The perpendicular distance between a line joining points F and B .

Question

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Answer 1

Question

Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance between a line joining points $F$ and $B$ .

Expert Solution & Answer

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Explanation of Solution

Refer the Problem 3.56P

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 $910 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(910 lb)(0.923i^−0.231j^+0.308k^)=(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$

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

$FB→=(30 in)i^+(15 in)j^−(10 in)k^$

The magnitude of the line joining points $F$ and $B$ is,

$FB=(30 in)2+(15 in)2+(−10 in)2=35.0 in$

The unit vector at line joining points $F$ and $B$ is,

$λFB=FB→FB$

Here, the unit vector at line joining points $F$ and $B$ is $λFB$ .

Substitute $(30 in)i^+(15 in)j^−(10 in)k^$ for $FB→$ and $35.0 in$ for $FB$ .

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

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$Pparallel=P→λFB$ (II)

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

Substitute $(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$ for $P→$ and $0.857i^+0.429j^−0.286k^$ for $λFB$ in equation (II).

$Pparallel=((840.0 lb)i^−(210.0 lb)j^+(280 lb)k^)(0.857i^+0.429j^−0.286k^)=720 lb−90.1 lb−80.1 lb=550 lb$

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 $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910 lb$ .

Substitute $550 lb$ for $Pparallel$ and $910 lb$ for $P$ .

$Pperpendicular=(910 lb)2−(550 lb)2828100−302500=724.98 lb$

Write the equation for the moment about a line joining points $F$ and $B$ .

$MFB=d(P)perpendicular$

Here, the moment about a line joining points $F$ and $B$ is $MFB$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MFB(P)perpendicular$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $MFB$ is $300 lb⋅ft$

Substitute $300 lb⋅ft$ for $MFB$ and $724.98 lb$ for $(P)perpendicular$ .

$d=300 lb⋅ft(12 lb⋅in1 lb⋅ft)724.98 lb=3600 lb⋅in724.98 lb=4.97 in$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

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Answer 2

Question

Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance between a line joining points $F$ and $B$ .

Expert Solution & Answer

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Explanation of Solution

Refer the Problem 3.56P

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 $910 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(910 lb)(0.923i^−0.231j^+0.308k^)=(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$

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

$FB→=(30 in)i^+(15 in)j^−(10 in)k^$

The magnitude of the line joining points $F$ and $B$ is,

$FB=(30 in)2+(15 in)2+(−10 in)2=35.0 in$

The unit vector at line joining points $F$ and $B$ is,

$λFB=FB→FB$

Here, the unit vector at line joining points $F$ and $B$ is $λFB$ .

Substitute $(30 in)i^+(15 in)j^−(10 in)k^$ for $FB→$ and $35.0 in$ for $FB$ .

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

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$Pparallel=P→λFB$ (II)

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

Substitute $(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$ for $P→$ and $0.857i^+0.429j^−0.286k^$ for $λFB$ in equation (II).

$Pparallel=((840.0 lb)i^−(210.0 lb)j^+(280 lb)k^)(0.857i^+0.429j^−0.286k^)=720 lb−90.1 lb−80.1 lb=550 lb$

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 $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910 lb$ .

Substitute $550 lb$ for $Pparallel$ and $910 lb$ for $P$ .

$Pperpendicular=(910 lb)2−(550 lb)2828100−302500=724.98 lb$

Write the equation for the moment about a line joining points $F$ and $B$ .

$MFB=d(P)perpendicular$

Here, the moment about a line joining points $F$ and $B$ is $MFB$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MFB(P)perpendicular$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $MFB$ is $300 lb⋅ft$

Substitute $300 lb⋅ft$ for $MFB$ and $724.98 lb$ for $(P)perpendicular$ .

$d=300 lb⋅ft(12 lb⋅in1 lb⋅ft)724.98 lb=3600 lb⋅in724.98 lb=4.97 in$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

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Answer 3

Question

Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance between a line joining points $F$ and $B$ .

Expert Solution & Answer

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Explanation of Solution

Refer the Problem 3.56P

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 $910 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(910 lb)(0.923i^−0.231j^+0.308k^)=(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$

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

$FB→=(30 in)i^+(15 in)j^−(10 in)k^$

The magnitude of the line joining points $F$ and $B$ is,

$FB=(30 in)2+(15 in)2+(−10 in)2=35.0 in$

The unit vector at line joining points $F$ and $B$ is,

$λFB=FB→FB$

Here, the unit vector at line joining points $F$ and $B$ is $λFB$ .

Substitute $(30 in)i^+(15 in)j^−(10 in)k^$ for $FB→$ and $35.0 in$ for $FB$ .

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

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$Pparallel=P→λFB$ (II)

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

Substitute $(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$ for $P→$ and $0.857i^+0.429j^−0.286k^$ for $λFB$ in equation (II).

