Cz

+ − += = .........Ans.

(b) ( )( ) ( )( ) ( )( )3.8197 10.0515 3.5 1.1168 0.25 2.0250

3.19 in.10.9597Gx

+ − += = .........Ans.

( )( ) ( )( ) ( )( )3.8197 10.0515 3 1.1168 3 2.0250

3.75 in.10.9597Gy+ − +

= = ...................Ans.

( )( ) ( )( ) ( )( )0.25 10.0515 0.25 1.1168 3.5 2.0250

0.851 in.10.9597Gz

+ − += = ............Ans.

5-106 A cylinder with a conical cavity and a hemispherical cap is shown in Fig. P5-106. Locate (a) The centroid of the composite volume if R = 200 mm, and h = 250 mm. (b) The center of gravity of the composite volume if the cylinder is made of brass (ρ = 8750 kg/m3) and the

cap is made of aluminum (ρ = 2770 kg/m3). SOLUTION

Cylinder: ( ) ( )2 6 31 200 250 31.416 10 mmV π= = ×

( )( )31 8750 31.416 10 274.89 kgm −= × =

Hemisphere: ( ) ( )3

6 32

4 3 20016.755 10 mm

2V

π= = ×

( )( )32 2770 16.755 10 46.412 kgm −= × =

Cavity: ( ) ( )2

6 33

200 25010.472 10 mm

3V

π= − = − ×

( )( )33 8750 10.472 10 91.630 kgm −= − × = −

Totals: 6 337.699 10 mmtotV = × 229.672 kgtotm =

( )

2 2

3 200250 325 mm

8C Gz z= = + =

3 3250 62.5 mm4C Gz z= = =


144

(a) ( )( ) ( )( ) ( )( )125 31.416 325 16.755 62.5 10.472

231 mm37.699Cz

+ + −= = ...............Ans.

0 mm (by symmetry)C Cx y= = ..................................................................................Ans.

(b) ( )( ) ( )( ) ( )( )125 274.89 325 46.412 62.5 91.630

190.4 mm229.672Gz

+ + −= = ...........Ans.

0 mm (by symmetry)G Gx y= = ..................................................................................Ans.

5-107 A slender rod is made of two materials; the segment along the x-axis is aluminum (γ = 0.100 lb/in3) and the remainder is made of brass (γ = 0.316 lb/in3). Locate the center of gravity of the slender rod.

SOLUTION Note that the curved portion of the wire is a half circle. Then,

1 16 in.L = ( )1 0.100 16 1.600 lbW = = 1 8 in.Gx = 1 0 in.Gy = 1 0 in.Gz =

2 14 in.L = ( )2 0.316 14 4.424 lbW = = 2 0 in.Gx = 2 7 in.Gy = 2 0 in.Gz =

( )3 9.9 31.102 in.2dL π π= = = ( )3 0.316 31.102 9.828 lbW = =

3 0 in.Gx =

( )

3

9.9sin 27 cos 45 11.457 in.

2Gyπ

π= + ° =

( )

3

9.9sin 27 sin 45 11.457 in.

2Gzπ

π= + ° =

15.852 lbtotW =

( )( ) ( )( ) ( )( )8 1.600 0 4.424 0 9.828

0.807 in.15.852Gx

+ += = ...........................................Ans.

( )( ) ( )( ) ( )( )0 1.600 7 4.424 11.457 9.828

9.06 in.15.852Gy

+ += = ..................................Ans.

( )( ) ( )( ) ( )( )0 1.600 0 4.424 11.457 9.828

7.10 in.15.852Gz

+ += = ...................................Ans.

5-108 The loads acting on a beam are distributed in a triangular manner as shown in Fig. P5-108. Determine and locate the resultant with respect to the left end of the beam.

SOLUTION

( )( )

1

2 750750 N

2R = =

( )( )

2

4 7501500 N

2R = =

750 1500 2250 N R = + = ↓ .........................................................................................Ans.

( )( ) ( )2250 750 2 3 2 1500 2 4 3AM d= = + +

2.67 md = .........................................................................................................................Ans.


145

5-109 Determine the resultant of the distributed loads acting on the beam shown in Fig. P5-109, and locate its line of action with respect to the support at A.

SOLUTION

( )( )1 200 3 600 lbR = =

( )( )2 150 3 450 lbR = =

( )( )3 100 3 300 lbR = =

600 450 300 1350 lb R = + + = ↓ ................................................................................Ans.

( ) ( ) ( )1350 600 1.5 450 4.5 300 7.5AM d= = + +

3.83 ftd = ..........................................................................................................................Ans.

5-110 A distributed load acts on the beam shown in Fig. P5-110. Determine the resultant of the distributed load and locate its line of action with respect to the support at A.

