A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at

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Answer 1

Textbook Question

Chapter 2.5, Problem 2.112P

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at

Expert Solution & Answer

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590 lb$ .

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $2000 lb_$ .

Explanation of Solution

Free body diagram at $A$ is shown in figure1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 1

Here, $TAB$ is the magnitude of tension in cable $AB$ , $TAC$ is the magnitude of tension in cable $AC$ , $TAD$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$ .

The tension in the cable $AB$ is $840 lb$ .

The sketch of plate supported by three cables is shown in figure2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 2

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y and z$ direction.

Write the equation of vector distance $AB$ .

$AB→=(x2−x1)i+(y2−y1)j+(z2−z1)k$ (I)

Here, $AB→$ is the vector distance of the cable $AB$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x2,y2 and z2$ are the coordinates of point $B$ .

Write the vector distance of the cable $AC$ .

$AC→=(x3−x1)i+(y3−y1)j+(z3−z1)k$ (II)

Here, $AC→$ is the vector distance of the cable $AC$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x3,y3 and z3$ are the coordinates of point $C$ .

Write the vector distance of the cable $AD$ .

$AD→=(x4−x1)i+(y4−y1)j+(z4−z1)k$ (III)

Here, $AD→$ is the vector distance of the cable $AD$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x4,y4 and z4$ are the coordinates of point $D$ .

Write the equation of tension in the cable $AB$ .

$TAB=λABTAB$ (IV)

Here, $TAB$ is the tension in the cable $AB$ , $TAB$ is the magnitude of the tension in the cable $AB$ and $λAB$ is the unit vector in the direction of $AB$ .

Write the equation of $λAB$ .

$λAB=AB→AB$ (V)

Write the equation of tension in the cable $AC$ .

$TAC=λACTAC$ (VI)

Here, $TAC$ is the tension in the cable $AC$ , $TAC$ is the magnitude of the tension in the cable $AC$ and $λAC$ is the unit vector in the direction of $AC$ .

Write the equation of $λAC$ .

$λAC=AC→AC$ (VII)

Write the equation of tension in the cable $AD$ .

$TAD=λADTAD$ (VIII)

Here, $TAD$ is the tension in the cable $AD$ , $TAD$ is the magnitude of the tension in the cable $AD$ and $λAD$ is the unit vector in the direction of $AD$ .

Write the equation of $λAD$ .

$λAD=AD→AD$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=P j$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$ .

Write the equilibrium condition for the forces at $A$ .

$∑F=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$ .

$TAB+TAC+TAD+P=0$

Conclusion:

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x2$ , $0 ft$ for $y2$ and $25 ft$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(20 ft)i−(100 ft)j+(25 ft)k$

Calculate the magnitude of $AB→$ .

$AB=(−20 ft)2+(−100 ft)2+(25 ft)2= 105 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $60 ft$ for $x3$ , $0 ft$ for $y3$ and $25 ft$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(60 ft)i−(100 ft)j+(18 ft)k$

Calculate the magnitude of $AC→$ .

$AC=(60 ft)2+(−100 ft)2+(18 ft)2=118 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x4$ , $0 ft$ for $y4$ and $−74 ft$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(−20 ft)i−(100 ft)j−(74 ft)k$

Calculate the magnitude of $AD→$ .

$AD=(−20 ft)2+(−100 ft)2+(−74 ft)2=126 ft$

Substitute $−(20 ft)i−(100 ft)j+(25 ft)k$ for $AB→$ and $105 ft$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(20 ft)i−(100 ft)j+(25 ft)k105 ft=−421i−2021j+521k$

Substitute $−421i−2021j+521k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−421i−2021j+521k)TAB$

Substitute $(60 ft)i−(100 ft)j+(18 ft)k$ for $AC→$ and $118 ft$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(60 ft)i−(100 ft)j+(18 ft)k118 ft=3059i−5059j+959k$

Substitute $3059i−5059j+959k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(3059i−5059j+959k)TAC$

Substitute $(−20 ft)i−(100 ft)j−(74 ft)k$ for $AD→$ and $126 ft$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(−20 ft)i−(100 ft)j−(74 ft)k126 ft=−1063i−5063j−3763k$

Substitute $−1063i−5063j−3763k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(−1063i−5063j−3763k)TAD$

Substitute $(−421i−2021j+521k)TAB$ for $TAB$ , $(3059i−5059j+959k)TAC$ for $TAC$ , $(−1063i−5063j−3763k)TAD$ , $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−421i−2021j+521k)TAB+(3059i−5059j+959k)TAC+(−1063i−5063j−3763k)TAD+Pj=0(−421TAB+3059TAC−1063TAD)i+(−2021TAB−5059TAC−5063TAD+P)j+(+521TAB+959TAC−3763TAD)k=0$

Since total force is zero. Equate force along each direction as zero.

$−421TAB+3059TAC−1063TAD=0$ (XII)

$−2021TAB−5059TAC−5063TAD+P=0$ (XIII)

$+521TAB+959TAC−3763TAD= 0$ (XIV)

Substitute $590 lb$ for $TAC$ in equation (XII) and (XIV) to modify the equation.