$Pparallel=((840.0 lb)i^−(210.0 lb)j^+(280 lb)k^)(0.857i^+0.429j^−0.286k^)=720 lb−90.1 lb−80.1 lb=550 lb$

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 $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910 lb$ .

Substitute $550 lb$ for $Pparallel$ and $910 lb$ for $P$ .

$Pperpendicular=(910 lb)2−(550 lb)2828100−302500=724.98 lb$

Write the equation for the moment about a line joining points $F$ and $B$ .

$MFB=d(P)perpendicular$

Here, the moment about a line joining points $F$ and $B$ is $MFB$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MFB(P)perpendicular$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $MFB$ is $300 lb⋅ft$

Substitute $300 lb⋅ft$ for $MFB$ and $724.98 lb$ for $(P)perpendicular$ .

$d=300 lb⋅ft(12 lb⋅in1 lb⋅ft)724.98 lb=3600 lb⋅in724.98 lb=4.97 in$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

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Answer 4

Question

Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance between a line joining points $F$ and $B$ .

Expert Solution & Answer

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Explanation of Solution

Refer the Problem 3.56P

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 $910 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(910 lb)(0.923i^−0.231j^+0.308k^)=(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$

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

$FB→=(30 in)i^+(15 in)j^−(10 in)k^$

The magnitude of the line joining points $F$ and $B$ is,

$FB=(30 in)2+(15 in)2+(−10 in)2=35.0 in$

The unit vector at line joining points $F$ and $B$ is,

$λFB=FB→FB$

Here, the unit vector at line joining points $F$ and $B$ is $λFB$ .

Substitute $(30 in)i^+(15 in)j^−(10 in)k^$ for $FB→$ and $35.0 in$ for $FB$ .

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

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$Pparallel=P→λFB$ (II)

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

Substitute $(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$ for $P→$ and $0.857i^+0.429j^−0.286k^$ for $λFB$ in equation (II).

$Pparallel=((840.0 lb)i^−(210.0 lb)j^+(280 lb)k^)(0.857i^+0.429j^−0.286k^)=720 lb−90.1 lb−80.1 lb=550 lb$

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 $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910 lb$ .

Substitute $550 lb$ for $Pparallel$ and $910 lb$ for $P$ .

$Pperpendicular=(910 lb)2−(550 lb)2828100−302500=724.98 lb$

Write the equation for the moment about a line joining points $F$ and $B$ .

$MFB=d(P)perpendicular$

Here, the moment about a line joining points $F$ and $B$ is $MFB$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MFB(P)perpendicular$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $MFB$ is $300 lb⋅ft$

Substitute $300 lb⋅ft$ for $MFB$ and $724.98 lb$ for $(P)perpendicular$ .

$d=300 lb⋅ft(12 lb⋅in1 lb⋅ft)724.98 lb=3600 lb⋅in724.98 lb=4.97 in$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

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Answer 5

Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance between a line joining points $F$ and $B$ .

Expert Solution & Answer

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Explanation of Solution

Refer the Problem 3.56P

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 $910 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(910 lb)(0.923i^−0.231j^+0.308k^)=(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$

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

$FB→=(30 in)i^+(15 in)j^−(10 in)k^$

The magnitude of the line joining points $F$ and $B$ is,

$FB=(30 in)2+(15 in)2+(−10 in)2=35.0 in$

The unit vector at line joining points $F$ and $B$ is,

$λFB=FB→FB$

Here, the unit vector at line joining points $F$ and $B$ is $λFB$ .

Substitute $(30 in)i^+(15 in)j^−(10 in)k^$ for $FB→$ and $35.0 in$ for $FB$ .

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

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$Pparallel=P→λFB$ (II)

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

Substitute $(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$ for $P→$ and $0.857i^+0.429j^−0.286k^$ for $λFB$ in equation (II).

$Pparallel=((840.0 lb)i^−(210.0 lb)j^+(280 lb)k^)(0.857i^+0.429j^−0.286k^)=720 lb−90.1 lb−80.1 lb=550 lb$

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 $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910 lb$ .

Substitute $550 lb$ for $Pparallel$ and $910 lb$ for $P$ .

$Pperpendicular=(910 lb)2−(550 lb)2828100−302500=724.98 lb$

Write the equation for the moment about a line joining points $F$ and $B$ .

$MFB=d(P)perpendicular$

Here, the moment about a line joining points $F$ and $B$ is $MFB$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MFB(P)perpendicular$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $MFB$ is $300 lb⋅ft$

Substitute $300 lb⋅ft$ for $MFB$ and $724.98 lb$ for $(P)perpendicular$ .

$d=300 lb⋅ft(12 lb⋅in1 lb⋅ft)724.98 lb=3600 lb⋅in724.98 lb=4.97 in$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Answer 6

Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance between a line joining points $F$ and $B$ .

