Engineering Mechanics : Statics-Practice Problems Workbook
Engineering Mechanics : Statics-Practice Problems Workbook
13th Edition
ISBN: 9780132915595
Author: HIBBELER
Publisher: PEARSON
Question
Chapter 7.4, Problem 94P
To determine

The tension in each segment of the cable and the cable’s total length.

Expert Solution & Answer
Check Mark

Answer to Problem 94P

  • The tension in the cable BD is TBD=78.19lb_.
  • The tension in the cable AC is TAC=74.75lb_.
  • The tension in the cable CD is TCD=43.72lb_.
  • The total length of the chain is l=15.75ft_.

Explanation of Solution

Given information:

  • Set P=80lb.

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 the member as tension where the force is pulling the member as compression where the force is pushing the member.
  • Consider the force indicating the right side as positive and left side as negative in horizontal components of forces.
  • Consider the force indicating upward is taken as positive and downward as negative in vertical components of forces.
  • Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable.
  • Apply Equation of equilibrium wherever applicable.

Segment ACD:

Show the free-b0dy diagram of the segment ACD as in Figure (1).

Engineering Mechanics : Statics-Practice Problems Workbook, Chapter 7.4, Problem 94P , additional homework tip  1

Using Figure (1),

Determine the value of θ using the relation:

tanθ=OppositesideAdjacentside (I)

Moment about point A:

Determine the tension in the BD by taking moment about point A.

MA=0TBDcosθ1(3)+TBDsinθ1(7)50(7)P(3)=0 (II)

Along the vertical direction:

Determine the vertical reaction at point A by resolving the vertical component of forces.

Fy=0TBDsinθ150P+Ay=0 (III)

Along the horizontal direction:

Determine the horizontal reaction at point A by resolving the horizontal component of forces.

Fx=0TBDcosθ1Ax=0 (IV)

Conclusion:

Substitute 5 ft for opposite side and 3 ft for adjacent side in Equation (I).

tanθ1=53θ1=59.04°

Substitute 59.04° for θ1 and 80 lb for P in Equation (II).

TBDcos59.04°(3)+TBDsin59.04°(7)35080(3)=01.543TBD+6.003TBD350240=07.546TBD=590TBD=78.19lb

Substitute 78.10 lb for TBD, 59.04° for θ1, and 80 lb for P in Equation (III).

78.10sin59.04°5080+Ay=0Ay=63.03lb

Substitute 78.10 lb for TBD and 59.04° for θ1 in Equation (IV).

78.10cos59.04°Ax=0Ax=40.18lb

Thus, the tension in the cable BD is TBD=78.19lb_.

Joint A:

Show the free-body diagram of the joint A as in Figure (2).

Engineering Mechanics : Statics-Practice Problems Workbook, Chapter 7.4, Problem 94P , additional homework tip  2

Using Figure (2),

Along the vertical direction:

Determine the tension in the cable AC and the angle θ2 by resolving the vertical component of forces.

Fy=0AyTACsinθ2=0 (V)

Along the horizontal direction:

Determine the tension in the cable AC and the angle θ2 by resolving the horizontal component of forces.

Fx=0TACcosθ2Ax=0 (VI)

Conclusion:

Substitute 63.03 lb for Ay in Equation (V).

63.03TACsinθ2=0TACsinθ2=63.03 (VII)

Substitute 40.18 lb for Ax in Equation (VI).

TACcosθ240.18=0TACcosθ2=40.18 (VIII)

Divide Equation (VII) by Equation (VIII).

TACsinθ2TACcosθ2=63.0340.18tanθ2=63.0340.18θ2=57.48°

Substitute 57.48° for θ2 in Equation (VII).

TACsin57.48°=63.03TAC=74.75lb

Thus, the tension in the cable AC is TAC=74.75lb_.

Joint D:

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

Engineering Mechanics : Statics-Practice Problems Workbook, Chapter 7.4, Problem 94P , additional homework tip  3

Using Figure (3),

Along the vertical direction:

Determine the tension in the cable CD and the angle θ3 by resolving the vertical component of forces.

Fy=0TBDsinθ1TCDsinθ350=0 (IX)

Along the horizontal direction:

Determine the tension in the cable CD and the angle θ3 by resolving the horizontal component of forces.

Fx=0TBDcosθ1TCDcosθ3=0 (X)

Conclusion:

Substitute 78.19 lb for TBD and 59.04° for θ1 in Equation (IX).

78.19sin59.04°TCDsinθ350=0TCDsinθ3=17.05 (XI)

Substitute 78.19 lb for TBD and 59.04° for θ1 in Equation (X).

78.19cos59.04TCDcosθ3=0TCDcosθ3=40.22 (XII)

Divide Equation (XI) by Equation (XII).

TCDsinθ3TCDcosθ3=17.0540.22tanθ3=17.0540.22θ3=22.97°

Substitute 22.97° for θ3 in Equation (XI).

TCDsin22.97°=17.05TCD=43.72lb

Thus, the tension in the cable CD is TCD=43.72lb_.

Entire segment:

Show the free-body diagram of the entire segment with the length of each cable as in Figure (4).

Engineering Mechanics : Statics-Practice Problems Workbook, Chapter 7.4, Problem 94P , additional homework tip  4

Using Figure (4),

Determine the entire length of the cable:

l=5sin59.04°+4cos22.97°+3cos57.48°=5.83+4.34+5.58=15.75ft

Thus, the total length of the chain is l=15.75ft_.

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Chapter 7 Solutions

Engineering Mechanics : Statics-Practice Problems Workbook

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