VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 4.2, Problem 4.71P

For the boom and loading shown, determine (a) the tension in cord BD, (b) the reaction at C.

Chapter 4.2, Problem 4.71P, For the boom and loading shown, determine (a) the tension in cord BD, (b) the reaction at C.

Fig.P4.71

(a)

Expert Solution
Check Mark
To determine

The tension in the cord BD.

Answer to Problem 4.71P

The tension in the cord BD is 5.63kips.

Explanation of Solution

Calculation:

If the rigid body subjected to three forces, such a body is commonly called a three-force body. If a two force body is in equilibrium the lines of action of the three forces must be either concurrent or parallel.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 4.2, Problem 4.71P , additional homework tip  1

The figure 1 represents the free body diagram of the system.

Write the expression for the angle α from the given figure.

tanα=EGAG (I)

Here, α is the angle.

Write the expression for the angle β.

tanβ=BHCF

Write the relations connecting the sides of the triangle.

BFAG=CFCG (II)

JEBH=DJDH (III)

Figure 2 represents a force triangle as shown in the above figure with its interior angles computed from the known directions of the forces. Then use the law of sines to find the unknown forces.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 4.2, Problem 4.71P , additional homework tip  2

Write the expression for the law of sines from the force triangle.

TBDsin94.76°=3kipssin32.106°=Csin(53.13°) (II)

Conclusion:

Substitute 12 for AG, 32 for CF, 48 for CG in equation II to find BH.

BF12=3248

BF=8in

BH=32BF=328=24in

Substitute 24 for BH, 48 for DJ, 48 for CG in equation III to find JE.

JE24=4832JE=36in

EG=JEJG=3632=4in

Substitute 4in for EG ,and 48in for AG in equation I to find α.

tanα=4in48in=0.083α=tan1(0.083)=4.763°

Substitute 24in for BH ,and 32in for CF in equation I to find β.

tanβ=24in32in=0.75β=tan(0.75)=36.86°

TBDsin94.76°=3kipssin32.106°=Csin(53.13°)

On rearranging the above equation to find T.

TBD×0.53=2.98TBD=2.980.53=5.63N=5.6N

Therefore, the tension in the cord BD is 5.63kips.

(b)

Expert Solution
Check Mark
To determine

The reaction at C.

Answer to Problem 4.71P

The reaction at C is 4.52kips.

Explanation of Solution

Calculation:

If the rigid body subjected to three forces, such a body is commonly called a three-force body. If a two force body is in equilibrium the lines of action of the three forces must be either concurrent or parallel.

Write the expression for the angle α from the given figure.

tanα=EGAG (I)

Here, α is the angle.

Write the expression for the angle β.

tanβ=BHCF

Write the relations connecting the sides of the triangle.

BFAG=CFCG (II)

JEBH=DJDH (III)

Figure 2 represents a force triangle as shown in the above figure with its interior angles computed from the known directions of the forces. Then use the law of sines to find the unknown forces.

Write the expression for the law of sines from the force triangle.

TBDsin94.76°=3kipssin32.106°=Csin(53.13°) (IV)

Conclusion:

Substitute 12 for AG, 32 for CF, 48 for CG in equation II to find BH.

BF12=3248

BF=8in

BH=32BF=328=24in

Substitute 24 for BH, 48 for DJ, 48 for CG in equation III to find JE.

JE24=4832JE=36in

EG=JEJG=3632=4in

Substitute 4in for EG ,and 48in for AG in equation I to find α.

tanα=4in48in=0.083α=tan1(0.083)=4.763°

Substitute 24in for BH ,and 32in for CF to find β.

tanβ=24in32in=0.75β=tan(0.75)=36.86°

TBDsin94.76°=3kipssin32.106°=Csin(53.13°)

On rearranging the above equation to find T.

TBD×0.53=2.98TBD=2.980.53=5.63N=5.6N

3kips×sin(53.13°)=C×sin(32.10°)C=4.52kips

Therefore, reaction at C is 4.52kips.

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Chapter 4 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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