$−421TAB+3059(590 lb)−1063TAD=0$

$−421TAB+300 lb−1063TAD=0$ (XV)

$+521TAB+959(590 lb)−3763TAD= 0$

$+521TAB+90 lb−3763TAD= 0$ (XVI)

Multiply equation (XV) by $5$ and XVI by $−10$ and add to get $TAC$ .

$−2021TAB+1500 lb−5063TAD+2021TAB+360 lb−14863TAD= 01860 lb−19863TAD= 0TAD= 591.82 lb$

Substitute $591.82 lb$ for $TAD$ in equation (XVI) to get $TAB$ .

$+521TAB+90 lb−3763(591.82 lb)= 0TAB= 1081.82 lb$

Substitute $1081.82 lb$ for $TAB$ , $590 lb$ for $TAC$ and $591.82 lb$ for $TAD$ in equation (XIII) to get $P$ .

$−2021(1081.82 lb)−5059(590 lb)−5063(591.82 lb)+P=0P= 2000 lb$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $2000 lb_$ .

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Answer 2

Textbook Question

Chapter 2.5, Problem 2.112P

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at

Expert Solution & Answer

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590 lb$ .

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $2000 lb_$ .

Explanation of Solution

Free body diagram at $A$ is shown in figure1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 1

Here, $TAB$ is the magnitude of tension in cable $AB$ , $TAC$ is the magnitude of tension in cable $AC$ , $TAD$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$ .

The tension in the cable $AB$ is $840 lb$ .

The sketch of plate supported by three cables is shown in figure2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 2

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y and z$ direction.

Write the equation of vector distance $AB$ .

$AB→=(x2−x1)i+(y2−y1)j+(z2−z1)k$ (I)

Here, $AB→$ is the vector distance of the cable $AB$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x2,y2 and z2$ are the coordinates of point $B$ .

Write the vector distance of the cable $AC$ .

$AC→=(x3−x1)i+(y3−y1)j+(z3−z1)k$ (II)

Here, $AC→$ is the vector distance of the cable $AC$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x3,y3 and z3$ are the coordinates of point $C$ .

Write the vector distance of the cable $AD$ .

$AD→=(x4−x1)i+(y4−y1)j+(z4−z1)k$ (III)

Here, $AD→$ is the vector distance of the cable $AD$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x4,y4 and z4$ are the coordinates of point $D$ .

Write the equation of tension in the cable $AB$ .

$TAB=λABTAB$ (IV)

Here, $TAB$ is the tension in the cable $AB$ , $TAB$ is the magnitude of the tension in the cable $AB$ and $λAB$ is the unit vector in the direction of $AB$ .

Write the equation of $λAB$ .

$λAB=AB→AB$ (V)

Write the equation of tension in the cable $AC$ .

$TAC=λACTAC$ (VI)

Here, $TAC$ is the tension in the cable $AC$ , $TAC$ is the magnitude of the tension in the cable $AC$ and $λAC$ is the unit vector in the direction of $AC$ .

Write the equation of $λAC$ .

$λAC=AC→AC$ (VII)

Write the equation of tension in the cable $AD$ .

$TAD=λADTAD$ (VIII)

Here, $TAD$ is the tension in the cable $AD$ , $TAD$ is the magnitude of the tension in the cable $AD$ and $λAD$ is the unit vector in the direction of $AD$ .

Write the equation of $λAD$ .

$λAD=AD→AD$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=P j$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$ .

Write the equilibrium condition for the forces at $A$ .

$∑F=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$ .

$TAB+TAC+TAD+P=0$

Conclusion:

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x2$ , $0 ft$ for $y2$ and $25 ft$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(20 ft)i−(100 ft)j+(25 ft)k$

Calculate the magnitude of $AB→$ .

$AB=(−20 ft)2+(−100 ft)2+(25 ft)2= 105 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $60 ft$ for $x3$ , $0 ft$ for $y3$ and $25 ft$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(60 ft)i−(100 ft)j+(18 ft)k$

Calculate the magnitude of $AC→$ .

$AC=(60 ft)2+(−100 ft)2+(18 ft)2=118 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x4$ , $0 ft$ for $y4$ and $−74 ft$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(−20 ft)i−(100 ft)j−(74 ft)k$

Calculate the magnitude of $AD→$ .

$AD=(−20 ft)2+(−100 ft)2+(−74 ft)2=126 ft$

Substitute $−(20 ft)i−(100 ft)j+(25 ft)k$ for $AB→$ and $105 ft$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(20 ft)i−(100 ft)j+(25 ft)k105 ft=−421i−2021j+521k$

Substitute $−421i−2021j+521k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−421i−2021j+521k)TAB$

Substitute $(60 ft)i−(100 ft)j+(18 ft)k$ for $AC→$ and $118 ft$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(60 ft)i−(100 ft)j+(18 ft)k118 ft=3059i−5059j+959k$

Substitute $3059i−5059j+959k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(3059i−5059j+959k)TAC$

