The perpendicular distance between a line joining points O and D .

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Answer 1

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance between a line joining points $O$ and $D$ .

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Explanation of Solution

Refer the Problem 3.55

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$ :

$EH→=(30 in)i^−(7.5 in)j^+(10 in)k^$

The magnitude of the line joining points $E$ and $H$ is,

$EH=(30 in)2+(−7.5 in)2+(10 in)2=32.5 in$

The unit vector at line joining points $E$ and $H$ is,

$λEH=EH→EH$

Here, the unit vector at line joining points $E$ and $H$ is $λEH$ .

Substitute $(30 in)i^−(7.5 in)j^+(10 in)k^$ for $EH→$ and $32.5 in$ for $EH$ .

$λEH=(30 in)i^−(7.5 in)j^+(10 in)k^32.5 in=0.923i^−0.231j^+0.308k^$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$P→=PλEH$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $520 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(520 lb)(0.923i^−0.231j^+0.308k^)=(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$

Write the expression for force $P$ in terms of unit vector at line joining points $O$ and $D$ :

$OD→=(30 in)i^+(15 in)j^+(10 in)k^$

The magnitude of the line joining points $O$ and $D$ is,

$OD=(30 in)2+(15 in)2+(10 in)2=35 in$

The unit vector at line joining points $O$ and $D$ is,

$λOD=OD→OD$

Here, the unit vector at line joining points $O$ and $D$ is $λOD$ .

Substitute $(30 in)i^+(15 in)j^+(10 in)k^$ for $OD→$ and $35 in$ for $OD$ .

$λOD=(30 in)i^+(15 in)j^+(10 in)k^35 in=0.857i^+0.429j^+0.286k^$

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

$Pparallel=P→λOD$ (II)

Here, the perpendicular component contribute to the moment of force about the line $OD$ is $P→$

Substitute $(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$ for $P→$ and $0.857i^+0.429j^+0.286k^$ for $λOD$ in equation (II).

$Pparallel=((480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^)(0.857i^+0.429j^+0.286k^)=411.4 lb−51.5 lb+45.8 lb=405.7 lb$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P2=Pparallel2+Pperpendicular2$ (III)

Here, the parallel component of moment is $Pparallel$ and perpendicular component of moment is $Pperpendicular$ .

Rewrite the equation (III) to get $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Substitute $405.7 lb$ for $Pparallel$ and $520 lb$ for $P$ .

$Pperpendicular=(520 lb)2−(405.7 lb)2=325.28 lb$

Write the equation for the moment about a line joining points $O$ and $D$ .

$MOD=d(P)perpendicular$

Here, the moment about a line joining points $O$ and $D$ is $MOD$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MOD(P)perpendicular$

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $MFB$ is $3090 lb⋅in$

Substitute $3090 lb⋅in$ for $MOD$ and $325.28 lb$ for $(P)perpendicular$ .

$d=3090 lb⋅in325.28 lb=9.50 in$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

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Answer 2

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance between a line joining points $O$ and $D$ .

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Explanation of Solution

Refer the Problem 3.55

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$ :

$EH→=(30 in)i^−(7.5 in)j^+(10 in)k^$

The magnitude of the line joining points $E$ and $H$ is,

$EH=(30 in)2+(−7.5 in)2+(10 in)2=32.5 in$

The unit vector at line joining points $E$ and $H$ is,

$λEH=EH→EH$

Here, the unit vector at line joining points $E$ and $H$ is $λEH$ .

Substitute $(30 in)i^−(7.5 in)j^+(10 in)k^$ for $EH→$ and $32.5 in$ for $EH$ .

$λEH=(30 in)i^−(7.5 in)j^+(10 in)k^32.5 in=0.923i^−0.231j^+0.308k^$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$P→=PλEH$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $520 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(520 lb)(0.923i^−0.231j^+0.308k^)=(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$

Write the expression for force $P$ in terms of unit vector at line joining points $O$ and $D$ :

$OD→=(30 in)i^+(15 in)j^+(10 in)k^$

The magnitude of the line joining points $O$ and $D$ is,

$OD=(30 in)2+(15 in)2+(10 in)2=35 in$

The unit vector at line joining points $O$ and $D$ is,

$λOD=OD→OD$

Here, the unit vector at line joining points $O$ and $D$ is $λOD$ .

