# SOLUTIONS OF ASSIGNMENT NO THE RESULTANT FORCE CONCURRENT

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```					                 SOLUTIONS OF ASSIGNMENT NO. (4)

THE RESULTANT FORCE

OF A NON-CONCURRENT FORCE SYSTEM
Problem 4.111

Replace the force system acting on the beam by an equivalent force and
couple moment.

1.5 KN
2.5 KN
3 KN
5                    30°
3
θ
A                                                             B

2m               4m                           2m

1.5 cos 30°
2.5 sin θ
3 KN

2.5 cos θ            1.5 sin 30°
A                                                                 B

(1)
FRy
≡

FRx
A                        (2)                            B MB

cos θ = 4/5 , sin θ = 3/5

FRx = ∑ Fx

١
= 1.5 sin 30° - 2.5 cos θ

= 1.5 sin 30° – 2.5 * (4/5)

= - 1.25 KN

FRx = 1.25 KN ←

FRy = ∑ Fy

= - 3 – 1.5 cos 30° – 2.5 sin θ

= - 3 – 1.5 cos 30° – 2.5 * (3/5)

= - 5.8 KN

FRy = 5.8 KN ↓

FR =   (1.252 + 5.82 )

5.8
FR = 5.93 KN     , tanθ =
1.25

θ = 77.8°

∑ MB = 1.5 cos 30 (2) + 2.5 *(3/5) (6)

∑ MB = 11.6 KN. m (counterclockwise)

٢
Problem 4.113

Replace the three forces acting on the shaft by a single resultant force. Specify
where the force acts, measured from end B.

2.5 m              1.5 m         1m             2m
A                                                                           B

α                                  β

3                       12   13
5
4                                  5
500 KN                           200 KN
260 KN
y

A                                                                                                    x
B
500 cos α                                      260 cos β

200 KN

500 sin α                 260 sin β

(1)
≡                                       y           FRy

x         FRx
y
x
A                                                                      B
(2)

cos α = 4/5   , sin α = 3/5 , cos β = 5/13 , sin β = 12/13

FRx = ∑ Fx| 1

= 260 cos β – 500 cos α

= 260 * (5/13) – 500 * (4/5)

= - 300 N

FRx = - 300 N

٣
FRy = ∑ Fy| 1

= - 260 sin β – 200 – 500 sin α

= -260 * (12/13) – 200 – 500 * (3/5)

FRy = - 740 N

FRx = 300 N ←          FRy = 740 N ↓
300 N
2        2                                              θ
FR =   FRx +    FRy
FR
FR = 300 2 + 74o 2
740 N

FR = 798.5 N

tan θ = 740 /300 =2.467

θ = 67.9°

MB| 1 = MB| 2

260 * (12/13) * 2 + 200 * 3 + 500 * (3/5) * 4.5 = FRy (x) – FRx (y)

2430 = - 740 x + 300 y (equation of the line of action of the resultant)

At y = 0

2430 = - 740 x

x =- 3.28 m

At x = 0

2460 = 300 y

y = 8.1 m

٤
Problem 4.114

Replace the loading on the frame by a single resultant force. Specify where its line
of action intersects member AB, measured from A.

y
y       FRy

200 N           400 N                       x         FRx
٣٠٠ N
0.3 m                     0.4 m
y
x                       B                                  x              B

A                                                0.2 m           A
60 N.m
200 N
(1)
(2)

0.7 m

C
C

∑ Fx| 1 = ∑ Fx| 2

FRx = ∑ Fx| 1

= - 200 N

FRx = - 200 N

∑ Fy| 1 = ∑ Fy| 2

٥
FRy =∑ Fy| 1

= - 400 – 200 – 300

= - 900 N

FRy = - 900 N
200 N
θ
FRx = 200 N ←             FRy = 900 N ↓
FR
2     2
FR = FRx + FRy
900 N
FR = 200 2 + 900 2

FR = 922 N

tan θ =900/200 = 4.5

θ = 77.47°

MA| 1 = MA| 2

60 – 200 * (0.3) – 400 * (0.7) – 200 * (0.2) = FRy (x) – FRx (y)

- 320 = -900 x + 200 y (equation of the line of action of the resultant)

At y = 0 (intersection with member AB)

-320 = - 900 x

x = 0.356 m

٦
Problem 4.115

Replace the loading acting on the beam by a single resultant force. Specify where
the force acts measured from end A.

300 N
450 N                                                            700 N
30°

A            60°                        B

1500 N.m
2m                     4m                      3m

300 N                 700 cos 30°
450 sin 60°

450 cos 60°                                      700 sin 30°
A                                        B

1500 N.m
2m                     4m                      3m

y                      ≡                            (1)

FRy

FRx                   (2)
x
y
A                                        B                                      x

∑ Fx| 1 = ∑ Fx| 2

FRx = ∑ Fx| 1

FRx = 450 cos 60° - 700 sin 30°

٧
FRx = - 125 N

∑ Fy| 1 = ∑ Fy| 2

FRy =∑ Fy| 1                                                 125 N
θ
= - 450 sin 60° - 300 – 700 cos 30°                     FR
1296 N
FRy = - 1296 N

FRx = 125 N ←               FRy = 1296 N ↓

2     2
FR = FRx + FRy
F R = 125 2 + 1296 2

FR = 1302 N

tan θ =1296/125 = 10.368
θ = 84.5°

MA| 1 = MA| 2

- 450 sin 60° *(2) – 300 *(6) – 700 cos 30° *(9) – 1500 = FRy (x) – FRx (y)

- 9535.4 = - 1296 x + 125 y

At y = 0

- 9535.4 = -1296 x

x = 7.36 m

٨
Problem 1.121

Replace the loading on the frame by a single resultant force. Specify where its line
of action intersects member CD measured from end C.

300 N                                250 N
1m        2m              3m                5
3
C               B
4

2m                             D
400 N.m

60°

3m

500 N

A

y

250 cos α
300 N
1m        2m              3m
C                   B
x
250 sin α
2m                             D
400 N.m

500 cos 60°

3m
500 sin 60°
(1)

A           ≡

٩
y
FRy

x                        FRx

y

C            B                                    x

D
(2)

A

sin α = 4/5 , cos α = 3/5

∑ Fx| 1 = ∑ Fx| 2

FRx = ∑ Fx| 1

FRx = - 250 sin α – 500 cos 60°

FRx = - 450 N

∑ Fy| 1 = ∑ Fy| 2

FRy =∑ Fy| 1

= - 300 – 250 cos α – 500 sin 60

FRy = - 883 N

١٠
FRx = 450 N ←              FRy = 883 N ↓
450 N
θ
FR
2        2
FR =   FRx +    FRy
833 N
F R = 450 2 + 833 2

FR = 991 N

tan θ =833/450 = 1.962

θ = 63°

MC| 1 = MC| 2

400 – 300 * (3) – 250 * (3/5) * (6) – 500 cos 60° * (2) – 500 sin 60° * (1)

= FRy (x) – FRx (y)

- 2333 = - 883 x + 450 y

At y = 0

- 2333 = - 883 x

x = 2.64 m (intersection of the line of action of the resultant with member
CD measured from end C)

١١

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 views: 186 posted: 4/10/2011 language: English pages: 11