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 3.3, Problem 3.76P

If P = 0 in the figure, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

Chapter 3.3, Problem 3.76P, If P = 0 in the figure, replace the two remaining couples with a single equivalent couple,

Expert Solution & Answer
Check Mark
To determine

The net equivalent couple vector shown in the triangular co-ordinates and its direction.

Answer to Problem 3.76P

The net equivalent couple vector is M=604lbin and the direction of the couple formed is θx=72.8°, θy=27.3°, θz=110.5°.

Explanation of Solution

Write the equation of moment of first couple about an arbitrary axis.

M1=rc×F1 (I)

Here, the first couple vector formed due to the force F1 is M1, perpendicular distance of the line of action of force is rc, and the vector force acting on the side C is F1.

Substitute (30in)i^ for rc and (16lb)j^ for F1 in equation (I).

M1=(30in)i^×(16lb)j^=(480lbin)k^ (II)

Write the equation of moment of second couple about an arbitrary axis.

M2=rE/B×F2 (III)

Here, the second couple vector formed due to the force F2 is M2, perpendicular distance of the vector line drawn from B to E is rE/B, and the vector force acting on the side EB is F2.

The vector line drawn from B to E is,

rE/B=r2i^r1j^

Here, the distance between the points DC is r1 and distance between the points EC is r2.

Substitute (5in) for r1 and (15in) for r2.

rE/B=(15in)i^(5in)j^

The distance of the line drawn from DE is,

dDE=(0)2+(5)2+(10)2=55in

The resultant magnitude of the each force is,

F2=MDEdDE (IV)

Here, moment of couple formed by the force at the points DE is MDE and perpendicular distance of the line drawn from DE is dDE.

Substitute 40lb for MDE and 55in for dDE in equation (IV).

F2=40lb55in(5j^10k^)=85[(1lb)j^(2lb)k^]

Write the vector form for the moment of second couple M2.

M2=85|i^        j^      k^15  5       00       1     2|=85[(10lbin)i^+(30lbin)j^+(15lbin)k^] (V)

Write the equation of net moment of couple about an arbitrary axis.

M=M1+M2 (VI)

Substitute equation (II) and (V) in equation (VI).

M=(480lbin)k^+85[(10lbin)i^+(30lbin)j^+(15lbin)k^]=(178.885lbin)i^+(536.66lbin)j^(211.67lbin)k^

The magnitude of the net moment of couple is,

|M|=(178.885lbin)2+(536.66lbin)2+(211.67lbin)2M=603.99lbin

Conclusion:

Write the equation for the unit vector of the couple.

λaxis=MM (VII)

Here, the unit vector is represented as λaxis.

Substitute (178.885lbin)i^+(536.66lbin)j^(211.67lbin)k^ for M and 603.99lbin for M in equation (VII).

λaxis=(178.885lbin)i^+(536.66lbin)j^(211.67lbin)k^603.99lbin=0.29617i^+0.88852j^0.35045k^

The resultant couple formed along the x direction.

cosθx=0.29617θx=cos1(0.29617)=72.8°

The resultant couple formed along the y direction.

cosθy=0.88852θy=cos1(0.88852)=27.3°

The resultant couple formed along the z direction.

cosθz=0.35045θz=cos1(0.35045)=110.5°

Therefore, the net equivalent couple vector is M=604lbin and the direction of the couple formed is θx=72.8°, θy=27.3°, θz=110.5°.

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Chapter 3 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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