Engineering Mechanics: Statics, Student Value Edition (14th Edition)
Engineering Mechanics: Statics, Student Value Edition (14th Edition)
14th Edition
ISBN: 9780134056388
Author: Russell C. Hibbeler
Publisher: PEARSON
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Textbook Question
Chapter 7.4, Problem 94P

The cable supports the three loads shown. Determine the sags yB and yD of B and D. Take P1 = 800 N, P2 = 500 N.

Chapter 7.4, Problem 94P, The cable supports the three loads shown. Determine the sags yB and yD of B and D. Take P1 = 800 N,

Probs. 7–94/95

Expert Solution & Answer
Check Mark
To determine

The sags yB and yD of points B and D.

Answer to Problem 94P

The sag yB at point B of the cable is yB=2.22m_.

The sag yD at point D of the cable is yD=1.55m_.

Explanation of Solution

Assumption:

  • The self-weight of the cable is ignored.
  • The cable is perfectly flexible and inextensible.
  • The tensile force acting in the cable is always tangent to the cable at points along its length.
  • Being inextensible, the cable has a constant length both before and after the load is applied.
  • The cable or a segment of the cable can be treated as a rigid body.
  • Method of joints is used to determine the force in the cable.
  • Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member.
  • Consider the force indicating right side as positive and left side as negative in the horizontal components of forces.
  • Consider the force indicating upward is taken as positive and downward as negative in the vertical components of forces.
  • Consider clockwise moment as negative and anticlockwise moment as positive wherever applicable.
  • Apply the Equation of equilibrium wherever applicable.

Given information:

  • Take the vertical external loads P1=800N and P2=500N.

Segment BCDE:

Show the free body diagram of segment BCDE as in Figure (1).

Engineering Mechanics: Statics, Student Value Edition (14th Edition), Chapter 7.4, Problem 94P , additional homework tip  1

Using Figure (1),

Determine the value of sinθ:

sinθ=OppositesideHypotenuse (I)

Determine the value of cosθ:

cosθ=AdjacentsideHypotenuse (II)

Moment about the point E:

Determine the force in cable segment AB and the sag yB by taking moment about point E.

ME=0P2(3)+P1(9)+P2(15)FABcosθ1(yB+1)FABsinθ1(15)=0 (III)

Conclusion:

Substitute yB for the opposite side and yB2+9 for hypotenuse in Equation (I).

sinθ1=yByB2+9 (IV)

Substitute 3 m for the adjacent side and yB2+9 for hypotenuse in Equation (II).

cosθ1=3yB2+9 (V)

Substitute 500 N for P2, 800 N for P1, yByB2+9 for sinθ1, and 3yB2+9 for cosθ1 Equation (III).

[500(3)+800(9)+500(15)FAB(3yB2+9)(yB+1)FAB(yByB2+9)(15)]=01,500+7,200+7,500FAB(3yB+3yB2+9)FAB(15yByB2+9)=016,200FAB(18yB+3yB2+9)=0FAB(18yB+3yB2+9)=16,200 (VI)

Segment BC:

Show the free body diagram of segment BC as in Figure (2).

Engineering Mechanics: Statics, Student Value Edition (14th Edition), Chapter 7.4, Problem 94P , additional homework tip  2

Using Figure (2),

Moment about the point C:

Determine the force in cable segment AB and the sag yB by taking moment about point C.

MC=0P2(6)+FABcosθ1(4yB)FABsinθ1(6)=0 (VII)

Conclusion:

Substitute 500 N for P2, yByB2+9 for sinθ1, and 3yB2+9 for cosθ1 Equation (VII).

500(6)+FAB(3yB2+9)(4yB)FAB(yByB2+9)(6)=03,000+FAB(123yByB2+9)FAB(6yByB2+9)=03,000FAB(9yB12yB2+9)=0FAB(9yB12yB2+9)=3,000 (VIII)

Divide Equation (VI) with Equation (VIII).

FAB(18yB+3yB2+9)FAB(9yB12yB2+9)=16,2003,00018yB+39yB12=16,2003,000

54,000yB+9,000=145,800yB194,40091,800yB=203,400yB=2.22m

Substitute 2.22 m for yB in Equation (VIII).

FAB(9(2.22)122.222+9)=3,000FAB(7.983.732)=3,000FAB=1,403.04N

Thus, the sag yB at point B of the cable is yB=2.22m_.

Joint B:

Show the free-body diagram of joint B as in Figure (3).

Engineering Mechanics: Statics, Student Value Edition (14th Edition), Chapter 7.4, Problem 94P , additional homework tip  3

Using Figure (3),

Determine the value of θ:

tanθ=OppositesideAdjacentside (IX)

Along the vertical direction:

Determine the force in cable segment BC by resolving the vertical component of forces.

Fy=0FABsinθ1P2FBCsinθ2=0 (X)

Conclusion:

Substitute 2.22 m for the opposite side and (2.22)2+9 for hypotenuse in Equation (I).

sinθ1=2.22(2.22)2+9sinθ1=0.5948θ1=36.5°

Substitute 1.78 m for the opposite side and 6 m for adjacent side in Equation (IX).

tanθ2=1.786θ2=16.52°

Substitute 1,403.04 N for FAB, 36.50° for θ1, 500 N for P2, and 16.52° for θ2 in Equation (X).

1,403.04sin36.50°500FBCsin16.52°=0834.565000.284FBC=00.284FBC=334.56FBC=1,178.03N

Joint C:

Show the free-body diagram of joint C as in Figure (4).

Engineering Mechanics: Statics, Student Value Edition (14th Edition), Chapter 7.4, Problem 94P , additional homework tip  4

Using Figure (4),

Along the vertical direction:

Determine sag yD at point D by resolving the vertical component of forces.

Fy=0FBCsinθ2+FCDsinθ3P1=0 (XI)

Along the horizontal direction:

Determine sag yD at point D by resolving the horizontal component of forces.

Fy=0FBCcosθ2+FCDcosθ3=0 (XII)

Conclusion:

Substitute (4yD) for the opposite side and 36+(4yD)2 for hypotenuse in Equation (I).

sinθ3=(4yD)36+(4yD)2

Substitute 6 m for the adjacent side and 36+(4yD)2 for hypotenuse in Equation (II).

cosθ3=636+(4yD)2

Substitute 1,178.03 N for FBC, 16.52° for θ2, (4yD)36+(4yD)2 for sinθ3, and 800 N for P1 in Equation (XI).

1,178.03sin16.52°+FCD((4yD)36+(4yD)2)800=0FCD((4yD)36+(4yD)2)=465.04 (XIII)

Substitute 1,178.03 N for FBC, 16.52° for θ2, and 636+(4yD)2 for cosθ3 in Equation (XII).

1,178.03cos16.52°+FCD(636+(4yD)2)=0FCD(636+(4yD)2)=1,129.40 (XIV)

Conclusion:

Divide Equation (XIII) with Equation (XIV).

FCD((4yD)36+(4yD)2)FCD(636+(4yD)2)=465.041,129.40(4yD)6=465.041,129.40

4,517.61,129.40yD=2,790.241,129.40yD=1,727.36yD=1.55m

Thus, sag yD at point D of the cable is yD=1.55m_.

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Engineering Mechanics: Statics, Student Value Edition (14th Edition)

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