SOLUTION

( )( )1 2.5 4 10 kNR = =

( )( )

2

2.5 45 kN

2R = =

10 5 15 kN R = + = ↓ ......................................................................................................Ans.

( ) ( )15 10 4 5 2 8 3AM d= = + +

4.22 md = .........................................................................................................................Ans.

5-111 A distributed load acts on a beam as shown in Fig. P5-111. Determine and locate the resultant of the distributed load with respect to the support at A.

SOLUTION

10 102 3

1 003 1000 lbR wdx x dx x = = = = ∫ ∫

10410 3

1 1 1 00

33 7500 lb ft4AxM R d xwdx x dx

= = = = = ⋅

∫ ∫

( )( )2 300 5 1500 lbR = =

1000 1500 2500 lb R = + = ↓ .......................................................................................Ans.

( )2500 7500 1500 10 2.5AM d= = + +

10.50 ftd = ........................................................................................................................Ans.

5-112 Determine the resultant of the distributed load acting on the beam shown in Fig. P5-112 and locate its line of action with respect to the support.

SOLUTION

232 2

1 00

100100 266.667 N3xR wdx x dx

= = = =

∫ ∫

2 23 4

1 1 1 00100 25 400 N mAM R d xwdx x dx x = = = = = ⋅ ∫ ∫


146

( )( )

2

400 2400 N

2R = =

266.667 400 666.667 667 N R = + = ≅ ↓ ................................................................Ans.

( )666.667 400 400 2 2 3AM d= = + +

2.20 md = .........................................................................................................................Ans.

5-113 Determine the resultant of the distributed load acting on the beam shown in Fig. P5-113 and locate its line of action with respect to the support.

SOLUTION

( )( )

1

200 6600 lb

2R = = ↓ ( )( )2 200 2 400 lb R = = ↓

( )( )3 100 6 600 lb R = = ↑

600 400 600 400 lb 400 lb R = − − + = − = ↓ ............................................................Ans.

( ) ( ) ( )400 600 4 400 7 600 3AM d= = + −

8.50 ftd = ..........................................................................................................................Ans.

5-114 A gate is used to hold water as shown in Fig. P5-114. If the width of the gate is 2 m. determine the resultant of the water pressure acting on the gate, and locate its line of action with respect to the bottom of the gate.

SOLUTION

( )( ) ( )( )9800 9 9 2

793,800 N 794 kN2

R = = = ....................................................Ans.

9 3 3 m (from bottom)d = = ........................................................................................Ans.

5-115 A flexible cable is used to tether the balloon shown in Fig. P5-115. The cable weighs 1.2 lb/ft along its length. Determine the magnitude of the resultant force and its location with respect to A.

SOLUTION

2 10y x= 5dy dx x=

( ) ( ) ( ) ( )2 2 2 2 2 211 1 5 55

dL dx dy dy dx dx x dx x dx= + = + = + = +

15 2 2

0

11.2 55

R wdL x dx = = + ∫ ∫

( ) 152 2 2 2 2

0

10.24 5 5 ln 5 33.916 lb2x x x x = + + + + =

...........................Ans.

( )

15 2 2

015

32 2

0

0.24 5

10.24 5 306.228 lb ft3

RRx xwdL x x dx

x

= = +

= + = ⋅

∫ ∫

306.228 9.029 ft33.916Rx = = .................................................................................................Ans.

2 210 9.029 10 8.15 ftR Ry x= = = ................................................................................Ans.


147

5-116 A loaded rivet causes a force distribution or pressure p on a plate, as shown in Fig. P5-116. The rivet has a diameter of 25 mm, and the plate is 15 mm thick. Determine the magnitude of the resultant force acting on the plate. Let po = 400 N/m2.

SOLUTION

( ) ( ) ( )40.025 0.015 1.875 102

dA r d t d dθ θ θ− = = = ×

( ) ( ) ( )4 2cos 400cos 1.875 10 cos 0.075cosydR p dA d dθ θ θ θ θ θ− = = × =

/ 2 2

/ 2/ 2

/ 2

/ 2/ 2

0.075cos

1 cos 2 sin 20.075 0.0752 2 4

y yR dR d

d

π

ππ

π

ππ

θ θ

θ θ θθ

−

−−

= =

+ = = +

∫ ∫

∫

0.1178 N = ↓ ............................................................................................................Ans.

( ) ( ) ( )4sin 400cos 1.875 10 sin 0.0375sin 2xdR p dA d dθ θ θ θ θ θ− = = × =

/ 2

/ 2

/ 2/ 2

cos 20.0375sin 2 0.03752x xR dR d

ππ

ππ

θθ θ−

−

= = = − ∫ ∫

0 N= ............................................................................................................................Ans.

5-117 The lift on the wing of an airplane due to aerodynamic forces is shown in Fig. P5-117. The lift L may be described by the function 25 lb/ftL x= . Determine the moment of the resultant lift force about point A.

