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Equivalent forces Three cables attached to a disk exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at A. (b) Determine the single force which is equivalent to the force couple system obtained in part a, and specify its point of application on a line drawn through points A and D. Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)

Equivalent force systems_2 [compatibility mode]

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Page 1: Equivalent force systems_2 [compatibility mode]

Equivalent forces

Three cables attached to a disk exert on it the forces shown.

(a) Replace the three forces with an equivalent force-couple system at A.

(b) Determine the single force which is equivalent to the force couple system obtained in part a, and specify its point of application on a line drawn through points A and D.

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Page 2: Equivalent force systems_2 [compatibility mode]

Equivalent forces

XB

Y

X

(-100,173.2)(141,141)

(0,-200)

-99i+99j 103.4i+37.6j

99i-99j

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Page 3: Equivalent force systems_2 [compatibility mode]

Equivalent forces

XB

Y

X

(-100,173.2)(100,173.2)

(0,-200)

-99i+99j 103.4i+37.6j

99i-99j

FFCC

Equivalent system

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Page 4: Equivalent force systems_2 [compatibility mode]

Equivalent forces

FFCC

,

99 99 103.4 37.6 99 99 103.4 37.6(141 141 ) 103.4 37.6

(0 200 ) 99 99 ( 100 173.2 ) 99 995317 14526 19800 9900 1714717838

A

i j i j i j i ji j i j

i j i j i j i jk k k k kk

Z

FM

103.4i+37.6j

17838k

XB

Y

X

(-100,173.2)(141,141)

(0,-200)

-99i+99j

99i-99j

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Page 5: Equivalent force systems_2 [compatibility mode]

Equivalent forces

XB

Y

X

(-100,173.2)(141,141)

(0,-200)

-99i+99j 103.4i+37.6j

99i-99j

FF

P (x,y)P (x,y)

Equivalent system

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Page 6: Equivalent force systems_2 [compatibility mode]

Equivalent forces

FFP (x,y)P (x,y)

, ( )x y

A x y y x

F i F j

xi yj F i F j xF yF k

Z

F

M

XB

Y

X

(-100,173.2)(141,141)

(0,-200)

-99i+99j

99i-99j

,

99 99 103.4 37.6 99 99 103.4 37.6(141 141 ) 103.4 37.6

(0 200 ) 99 99 ( 100 173.2 ) 99 995317 14526 19800 9900 1714717838

A

i j i j i j i ji j i j

i j i j i j i jk k k k kk

Z

FM

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Page 7: Equivalent force systems_2 [compatibility mode]

Equivalent forces

FFP (x,y)P (x,y)

103 4 37 6 103 4 37 6

17838 103 4 37 6 17838

17838

17838

x y x y

y x

F i F j . i . j F . ,F .xF yF . x . y

Equation of line of action Equation of line AD

103.4x-37.6y=x=0

37.6y=- y=-474mm

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Page 8: Equivalent force systems_2 [compatibility mode]

Equivalent forces

The steel plate shown will support six 50 mm diameter idler rollers mounted on the plate as shown. Two flat belts pass around the rollers, and rollers A and D will be adjusted so that the tension in each belt is 45 N. Determine

(a) the resultant couple acting on the plate if a = 0.2 m,

(b) The value of a so that the resultant couple acting on the plate is 54 Nm clockwise.

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Page 9: Equivalent force systems_2 [compatibility mode]

Equivalent forces

Y

X

(2a+c,0)

(-200,-a)

(-200+c,a+200)

c

c

-45i+0j

45i+0j

-31.82i+31.82j

i aj

a2

45 200

40000

(2a,2a)

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Page 10: Equivalent force systems_2 [compatibility mode]

Equivalent forces

Y

X(2a+c,0)

(-200,-a)

(-200+c,a+200)

c

c

-45i+0j

45i+0j

-31.82i+31.82j

i aj. i . j

a 2

45 20031 82 31 82

40000

2

2

45 20045 0 45 0 31.82 31.82

4000045 200

31.82 31.8240000

0

(0 0 ) (31.82 31.82 ) ( 400 0 ) 45 0

( 200 400 ) 45 0 (2 2 ) 31.82 31.820 0 180

c

i aji j i j i j

ai aj

i ja

for a = 200mm

i j i j c i j i j

c i j i j ai aj i jk k

F

M

00 63.86 63.86 18000 255447544

k ak ak k kk

a=200mm

(2a,2a)

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Page 11: Equivalent force systems_2 [compatibility mode]

Equivalent forces

Y

X(2a+c,0)

(-200,-a)

(-200+c,2a)

c

c

-45i+0j

45i+0j

-31.82i+31.82j

2

45 200

40000

i aj

a

2

2

2

45 20045 0 45 0 31.82 31.82

4000045 200

31.82 31.8240000

45 200(0 0 ) 2 0 45 0

40000200 2 45 0 (2 2 ) 31.82 31.82

0 0 90 63.86 63.86

c

i aji j i j i j

ai aj

i ja

i aji j a c i j i j

ac i aj i j ai aj i j

k k ak ak ak

F

M

90 127.7290 127.72 5490 73.72

0.82

ak kak k ka

a

(2a,2a)

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