Substitute $(−20 ft)i−(100 ft)j−(74 ft)k$ for $AD→$ and $126 ft$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(−20 ft)i−(100 ft)j−(74 ft)k126 ft=−1063i−5063j−3763k$

Substitute $−1063i−5063j−3763k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(−1063i−5063j−3763k)TAD$

Substitute $(−421i−2021j+521k)TAB$ for $TAB$ , $(3059i−5059j+959k)TAC$ for $TAC$ , $(−1063i−5063j−3763k)TAD$ , $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−421i−2021j+521k)TAB+(3059i−5059j+959k)TAC+(−1063i−5063j−3763k)TAD+Pj=0(−421TAB+3059TAC−1063TAD)i+(−2021TAB−5059TAC−5063TAD+P)j+(+521TAB+959TAC−3763TAD)k=0$

Since total force is zero. Equate force along each direction as zero.

$−421TAB+3059TAC−1063TAD=0$ (XII)

$−2021TAB−5059TAC−5063TAD+P=0$ (XIII)

$+521TAB+959TAC−3763TAD= 0$ (XIV)

Substitute $590 lb$ for $TAC$ in equation (XII) and (XIV) to modify the equation.

$−421TAB+3059(590 lb)−1063TAD=0$

$−421TAB+300 lb−1063TAD=0$ (XV)

$+521TAB+959(590 lb)−3763TAD= 0$

$+521TAB+90 lb−3763TAD= 0$ (XVI)

Multiply equation (XV) by $5$ and XVI by $−10$ and add to get $TAC$ .

$−2021TAB+1500 lb−5063TAD+2021TAB+360 lb−14863TAD= 01860 lb−19863TAD= 0TAD= 591.82 lb$

Substitute $591.82 lb$ for $TAD$ in equation (XVI) to get $TAB$ .

$+521TAB+90 lb−3763(591.82 lb)= 0TAB= 1081.82 lb$

Substitute $1081.82 lb$ for $TAB$ , $590 lb$ for $TAC$ and $591.82 lb$ for $TAD$ in equation (XIII) to get $P$ .

$−2021(1081.82 lb)−5059(590 lb)−5063(591.82 lb)+P=0P= 2000 lb$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $2000 lb_$ .

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Answer 3

Textbook Question

Chapter 2.5, Problem 2.112P

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at

Expert Solution & Answer

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590 lb$ .

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $2000 lb_$ .

Explanation of Solution

Free body diagram at $A$ is shown in figure1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 1

Here, $TAB$ is the magnitude of tension in cable $AB$ , $TAC$ is the magnitude of tension in cable $AC$ , $TAD$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$ .

The tension in the cable $AB$ is $840 lb$ .

The sketch of plate supported by three cables is shown in figure2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 2

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y and z$ direction.

Write the equation of vector distance $AB$ .

$AB→=(x2−x1)i+(y2−y1)j+(z2−z1)k$ (I)

Here, $AB→$ is the vector distance of the cable $AB$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x2,y2 and z2$ are the coordinates of point $B$ .

Write the vector distance of the cable $AC$ .

$AC→=(x3−x1)i+(y3−y1)j+(z3−z1)k$ (II)

Here, $AC→$ is the vector distance of the cable $AC$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x3,y3 and z3$ are the coordinates of point $C$ .

Write the vector distance of the cable $AD$ .

$AD→=(x4−x1)i+(y4−y1)j+(z4−z1)k$ (III)

Here, $AD→$ is the vector distance of the cable $AD$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x4,y4 and z4$ are the coordinates of point $D$ .

Write the equation of tension in the cable $AB$ .

$TAB=λABTAB$ (IV)

Here, $TAB$ is the tension in the cable $AB$ , $TAB$ is the magnitude of the tension in the cable $AB$ and $λAB$ is the unit vector in the direction of $AB$ .

Write the equation of $λAB$ .

$λAB=AB→AB$ (V)

Write the equation of tension in the cable $AC$ .

$TAC=λACTAC$ (VI)

Here, $TAC$ is the tension in the cable $AC$ , $TAC$ is the magnitude of the tension in the cable $AC$ and $λAC$ is the unit vector in the direction of $AC$ .

Write the equation of $λAC$ .

$λAC=AC→AC$ (VII)

Write the equation of tension in the cable $AD$ .

$TAD=λADTAD$ (VIII)

Here, $TAD$ is the tension in the cable $AD$ , $TAD$ is the magnitude of the tension in the cable $AD$ and $λAD$ is the unit vector in the direction of $AD$ .

Write the equation of $λAD$ .

$λAD=AD→AD$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=P j$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$ .

Write the equilibrium condition for the forces at $A$ .

$∑F=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$ .

$TAB+TAC+TAD+P=0$

Conclusion:

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x2$ , $0 ft$ for $y2$ and $25 ft$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(20 ft)i−(100 ft)j+(25 ft)k$

Calculate the magnitude of $AB→$ .

$AB=(−20 ft)2+(−100 ft)2+(25 ft)2= 105 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $60 ft$ for $x3$ , $0 ft$ for $y3$ and $25 ft$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(60 ft)i−(100 ft)j+(18 ft)k$

Calculate the magnitude of $AC→$ .