Substitute $(30 in)i^+(15 in)j^+(10 in)k^$ for $OD→$ and $35 in$ for $OD$ .

$λOD=(30 in)i^+(15 in)j^+(10 in)k^35 in=0.857i^+0.429j^+0.286k^$

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

$Pparallel=P→λOD$ (II)

Here, the perpendicular component contribute to the moment of force about the line $OD$ is $P→$

Substitute $(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$ for $P→$ and $0.857i^+0.429j^+0.286k^$ for $λOD$ in equation (II).

$Pparallel=((480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^)(0.857i^+0.429j^+0.286k^)=411.4 lb−51.5 lb+45.8 lb=405.7 lb$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P2=Pparallel2+Pperpendicular2$ (III)

Here, the parallel component of moment is $Pparallel$ and perpendicular component of moment is $Pperpendicular$ .

Rewrite the equation (III) to get $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Substitute $405.7 lb$ for $Pparallel$ and $520 lb$ for $P$ .

$Pperpendicular=(520 lb)2−(405.7 lb)2=325.28 lb$

Write the equation for the moment about a line joining points $O$ and $D$ .

$MOD=d(P)perpendicular$

Here, the moment about a line joining points $O$ and $D$ is $MOD$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MOD(P)perpendicular$

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $MFB$ is $3090 lb⋅in$

Substitute $3090 lb⋅in$ for $MOD$ and $325.28 lb$ for $(P)perpendicular$ .

$d=3090 lb⋅in325.28 lb=9.50 in$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

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Answer 3

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance between a line joining points $O$ and $D$ .

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Explanation of Solution

Refer the Problem 3.55

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$ :

$EH→=(30 in)i^−(7.5 in)j^+(10 in)k^$

The magnitude of the line joining points $E$ and $H$ is,

$EH=(30 in)2+(−7.5 in)2+(10 in)2=32.5 in$

The unit vector at line joining points $E$ and $H$ is,

$λEH=EH→EH$

Here, the unit vector at line joining points $E$ and $H$ is $λEH$ .

Substitute $(30 in)i^−(7.5 in)j^+(10 in)k^$ for $EH→$ and $32.5 in$ for $EH$ .

$λEH=(30 in)i^−(7.5 in)j^+(10 in)k^32.5 in=0.923i^−0.231j^+0.308k^$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$P→=PλEH$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $520 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(520 lb)(0.923i^−0.231j^+0.308k^)=(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$

Write the expression for force $P$ in terms of unit vector at line joining points $O$ and $D$ :

$OD→=(30 in)i^+(15 in)j^+(10 in)k^$

The magnitude of the line joining points $O$ and $D$ is,

$OD=(30 in)2+(15 in)2+(10 in)2=35 in$

The unit vector at line joining points $O$ and $D$ is,

$λOD=OD→OD$

Here, the unit vector at line joining points $O$ and $D$ is $λOD$ .

Substitute $(30 in)i^+(15 in)j^+(10 in)k^$ for $OD→$ and $35 in$ for $OD$ .

$λOD=(30 in)i^+(15 in)j^+(10 in)k^35 in=0.857i^+0.429j^+0.286k^$

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

$Pparallel=P→λOD$ (II)

Here, the perpendicular component contribute to the moment of force about the line $OD$ is $P→$

Substitute $(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$ for $P→$ and $0.857i^+0.429j^+0.286k^$ for $λOD$ in equation (II).

$Pparallel=((480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^)(0.857i^+0.429j^+0.286k^)=411.4 lb−51.5 lb+45.8 lb=405.7 lb$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P2=Pparallel2+Pperpendicular2$ (III)

Here, the parallel component of moment is $Pparallel$ and perpendicular component of moment is $Pperpendicular$ .

Rewrite the equation (III) to get $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Substitute $405.7 lb$ for $Pparallel$ and $520 lb$ for $P$ .