SOLUTION

( )25dM xdL x x dx= =

155/ 215 3/ 2

00

225 25 8710 lb ft5xM dM x dx

= = = = ⋅

∫ ∫ .........................................Ans.

5-118 Determine the moment of the 1650-N force shown in Fig. P5-118 about point O. SOLUTION

( )2 2 2

180 180 801650 1113.055 1113.055 494.691 N180 180 80

− + += = − + ++ +

i j kF i j k

( ) ( )0.36 0.24 0.4 1113.055 1113.055 494.691O = + + − + +M i j k × i j k

( )326 623 668 N m= − − + ⋅i j k ..........................................................................Ans.

5-119 The driving wheel of a truck is subjected to the force-couple system shown in Fig. P5-119. Replace this system by an equivalent single force and determine the point of application of the force along the vertical diameter of the wheel.

SOLUTION

750 lbxR = 1800 1800 0 lbyR = − =

750 lb = →R ...................................................................................................................Ans.

600 lb ft 750OM y= ⋅ =

0.8 ft 9.60 in. up from the groundy = = ..................................................................Ans.


148

5-120 A 200-N force is applied at corner B of a rectangular plate as shown in Fig. P5-120. Determine the moment of the force

(a) About point O. (b) About line OD. SOLUTION

(a) ( )2 2 2

0.46 0.5 0.9200 81.5854 88.6798 159.6237 N0.46 0.5 0.9

+ −= = + −+ +

i j kF i j k

( ) ( )0.8 2 81.5854 88.6798 159.6237O = − + + −M i k × i j k

( )177.360 35.472 70.944 N m= − + − ⋅i j k .....................................................Ans.

(b) ( )2 2

0.8 0.5 0.84800 0.530000.8 0.5

OD− += = − +

+i je i j

( )( ) ( )( )177.360 0.84800 35.472 0.53000OD O ODM = = − − +M e

169.2 N m= ⋅ ..........................................................................................................Ans.

( )169.2 0.84800 0.53000OD OD ODM= = − +M e i j

( )143.5 89.7 N m= − + ⋅i j ..................................................................................Ans.

5-121 A 2500-lb jet engine is suspended from the wing of an airplane as shown in Fig. P5-121. Determine the moment produced by the engine at point A in the wing when the plane is

(a) On the ground with the engine not operating. (b) In flight with the engine developing a thrust T of 15,000 lb. SOLUTION

(a) ( )2500 8sin 35 11,470 lb ft = ° = ⋅M ..................................................................Ans.

(b) ( ) ( )2500 8sin 35 15,000 8cos35= ° − °M

86,800 lb ft 86,800 lb ft = − ⋅ = ⋅ ...............................................................Ans.

5-122 Two forces and a couple act on a beam as shown in Fig. P5-122. Determine the resultant R and its location x.

SOLUTION

1.8 1 2.8 kN = + = ↓R ...................................................................................................Ans.

( )( ) ( )( )1.8 1 1 2 10 13.800 kN m 2.8OM x= + + = ⋅ =

4.92 mx = .........................................................................................................................Ans.

5-123 Replace the force and couple acting on the wall bracket shown in Fig. P5-123 by a force-couple system at point A.

SOLUTION

0 lbxR = 40 lbyR = −

40 lb = ↓R .......................................................................................................................Ans.

( )40 8 30 350 lb in.AM = + = ⋅ ..................................................................................Ans.


149

5-124 Determine the resultant of the parallel force system shown in Fig. P5-124 and locate the intersection of its line of action with the xy-plane.

SOLUTION

( ) ( ) ( ) ( ) ( ) ( )125 80 100 200 75 130 N= + − + + − + − = −R k k k k k k ................Ans.

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )3 1.5 125 1 1.5 80 2 3.5 100

3 5.5 200 1 5.5 75O = + + + − + +

+ + − + + −

M i j × k i j × k i j × k

i j × k i j × k

( )1095 180 N m= − + ⋅i j

( ) ( ) ( )130 130 130x y y x= + − = − +i j × k i j

180 1.385 m130

x = = 1095 8.42 m130

y = = ..........................................Ans.

5-125 The concrete floor of a building supports four wood columns as shown in Fig. P5-125. The resultant of the forces transmitted through the columns to the floor is R = 75 kip at the location shown in the figure. Determine the magnitude of the forces F1 and F2.

SOLUTION

1 215 20 75 kip R F F= + + + = ↓

( ) ( ) ( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )

1 2

1 2 1

20 15 20 30 30

30 300 20

13 16 75 1200 975 kip ft

O F F

F F F

= − + + − + −

= − + + +

= + − = − + ⋅

M i × k i j × k j × k

i j

i j × k i j

1 33.7 kip F = ↓ ................................................................................................................Ans.

2 6.25 kip F = ↓ ................................................................................................................Ans.