$AC=(60 ft)2+(−100 ft)2+(18 ft)2=118 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x4$ , $0 ft$ for $y4$ and $−74 ft$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(−20 ft)i−(100 ft)j−(74 ft)k$

Calculate the magnitude of $AD→$ .

$AD=(−20 ft)2+(−100 ft)2+(−74 ft)2=126 ft$

Substitute $−(20 ft)i−(100 ft)j+(25 ft)k$ for $AB→$ and $105 ft$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(20 ft)i−(100 ft)j+(25 ft)k105 ft=−421i−2021j+521k$

Substitute $−421i−2021j+521k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−421i−2021j+521k)TAB$

Substitute $(60 ft)i−(100 ft)j+(18 ft)k$ for $AC→$ and $118 ft$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(60 ft)i−(100 ft)j+(18 ft)k118 ft=3059i−5059j+959k$

Substitute $3059i−5059j+959k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(3059i−5059j+959k)TAC$

Substitute $(−20 ft)i−(100 ft)j−(74 ft)k$ for $AD→$ and $126 ft$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(−20 ft)i−(100 ft)j−(74 ft)k126 ft=−1063i−5063j−3763k$

Substitute $−1063i−5063j−3763k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(−1063i−5063j−3763k)TAD$

Substitute $(−421i−2021j+521k)TAB$ for $TAB$ , $(3059i−5059j+959k)TAC$ for $TAC$ , $(−1063i−5063j−3763k)TAD$ , $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−421i−2021j+521k)TAB+(3059i−5059j+959k)TAC+(−1063i−5063j−3763k)TAD+Pj=0(−421TAB+3059TAC−1063TAD)i+(−2021TAB−5059TAC−5063TAD+P)j+(+521TAB+959TAC−3763TAD)k=0$

Since total force is zero. Equate force along each direction as zero.

$−421TAB+3059TAC−1063TAD=0$ (XII)

$−2021TAB−5059TAC−5063TAD+P=0$ (XIII)

$+521TAB+959TAC−3763TAD= 0$ (XIV)

Substitute $590 lb$ for $TAC$ in equation (XII) and (XIV) to modify the equation.

$−421TAB+3059(590 lb)−1063TAD=0$

$−421TAB+300 lb−1063TAD=0$ (XV)

$+521TAB+959(590 lb)−3763TAD= 0$

$+521TAB+90 lb−3763TAD= 0$ (XVI)

Multiply equation (XV) by $5$ and XVI by $−10$ and add to get $TAC$ .

$−2021TAB+1500 lb−5063TAD+2021TAB+360 lb−14863TAD= 01860 lb−19863TAD= 0TAD= 591.82 lb$

Substitute $591.82 lb$ for $TAD$ in equation (XVI) to get $TAB$ .

$+521TAB+90 lb−3763(591.82 lb)= 0TAB= 1081.82 lb$

Substitute $1081.82 lb$ for $TAB$ , $590 lb$ for $TAC$ and $591.82 lb$ for $TAD$ in equation (XIII) to get $P$ .

$−2021(1081.82 lb)−5059(590 lb)−5063(591.82 lb)+P=0P= 2000 lb$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $2000 lb_$ .

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Answer 4

Textbook Question

Chapter 2.5, Problem 2.112P

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at

Expert Solution & Answer

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590 lb$ .

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $2000 lb_$ .

Explanation of Solution

Free body diagram at $A$ is shown in figure1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 1

Here, $TAB$ is the magnitude of tension in cable $AB$ , $TAC$ is the magnitude of tension in cable $AC$ , $TAD$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$ .

The tension in the cable $AB$ is $840 lb$ .

The sketch of plate supported by three cables is shown in figure2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 2

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y and z$ direction.

Write the equation of vector distance $AB$ .

$AB→=(x2−x1)i+(y2−y1)j+(z2−z1)k$ (I)

Here, $AB→$ is the vector distance of the cable $AB$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x2,y2 and z2$ are the coordinates of point $B$ .

Write the vector distance of the cable $AC$ .

$AC→=(x3−x1)i+(y3−y1)j+(z3−z1)k$ (II)

Here, $AC→$ is the vector distance of the cable $AC$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x3,y3 and z3$ are the coordinates of point $C$ .

Write the vector distance of the cable $AD$ .

$AD→=(x4−x1)i+(y4−y1)j+(z4−z1)k$ (III)

Here, $AD→$ is the vector distance of the cable $AD$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x4,y4 and z4$ are the coordinates of point $D$ .

Write the equation of tension in the cable $AB$ .

$TAB=λABTAB$ (IV)

Here, $TAB$ is the tension in the cable $AB$ , $TAB$ is the magnitude of the tension in the cable $AB$ and $λAB$ is the unit vector in the direction of $AB$ .

Write the equation of $λAB$ .

$λAB=AB→AB$ (V)

Write the equation of tension in the cable $AC$ .

$TAC=λACTAC$ (VI)

Here, $TAC$ is the tension in the cable $AC$ , $TAC$ is the magnitude of the tension in the cable $AC$ and $λAC$ is the unit vector in the direction of $AC$ .