$Pperpendicular=(520 lb)2−(405.7 lb)2=325.28 lb$

Write the equation for the moment about a line joining points $O$ and $D$ .

$MOD=d(P)perpendicular$

Here, the moment about a line joining points $O$ and $D$ is $MOD$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MOD(P)perpendicular$

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $MFB$ is $3090 lb⋅in$

Substitute $3090 lb⋅in$ for $MOD$ and $325.28 lb$ for $(P)perpendicular$ .

$d=3090 lb⋅in325.28 lb=9.50 in$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

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Answer 4

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance between a line joining points $O$ and $D$ .

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Explanation of Solution

Refer the Problem 3.55

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$ :

$EH→=(30 in)i^−(7.5 in)j^+(10 in)k^$

The magnitude of the line joining points $E$ and $H$ is,

$EH=(30 in)2+(−7.5 in)2+(10 in)2=32.5 in$

The unit vector at line joining points $E$ and $H$ is,

$λEH=EH→EH$

Here, the unit vector at line joining points $E$ and $H$ is $λEH$ .

Substitute $(30 in)i^−(7.5 in)j^+(10 in)k^$ for $EH→$ and $32.5 in$ for $EH$ .

$λEH=(30 in)i^−(7.5 in)j^+(10 in)k^32.5 in=0.923i^−0.231j^+0.308k^$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$P→=PλEH$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $520 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(520 lb)(0.923i^−0.231j^+0.308k^)=(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$

Write the expression for force $P$ in terms of unit vector at line joining points $O$ and $D$ :

$OD→=(30 in)i^+(15 in)j^+(10 in)k^$

The magnitude of the line joining points $O$ and $D$ is,

$OD=(30 in)2+(15 in)2+(10 in)2=35 in$

The unit vector at line joining points $O$ and $D$ is,

$λOD=OD→OD$

Here, the unit vector at line joining points $O$ and $D$ is $λOD$ .

Substitute $(30 in)i^+(15 in)j^+(10 in)k^$ for $OD→$ and $35 in$ for $OD$ .

$λOD=(30 in)i^+(15 in)j^+(10 in)k^35 in=0.857i^+0.429j^+0.286k^$

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

$Pparallel=P→λOD$ (II)

Here, the perpendicular component contribute to the moment of force about the line $OD$ is $P→$

Substitute $(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$ for $P→$ and $0.857i^+0.429j^+0.286k^$ for $λOD$ in equation (II).

$Pparallel=((480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^)(0.857i^+0.429j^+0.286k^)=411.4 lb−51.5 lb+45.8 lb=405.7 lb$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P2=Pparallel2+Pperpendicular2$ (III)

Here, the parallel component of moment is $Pparallel$ and perpendicular component of moment is $Pperpendicular$ .

Rewrite the equation (III) to get $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Substitute $405.7 lb$ for $Pparallel$ and $520 lb$ for $P$ .

$Pperpendicular=(520 lb)2−(405.7 lb)2=325.28 lb$

Write the equation for the moment about a line joining points $O$ and $D$ .

$MOD=d(P)perpendicular$

Here, the moment about a line joining points $O$ and $D$ is $MOD$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MOD(P)perpendicular$

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $MFB$ is $3090 lb⋅in$

Substitute $3090 lb⋅in$ for $MOD$ and $325.28 lb$ for $(P)perpendicular$ .

$d=3090 lb⋅in325.28 lb=9.50 in$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

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Answer 5

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance between a line joining points $O$ and $D$ .

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Explanation of Solution

Refer the Problem 3.55

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$ :

$EH→=(30 in)i^−(7.5 in)j^+(10 in)k^$

The magnitude of the line joining points $E$ and $H$ is,

$EH=(30 in)2+(−7.5 in)2+(10 in)2=32.5 in$

The unit vector at line joining points $E$ and $H$ is,

$λEH=EH→EH$

Here, the unit vector at line joining points $E$ and $H$ is $λEH$ .

Substitute $(30 in)i^−(7.5 in)j^+(10 in)k^$ for $EH→$ and $32.5 in$ for $EH$ .