Expert Solution & Answer

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Explanation of Solution

Refer the Problem 3.56P

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 $910 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(910 lb)(0.923i^−0.231j^+0.308k^)=(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$

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

$FB→=(30 in)i^+(15 in)j^−(10 in)k^$

The magnitude of the line joining points $F$ and $B$ is,

$FB=(30 in)2+(15 in)2+(−10 in)2=35.0 in$

The unit vector at line joining points $F$ and $B$ is,

$λFB=FB→FB$

Here, the unit vector at line joining points $F$ and $B$ is $λFB$ .

Substitute $(30 in)i^+(15 in)j^−(10 in)k^$ for $FB→$ and $35.0 in$ for $FB$ .

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

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$Pparallel=P→λFB$ (II)

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

Substitute $(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$ for $P→$ and $0.857i^+0.429j^−0.286k^$ for $λFB$ in equation (II).

$Pparallel=((840.0 lb)i^−(210.0 lb)j^+(280 lb)k^)(0.857i^+0.429j^−0.286k^)=720 lb−90.1 lb−80.1 lb=550 lb$

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 $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910 lb$ .

Substitute $550 lb$ for $Pparallel$ and $910 lb$ for $P$ .

$Pperpendicular=(910 lb)2−(550 lb)2828100−302500=724.98 lb$

Write the equation for the moment about a line joining points $F$ and $B$ .

$MFB=d(P)perpendicular$

Here, the moment about a line joining points $F$ and $B$ is $MFB$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MFB(P)perpendicular$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $MFB$ is $300 lb⋅ft$

Substitute $300 lb⋅ft$ for $MFB$ and $724.98 lb$ for $(P)perpendicular$ .

$d=300 lb⋅ft(12 lb⋅in1 lb⋅ft)724.98 lb=3600 lb⋅in724.98 lb=4.97 in$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Answer 7

Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance between a line joining points $F$ and $B$ .

Answer 8

Expert Solution & Answer

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Refer the Problem 3.56P

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 $910 lb$ for $P$ and $0.923i^−0.231j^+0.308k^$ for $λEH$ in equation (I).

$P→=(910 lb)(0.923i^−0.231j^+0.308k^)=(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$

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

$FB→=(30 in)i^+(15 in)j^−(10 in)k^$

The magnitude of the line joining points $F$ and $B$ is,

$FB=(30 in)2+(15 in)2+(−10 in)2=35.0 in$

The unit vector at line joining points $F$ and $B$ is,

$λFB=FB→FB$

Here, the unit vector at line joining points $F$ and $B$ is $λFB$ .

Substitute $(30 in)i^+(15 in)j^−(10 in)k^$ for $FB→$ and $35.0 in$ for $FB$ .

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

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$Pparallel=P→λFB$ (II)

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

Substitute $(840.0 lb)i^−(210.0 lb)j^+(280 lb)k^$ for $P→$ and $0.857i^+0.429j^−0.286k^$ for $λFB$ in equation (II).

$Pparallel=((840.0 lb)i^−(210.0 lb)j^+(280 lb)k^)(0.857i^+0.429j^−0.286k^)=720 lb−90.1 lb−80.1 lb=550 lb$

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 $Pperpendicular$ .

$Pperpendicular=P2−Pparallel2$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910 lb$ .

Substitute $550 lb$ for $Pparallel$ and $910 lb$ for $P$ .

$Pperpendicular=(910 lb)2−(550 lb)2828100−302500=724.98 lb$

Write the equation for the moment about a line joining points $F$ and $B$ .

$MFB=d(P)perpendicular$

Here, the moment about a line joining points $F$ and $B$ is $MFB$ and the perpendicular distance is $d$ .

Rewrite the relation in terms of $d$ .

$d=MFB(P)perpendicular$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $MFB$ is $300 lb⋅ft$

Substitute $300 lb⋅ft$ for $MFB$ and $724.98 lb$ for $(P)perpendicular$ .

$d=300 lb⋅ft(12 lb⋅in1 lb⋅ft)724.98 lb=3600 lb⋅in724.98 lb=4.97 in$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97 in$ .

Answer 12

Ch. 3.1 - A foot valve for a pneumatic system is hinged at...Ch. 3.1 - 3.2A foot valve for a pneumatic system is hinged...Ch. 3.1 - It is known that a vertical force of 200 lb is...Ch. 3.1 - A 300-N force is applied at A as shown. Determine...Ch. 3.1 - A 300-N force is applied at A as shown. Determine...Ch. 3.1 - An 8-lb force P is applied to a shift lever....Ch. 3.1 - Prob. 3.7P Ch. 3.1 - An 11-lb force P is applied to a shift lever. The...Ch. 3.1 - Rod AB is held in place by the cord AC. Knowing...Ch. 3.1 - Rod AB is held in place by the cord AC. Knowing...

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Answer to Problem 3.65P

Explanation of Solution

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