Write the equation of $λAC$ .

$λAC=AC→AC$ (VII)

Write the equation of tension in the cable $AD$ .

$TAD=λADTAD$ (VIII)

Here, $TAD$ is the tension in the cable $AD$ , $TAD$ is the magnitude of the tension in the cable $AD$ and $λAD$ is the unit vector in the direction of $AD$ .

Write the equation of $λAD$ .

$λAD=AD→AD$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=P j$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$ .

Write the equilibrium condition for the forces at $A$ .

$∑F=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$ .

$TAB+TAC+TAD+P=0$

Conclusion:

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x2$ , $0 ft$ for $y2$ and $25 ft$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(20 ft)i−(100 ft)j+(25 ft)k$

Calculate the magnitude of $AB→$ .

$AB=(−20 ft)2+(−100 ft)2+(25 ft)2= 105 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $60 ft$ for $x3$ , $0 ft$ for $y3$ and $25 ft$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(60 ft)i−(100 ft)j+(18 ft)k$

Calculate the magnitude of $AC→$ .

$AC=(60 ft)2+(−100 ft)2+(18 ft)2=118 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x4$ , $0 ft$ for $y4$ and $−74 ft$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(−20 ft)i−(100 ft)j−(74 ft)k$

Calculate the magnitude of $AD→$ .

$AD=(−20 ft)2+(−100 ft)2+(−74 ft)2=126 ft$

Substitute $−(20 ft)i−(100 ft)j+(25 ft)k$ for $AB→$ and $105 ft$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(20 ft)i−(100 ft)j+(25 ft)k105 ft=−421i−2021j+521k$

Substitute $−421i−2021j+521k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−421i−2021j+521k)TAB$

Substitute $(60 ft)i−(100 ft)j+(18 ft)k$ for $AC→$ and $118 ft$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(60 ft)i−(100 ft)j+(18 ft)k118 ft=3059i−5059j+959k$

Substitute $3059i−5059j+959k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(3059i−5059j+959k)TAC$

Substitute $(−20 ft)i−(100 ft)j−(74 ft)k$ for $AD→$ and $126 ft$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(−20 ft)i−(100 ft)j−(74 ft)k126 ft=−1063i−5063j−3763k$

Substitute $−1063i−5063j−3763k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(−1063i−5063j−3763k)TAD$

Substitute $(−421i−2021j+521k)TAB$ for $TAB$ , $(3059i−5059j+959k)TAC$ for $TAC$ , $(−1063i−5063j−3763k)TAD$ , $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−421i−2021j+521k)TAB+(3059i−5059j+959k)TAC+(−1063i−5063j−3763k)TAD+Pj=0(−421TAB+3059TAC−1063TAD)i+(−2021TAB−5059TAC−5063TAD+P)j+(+521TAB+959TAC−3763TAD)k=0$

Since total force is zero. Equate force along each direction as zero.

$−421TAB+3059TAC−1063TAD=0$ (XII)

$−2021TAB−5059TAC−5063TAD+P=0$ (XIII)

$+521TAB+959TAC−3763TAD= 0$ (XIV)

Substitute $590 lb$ for $TAC$ in equation (XII) and (XIV) to modify the equation.

$−421TAB+3059(590 lb)−1063TAD=0$

$−421TAB+300 lb−1063TAD=0$ (XV)

$+521TAB+959(590 lb)−3763TAD= 0$

$+521TAB+90 lb−3763TAD= 0$ (XVI)

Multiply equation (XV) by $5$ and XVI by $−10$ and add to get $TAC$ .

$−2021TAB+1500 lb−5063TAD+2021TAB+360 lb−14863TAD= 01860 lb−19863TAD= 0TAD= 591.82 lb$

Substitute $591.82 lb$ for $TAD$ in equation (XVI) to get $TAB$ .

$+521TAB+90 lb−3763(591.82 lb)= 0TAB= 1081.82 lb$

Substitute $1081.82 lb$ for $TAB$ , $590 lb$ for $TAC$ and $591.82 lb$ for $TAD$ in equation (XIII) to get $P$ .

$−2021(1081.82 lb)−5059(590 lb)−5063(591.82 lb)+P=0P= 2000 lb$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $2000 lb_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590 lb$ .

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $2000 lb_$ .

Explanation of Solution

Free body diagram at $A$ is shown in figure1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 1

Here, $TAB$ is the magnitude of tension in cable $AB$ , $TAC$ is the magnitude of tension in cable $AC$ , $TAD$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$ .

The tension in the cable $AB$ is $840 lb$ .

The sketch of plate supported by three cables is shown in figure2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 2

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y and z$ direction.

Write the equation of vector distance $AB$ .

$AB→=(x2−x1)i+(y2−y1)j+(z2−z1)k$ (I)

Here, $AB→$ is the vector distance of the cable $AB$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x2,y2 and z2$ are the coordinates of point $B$ .

Write the vector distance of the cable $AC$ .

$AC→=(x3−x1)i+(y3−y1)j+(z3−z1)k$ (II)

Here, $AC→$ is the vector distance of the cable $AC$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x3,y3 and z3$ are the coordinates of point $C$ .