$λEH=(30 in)i^−(7.5 in)j^+(10 in)k^32.5 in=0.923i^−0.231j^+0.308k^$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$P→=PλEH$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $520 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(520 lb)(0.923i^−0.231j^+0.308k^)=(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$

Write the expression for force $P$ in terms of unit vector at line joining points $O$ and $D$ :

$OD→=(30 in)i^+(15 in)j^+(10 in)k^$

The magnitude of the line joining points $O$ and $D$ is,

$OD=(30 in)2+(15 in)2+(10 in)2=35 in$

The unit vector at line joining points $O$ and $D$ is,

$λOD=OD→OD$

Here, the unit vector at line joining points $O$ and $D$ is $λOD$ .

Substitute $(30 in)i^+(15 in)j^+(10 in)k^$ for $OD→$ and $35 in$ for $OD$ .

$λOD=(30 in)i^+(15 in)j^+(10 in)k^35 in=0.857i^+0.429j^+0.286k^$

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

$Pparallel=P→λOD$ (II)

Here, the perpendicular component contribute to the moment of force about the line $OD$ is $P→$

Substitute $(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$ for $P→$ and $0.857i^+0.429j^+0.286k^$ for $λOD$ in equation (II).

$Pparallel=((480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^)(0.857i^+0.429j^+0.286k^)=411.4 lb−51.5 lb+45.8 lb=405.7 lb$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P2=Pparallel2+Pperpendicular2$ (III)

Here, the parallel component of moment is $Pparallel$ and perpendicular component of moment is $Pperpendicular$ .

Rewrite the equation (III) to get $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Substitute $405.7 lb$ for $Pparallel$ and $520 lb$ for $P$ .

$Pperpendicular=(520 lb)2−(405.7 lb)2=325.28 lb$

Write the equation for the moment about a line joining points $O$ and $D$ .

$MOD=d(P)perpendicular$

Here, the moment about a line joining points $O$ and $D$ is $MOD$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MOD(P)perpendicular$

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $MFB$ is $3090 lb⋅in$

Substitute $3090 lb⋅in$ for $MOD$ and $325.28 lb$ for $(P)perpendicular$ .

$d=3090 lb⋅in325.28 lb=9.50 in$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Answer 6

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance between a line joining points $O$ and $D$ .

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Explanation of Solution

Refer the Problem 3.55

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$ :

$EH→=(30 in)i^−(7.5 in)j^+(10 in)k^$

The magnitude of the line joining points $E$ and $H$ is,

$EH=(30 in)2+(−7.5 in)2+(10 in)2=32.5 in$

The unit vector at line joining points $E$ and $H$ is,

$λEH=EH→EH$

Here, the unit vector at line joining points $E$ and $H$ is $λEH$ .

Substitute $(30 in)i^−(7.5 in)j^+(10 in)k^$ for $EH→$ and $32.5 in$ for $EH$ .

$λEH=(30 in)i^−(7.5 in)j^+(10 in)k^32.5 in=0.923i^−0.231j^+0.308k^$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$P→=PλEH$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $520 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(520 lb)(0.923i^−0.231j^+0.308k^)=(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$

Write the expression for force $P$ in terms of unit vector at line joining points $O$ and $D$ :

$OD→=(30 in)i^+(15 in)j^+(10 in)k^$

The magnitude of the line joining points $O$ and $D$ is,

$OD=(30 in)2+(15 in)2+(10 in)2=35 in$

The unit vector at line joining points $O$ and $D$ is,

$λOD=OD→OD$

Here, the unit vector at line joining points $O$ and $D$ is $λOD$ .

Substitute $(30 in)i^+(15 in)j^+(10 in)k^$ for $OD→$ and $35 in$ for $OD$ .

$λOD=(30 in)i^+(15 in)j^+(10 in)k^35 in=0.857i^+0.429j^+0.286k^$

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

$Pparallel=P→λOD$ (II)

Here, the perpendicular component contribute to the moment of force about the line $OD$ is $P→$

Substitute $(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$ for $P→$ and $0.857i^+0.429j^+0.286k^$ for $λOD$ in equation (II).