5-126 A bent rod supports a 450-kN force F as shown in Fig. P5-126. (a) Replace the 450-N force with a force R through the coordinate origin O and a couple C. (b) Determine the twisting moments produced by force F in the three different segments of the rod. SOLUTION

425cos 45 750 449.48 mmDx = °− = −

300 150 150 mmDy = − = 425sin 45 300.52 mmDz = − ° = −

(a) ( )450 N=R k ..................................................................................................................Ans.

( ) ( )0.44948 0.150 0.30052 450= − + −C i j k × k

( )67.5 202.3 N m= + ⋅i j ..........................................................................................Ans.

(b) 202.3 N mOA OM = = ⋅M j ............................................................................................Ans.

( ) ( ) ( )0.75 0.15 450 67.5 337.5 N mB = − − = − + ⋅M i j × k i j

( ) ( ) ( )425cos 45 425sin 45

0.70711 0.70711425AB

° − °= = −

i ke i k

( )( )67.5 0.70711 47.7 N mAB B ABM = = − = − ⋅M e ................................................Ans.

67.5 N mBC BM = = − ⋅M i ...........................................................................................Ans.


150


( )22 0.5VdA x x dx= −

( )22 0.5HdA y y dy= −

( )2 2

0

23/ 2 32

0

2 0.5

2 4 in.3 2 6 3

VA dA x x dx

x x

= = −

= − =

∫ ∫

( )25/ 2 42 2 3

00

2 62 0.5 in.5 2 8 5y Vx xM xdA x x x dx

= = − = − =

∫ ∫

6 5 18 0.9 in.4 3 20

yC

Mx

A= = = = ......................................................................................Ans.

( )25/ 2 42 2 3

00

2 62 0.5 in.5 2 8 5x Hy yM y dA y y y dy

= = − = − =

∫ ∫

6 5 18 0.9 in.4 3 20

xC

MyA

= = = = ......................................................................................Ans.

5-128 Two channel sections and a plate are used to form the cross section shown in Fig. P5-128. Each of the channels has a cross-sectional area of 2605 mm2. Locate the y-coordinate of the centroid of the composite section with respect to the top surface of the plate.

SOLUTION

737,190 64.3 mm11,460Cy−= = − ........................Ans.

5-129 Locate the centroid of the volume shown in Fig. P5-129 if R = 10 in. and h = 32 in. SOLUTION

22

22

RydV r dy dyh

π π

= =

2 4 2 5

4 400

2

5

5

hh R y R yV dV dyh h

R h

π π

π

= = =

=

∫ ∫


151

22 2 6 2 2

2 4006 6

hh

xzRy R y R hM y dV y dyh h

π ππ

= = = =

∫ ∫

( )2 2

2

5 326 5 26.7 in.5 6 6

xzC

M R h hyV R h

ππ

= = = = = ........................................................Ans.

0 in. (by symmetry)C Cx z= = .....................................................................................Ans.

5-130 Locate the centroid and the mass center of the volume shown in Fig. P5-130, which consists of an aluminum cylinder (ρ= 2770 kg/m3) and a steel (ρ = 7870 kg/m3) cylinder and sphere.

SOLUTION

( ) ( )2 3 31 0.1 0.3 9.4248 10 mV π −= = ×

( )31 2770 9.4248 10 26.1066 kgm −= × =

( ) ( )2 3 32 0.05 0.175 1.3744 10 mV π −= = ×

( )32 7870 1.3744 10 10.8169 kgm −= × =

( )3

3 33

4 0.14.1888 10 m

3V

π −= = ×

( )33 7870 4.1888 10 32.9658 kgm −= × =

3 314.9880 10 mtotV−= × 69.8893 kgtotm =

( )( ) ( )( ) ( )( )0 9.4248 187.5 1.3744 375 4.1888

122.0 mm14.9880Cy

+ += = .................Ans.

( )( ) ( )( ) ( )( )0 26.1066 187.5 10.8169 375 32.9658

205.9 mm69.8893Gy

+ += = .........Ans.

0 mm (by symmetry)C G C Gx x z z= = = = ................................................................Ans.

5-131 Determine the resultant R of the system of distributed loads on the beam of Fig. P5-131, and locate its line of action with respect to the left support of the beam.

SOLUTION

( )( ) ( )( )4

0

400 4200 400 4

2R x dx= + +∫

4

3/ 2

0

2200 1600 800 3466.667 lb 3470 lb 3x = + + = ≅ ↓

....................................Ans.

( ) ( )( ) ( ) ( )( ) ( )4

0

45/ 2

0

400 4200 400 4 6 8 4 3

2

2200 9600 7466.667 19,626.667 lb ft5

OM x x dx

x

= + + +

= + + = ⋅

∫

19,626.667 5.66 ft3466.667

x = = ...............................................................................................Ans.

ch05

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