Write the vector distance of the cable $AD$ .

$AD→=(x4−x1)i+(y4−y1)j+(z4−z1)k$ (III)

Here, $AD→$ is the vector distance of the cable $AD$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x4,y4 and z4$ are the coordinates of point $D$ .

Write the equation of tension in the cable $AB$ .

$TAB=λABTAB$ (IV)

Here, $TAB$ is the tension in the cable $AB$ , $TAB$ is the magnitude of the tension in the cable $AB$ and $λAB$ is the unit vector in the direction of $AB$ .

Write the equation of $λAB$ .

$λAB=AB→AB$ (V)

Write the equation of tension in the cable $AC$ .

$TAC=λACTAC$ (VI)

Here, $TAC$ is the tension in the cable $AC$ , $TAC$ is the magnitude of the tension in the cable $AC$ and $λAC$ is the unit vector in the direction of $AC$ .

Write the equation of $λAC$ .

$λAC=AC→AC$ (VII)

Write the equation of tension in the cable $AD$ .

$TAD=λADTAD$ (VIII)

Here, $TAD$ is the tension in the cable $AD$ , $TAD$ is the magnitude of the tension in the cable $AD$ and $λAD$ is the unit vector in the direction of $AD$ .

Write the equation of $λAD$ .

$λAD=AD→AD$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=P j$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$ .

Write the equilibrium condition for the forces at $A$ .

$∑F=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$ .

$TAB+TAC+TAD+P=0$

Conclusion:

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x2$ , $0 ft$ for $y2$ and $25 ft$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(20 ft)i−(100 ft)j+(25 ft)k$

Calculate the magnitude of $AB→$ .

$AB=(−20 ft)2+(−100 ft)2+(25 ft)2= 105 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $60 ft$ for $x3$ , $0 ft$ for $y3$ and $25 ft$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(60 ft)i−(100 ft)j+(18 ft)k$

Calculate the magnitude of $AC→$ .

$AC=(60 ft)2+(−100 ft)2+(18 ft)2=118 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x4$ , $0 ft$ for $y4$ and $−74 ft$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(−20 ft)i−(100 ft)j−(74 ft)k$

Calculate the magnitude of $AD→$ .

$AD=(−20 ft)2+(−100 ft)2+(−74 ft)2=126 ft$

Substitute $−(20 ft)i−(100 ft)j+(25 ft)k$ for $AB→$ and $105 ft$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(20 ft)i−(100 ft)j+(25 ft)k105 ft=−421i−2021j+521k$

Substitute $−421i−2021j+521k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−421i−2021j+521k)TAB$

Substitute $(60 ft)i−(100 ft)j+(18 ft)k$ for $AC→$ and $118 ft$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(60 ft)i−(100 ft)j+(18 ft)k118 ft=3059i−5059j+959k$

Substitute $3059i−5059j+959k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(3059i−5059j+959k)TAC$

Substitute $(−20 ft)i−(100 ft)j−(74 ft)k$ for $AD→$ and $126 ft$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(−20 ft)i−(100 ft)j−(74 ft)k126 ft=−1063i−5063j−3763k$

Substitute $−1063i−5063j−3763k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(−1063i−5063j−3763k)TAD$

Substitute $(−421i−2021j+521k)TAB$ for $TAB$ , $(3059i−5059j+959k)TAC$ for $TAC$ , $(−1063i−5063j−3763k)TAD$ , $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−421i−2021j+521k)TAB+(3059i−5059j+959k)TAC+(−1063i−5063j−3763k)TAD+Pj=0(−421TAB+3059TAC−1063TAD)i+(−2021TAB−5059TAC−5063TAD+P)j+(+521TAB+959TAC−3763TAD)k=0$

Since total force is zero. Equate force along each direction as zero.

$−421TAB+3059TAC−1063TAD=0$ (XII)

$−2021TAB−5059TAC−5063TAD+P=0$ (XIII)

$+521TAB+959TAC−3763TAD= 0$ (XIV)

Substitute $590 lb$ for $TAC$ in equation (XII) and (XIV) to modify the equation.

$−421TAB+3059(590 lb)−1063TAD=0$

$−421TAB+300 lb−1063TAD=0$ (XV)

$+521TAB+959(590 lb)−3763TAD= 0$

$+521TAB+90 lb−3763TAD= 0$ (XVI)

Multiply equation (XV) by $5$ and XVI by $−10$ and add to get $TAC$ .

$−2021TAB+1500 lb−5063TAD+2021TAB+360 lb−14863TAD= 01860 lb−19863TAD= 0TAD= 591.82 lb$

Substitute $591.82 lb$ for $TAD$ in equation (XVI) to get $TAB$ .

$+521TAB+90 lb−3763(591.82 lb)= 0TAB= 1081.82 lb$

Substitute $1081.82 lb$ for $TAB$ , $590 lb$ for $TAC$ and $591.82 lb$ for $TAD$ in equation (XIII) to get $P$ .