$Pparallel=((480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^)(0.857i^+0.429j^+0.286k^)=411.4 lb−51.5 lb+45.8 lb=405.7 lb$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P2=Pparallel2+Pperpendicular2$ (III)

Here, the parallel component of moment is $Pparallel$ and perpendicular component of moment is $Pperpendicular$ .

Rewrite the equation (III) to get $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Substitute $405.7 lb$ for $Pparallel$ and $520 lb$ for $P$ .

$Pperpendicular=(520 lb)2−(405.7 lb)2=325.28 lb$

Write the equation for the moment about a line joining points $O$ and $D$ .

$MOD=d(P)perpendicular$

Here, the moment about a line joining points $O$ and $D$ is $MOD$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MOD(P)perpendicular$

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $MFB$ is $3090 lb⋅in$

Substitute $3090 lb⋅in$ for $MOD$ and $325.28 lb$ for $(P)perpendicular$ .

$d=3090 lb⋅in325.28 lb=9.50 in$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Answer 7

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance between a line joining points $O$ and $D$ .

Answer 8

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Refer the Problem 3.55

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$ :

$EH→=(30 in)i^−(7.5 in)j^+(10 in)k^$

The magnitude of the line joining points $E$ and $H$ is,

$EH=(30 in)2+(−7.5 in)2+(10 in)2=32.5 in$

The unit vector at line joining points $E$ and $H$ is,

$λEH=EH→EH$

Here, the unit vector at line joining points $E$ and $H$ is $λEH$ .

Substitute $(30 in)i^−(7.5 in)j^+(10 in)k^$ for $EH→$ and $32.5 in$ for $EH$ .

$λEH=(30 in)i^−(7.5 in)j^+(10 in)k^32.5 in=0.923i^−0.231j^+0.308k^$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$P→=PλEH$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $520 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(520 lb)(0.923i^−0.231j^+0.308k^)=(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$

Write the expression for force $P$ in terms of unit vector at line joining points $O$ and $D$ :

$OD→=(30 in)i^+(15 in)j^+(10 in)k^$

The magnitude of the line joining points $O$ and $D$ is,

$OD=(30 in)2+(15 in)2+(10 in)2=35 in$

The unit vector at line joining points $O$ and $D$ is,

$λOD=OD→OD$

Here, the unit vector at line joining points $O$ and $D$ is $λOD$ .

Substitute $(30 in)i^+(15 in)j^+(10 in)k^$ for $OD→$ and $35 in$ for $OD$ .

$λOD=(30 in)i^+(15 in)j^+(10 in)k^35 in=0.857i^+0.429j^+0.286k^$

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

$Pparallel=P→λOD$ (II)

Here, the perpendicular component contribute to the moment of force about the line $OD$ is $P→$

Substitute $(480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^$ for $P→$ and $0.857i^+0.429j^+0.286k^$ for $λOD$ in equation (II).

$Pparallel=((480.0 lb)i^−(120.0 lb)j^+(160.0 lb)k^)(0.857i^+0.429j^+0.286k^)=411.4 lb−51.5 lb+45.8 lb=405.7 lb$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P2=Pparallel2+Pperpendicular2$ (III)

Here, the parallel component of moment is $Pparallel$ and perpendicular component of moment is $Pperpendicular$ .

Rewrite the equation (III) to get $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Substitute $405.7 lb$ for $Pparallel$ and $520 lb$ for $P$ .

$Pperpendicular=(520 lb)2−(405.7 lb)2=325.28 lb$

Write the equation for the moment about a line joining points $O$ and $D$ .

$MOD=d(P)perpendicular$

Here, the moment about a line joining points $O$ and $D$ is $MOD$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MOD(P)perpendicular$

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $MFB$ is $3090 lb⋅in$

Substitute $3090 lb⋅in$ for $MOD$ and $325.28 lb$ for $(P)perpendicular$ .

$d=3090 lb⋅in325.28 lb=9.50 in$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50 in$ .

Answer 12

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Answer to Problem 3.64P

Explanation of Solution

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The perpendicular distance between a line joining points $O$ and $D$ .