Mecãnica dos sólidos

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PROBLEM 7.98
Knowing that WB = 150 lb and WC = 50 lb, determine the magnitude of the force P required to maintain equilibrium.

SOLUTION FBD CD:

ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0 3Dx = 4 Dy

(1)

ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 50 lb ) = 0

FBD BCD: Solving (1) and (2)

2D y − Dx = 60 lb
D x = 120 lb D y = 90 lb

(2)

FBD Cable:

ΣM A = 0: ( 42 ft )( 90 lb ) − ( 30 ft )( 50 lb ) − (12 ft )(150 lb ) − (15 ft ) P = 0

P = 32.0 lb

PROBLEM 7.99
Knowing that WB = 40 lb and WC = 22 lb, determine the magnitude of the force P required to maintain equilibrium.

SOLUTION FBD CD:

ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0 4D y = 3Dx

(1)

ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 22 lb ) = 0

FBD BCD:
10 Dy − 5Dx = 132 lb

(2)
D y = 39.6 lb

Solving (1) and (2)

D x = 52.8 lb

FBD Whole:
ΣM A = 0: ( 42 ft )( 39.6 lb ) − ( 30 ft )( 22 lb ) − (12 ft )( 40 lb ) − (15 ft ) P = 0

P = 34.9 lb

PROBLEM 7.100
If a = 4.5 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.

SOLUTION

By symmetry:
FBD pt C:

TBC = TCD = T
T = 90 5 kN

 1  ΣFy = 0: 2  T  − 180 kN = 0  5 

Tx = 180 kN

Ty = 90 kN

Segment DE:
ΣM E = 0: ( 7.5 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0

P = 108.0 kN

Segment AB:

ΣM A = 0: ( 4.5 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0

Q = 45.0 kN

PROBLEM 7.101
If a = 6 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.

SOLUTION
FBD pt C: By symmetry:

TBC = TCD = T
T = 90 5 kN

 1  ΣFy = 0: 2  T  − 180 kN = 0  5 

Tx = 180 kN

Ty = 90 kN

FBD DE:

ΣM E = 0: ( 9 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0

P = 120.0 kN

FBD AB:

ΣM A = 0: ( 6 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0 Q = 90.0 kN

PROBLEM 7.102
A transmission cable having a mass per unit length of 1 kg/m is strung