$−2021(1081.82 lb)−5059(590 lb)−5063(591.82 lb)+P=0P= 2000 lb$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $2000 lb_$ .

Answer 8

Expert Solution & Answer

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590 lb$ .

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $2000 lb_$ .

Explanation of Solution

Free body diagram at $A$ is shown in figure1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 1

Here, $TAB$ is the magnitude of tension in cable $AB$ , $TAC$ is the magnitude of tension in cable $AC$ , $TAD$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$ .

The tension in the cable $AB$ is $840 lb$ .

The sketch of plate supported by three cables is shown in figure2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 2

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y and z$ direction.

Write the equation of vector distance $AB$ .

$AB→=(x2−x1)i+(y2−y1)j+(z2−z1)k$ (I)

Here, $AB→$ is the vector distance of the cable $AB$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x2,y2 and z2$ are the coordinates of point $B$ .

Write the vector distance of the cable $AC$ .

$AC→=(x3−x1)i+(y3−y1)j+(z3−z1)k$ (II)

Here, $AC→$ is the vector distance of the cable $AC$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x3,y3 and z3$ are the coordinates of point $C$ .

Write the vector distance of the cable $AD$ .

$AD→=(x4−x1)i+(y4−y1)j+(z4−z1)k$ (III)

Here, $AD→$ is the vector distance of the cable $AD$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x4,y4 and z4$ are the coordinates of point $D$ .

Write the equation of tension in the cable $AB$ .

$TAB=λABTAB$ (IV)

Here, $TAB$ is the tension in the cable $AB$ , $TAB$ is the magnitude of the tension in the cable $AB$ and $λAB$ is the unit vector in the direction of $AB$ .

Write the equation of $λAB$ .

$λAB=AB→AB$ (V)

Write the equation of tension in the cable $AC$ .

$TAC=λACTAC$ (VI)

Here, $TAC$ is the tension in the cable $AC$ , $TAC$ is the magnitude of the tension in the cable $AC$ and $λAC$ is the unit vector in the direction of $AC$ .

Write the equation of $λAC$ .

$λAC=AC→AC$ (VII)

Write the equation of tension in the cable $AD$ .

$TAD=λADTAD$ (VIII)

Here, $TAD$ is the tension in the cable $AD$ , $TAD$ is the magnitude of the tension in the cable $AD$ and $λAD$ is the unit vector in the direction of $AD$ .

Write the equation of $λAD$ .

$λAD=AD→AD$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=P j$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$ .

Write the equilibrium condition for the forces at $A$ .

$∑F=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$ .

$TAB+TAC+TAD+P=0$

Conclusion:

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x2$ , $0 ft$ for $y2$ and $25 ft$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(20 ft)i−(100 ft)j+(25 ft)k$

Calculate the magnitude of $AB→$ .

$AB=(−20 ft)2+(−100 ft)2+(25 ft)2= 105 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $60 ft$ for $x3$ , $0 ft$ for $y3$ and $25 ft$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(60 ft)i−(100 ft)j+(18 ft)k$

Calculate the magnitude of $AC→$ .

$AC=(60 ft)2+(−100 ft)2+(18 ft)2=118 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x4$ , $0 ft$ for $y4$ and $−74 ft$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(−20 ft)i−(100 ft)j−(74 ft)k$

Calculate the magnitude of $AD→$ .

$AD=(−20 ft)2+(−100 ft)2+(−74 ft)2=126 ft$

Substitute $−(20 ft)i−(100 ft)j+(25 ft)k$ for $AB→$ and $105 ft$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(20 ft)i−(100 ft)j+(25 ft)k105 ft=−421i−2021j+521k$

Substitute $−421i−2021j+521k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−421i−2021j+521k)TAB$

Substitute $(60 ft)i−(100 ft)j+(18 ft)k$ for $AC→$ and $118 ft$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(60 ft)i−(100 ft)j+(18 ft)k118 ft=3059i−5059j+959k$

Substitute $3059i−5059j+959k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(3059i−5059j+959k)TAC$

Substitute $(−20 ft)i−(100 ft)j−(74 ft)k$ for $AD→$ and $126 ft$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(−20 ft)i−(100 ft)j−(74 ft)k126 ft=−1063i−5063j−3763k$

Substitute $−1063i−5063j−3763k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(−1063i−5063j−3763k)TAD$

Substitute $(−421i−2021j+521k)TAB$ for $TAB$ , $(3059i−5059j+959k)TAC$ for $TAC$ , $(−1063i−5063j−3763k)TAD$ , $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−421i−2021j+521k)TAB+(3059i−5059j+959k)TAC+(−1063i−5063j−3763k)TAD+Pj=0(−421TAB+3059TAC−1063TAD)i+(−2021TAB−5059TAC−5063TAD+P)j+(+521TAB+959TAC−3763TAD)k=0$

Since total force is zero. Equate force along each direction as zero.

$−421TAB+3059TAC−1063TAD=0$ (XII)

$−2021TAB−5059TAC−5063TAD+P=0$ (XIII)

$+521TAB+959TAC−3763TAD= 0$ (XIV)

Substitute $590 lb$ for $TAC$ in equation (XII) and (XIV) to modify the equation.

$−421TAB+3059(590 lb)−1063TAD=0$

$−421TAB+300 lb−1063TAD=0$ (XV)

$+521TAB+959(590 lb)−3763TAD= 0$

$+521TAB+90 lb−3763TAD= 0$ (XVI)

Multiply equation (XV) by $5$ and XVI by $−10$ and add to get $TAC$ .

$−2021TAB+1500 lb−5063TAD+2021TAB+360 lb−14863TAD= 01860 lb−19863TAD= 0TAD= 591.82 lb$

Substitute $591.82 lb$ for $TAD$ in equation (XVI) to get $TAB$ .

$+521TAB+90 lb−3763(591.82 lb)= 0TAB= 1081.82 lb$

Substitute $1081.82 lb$ for $TAB$ , $590 lb$ for $TAC$ and $591.82 lb$ for $TAD$ in equation (XIII) to get $P$ .

$−2021(1081.82 lb)−5059(590 lb)−5063(591.82 lb)+P=0P= 2000 lb$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $2000 lb_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590 lb$ .

Answer 12

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $2000 lb_$ .

Answer 13

Explanation of Solution

Free body diagram at $A$ is shown in figure1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 1

Here, $TAB$ is the magnitude of tension in cable $AB$ , $TAC$ is the magnitude of tension in cable $AC$ , $TAD$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$ .

The tension in the cable $AB$ is $840 lb$ .

The sketch of plate supported by three cables is shown in figure2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2.5, Problem 2.112P , additional homework tip 2

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y and z$ direction.

Write the equation of vector distance $AB$ .

$AB→=(x2−x1)i+(y2−y1)j+(z2−z1)k$ (I)

Here, $AB→$ is the vector distance of the cable $AB$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x2,y2 and z2$ are the coordinates of point $B$ .

Write the vector distance of the cable $AC$ .

$AC→=(x3−x1)i+(y3−y1)j+(z3−z1)k$ (II)

Here, $AC→$ is the vector distance of the cable $AC$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x3,y3 and z3$ are the coordinates of point $C$ .

Write the vector distance of the cable $AD$ .

$AD→=(x4−x1)i+(y4−y1)j+(z4−z1)k$ (III)

Here, $AD→$ is the vector distance of the cable $AD$ , the variables $x1,y1 and z1$ are the coordinates of point $A$ and $x4,y4 and z4$ are the coordinates of point $D$ .

Write the equation of tension in the cable $AB$ .

$TAB=λABTAB$ (IV)

Here, $TAB$ is the tension in the cable $AB$ , $TAB$ is the magnitude of the tension in the cable $AB$ and $λAB$ is the unit vector in the direction of $AB$ .

Write the equation of $λAB$ .

$λAB=AB→AB$ (V)

Write the equation of tension in the cable $AC$ .

$TAC=λACTAC$ (VI)

Here, $TAC$ is the tension in the cable $AC$ , $TAC$ is the magnitude of the tension in the cable $AC$ and $λAC$ is the unit vector in the direction of $AC$ .

Write the equation of $λAC$ .

$λAC=AC→AC$ (VII)

Write the equation of tension in the cable $AD$ .

$TAD=λADTAD$ (VIII)

Here, $TAD$ is the tension in the cable $AD$ , $TAD$ is the magnitude of the tension in the cable $AD$ and $λAD$ is the unit vector in the direction of $AD$ .

Write the equation of $λAD$ .

$λAD=AD→AD$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=P j$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$ .

Write the equilibrium condition for the forces at $A$ .

$∑F=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$ .

$TAB+TAC+TAD+P=0$

Conclusion:

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x2$ , $0 ft$ for $y2$ and $25 ft$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(20 ft)i−(100 ft)j+(25 ft)k$

Calculate the magnitude of $AB→$ .

$AB=(−20 ft)2+(−100 ft)2+(25 ft)2= 105 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $60 ft$ for $x3$ , $0 ft$ for $y3$ and $25 ft$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(60 ft)i−(100 ft)j+(18 ft)k$

Calculate the magnitude of $AC→$ .

$AC=(60 ft)2+(−100 ft)2+(18 ft)2=118 ft$

Substitute $0 ft$ for $x1$ , $100 ft$ for $y1$ , $0 ft$ for $z1$ , $−20 ft$ for $x4$ , $0 ft$ for $y4$ and $−74 ft$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(−20 ft)i−(100 ft)j−(74 ft)k$

Calculate the magnitude of $AD→$ .

$AD=(−20 ft)2+(−100 ft)2+(−74 ft)2=126 ft$

Substitute $−(20 ft)i−(100 ft)j+(25 ft)k$ for $AB→$ and $105 ft$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(20 ft)i−(100 ft)j+(25 ft)k105 ft=−421i−2021j+521k$

Substitute $−421i−2021j+521k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−421i−2021j+521k)TAB$

Substitute $(60 ft)i−(100 ft)j+(18 ft)k$ for $AC→$ and $118 ft$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(60 ft)i−(100 ft)j+(18 ft)k118 ft=3059i−5059j+959k$

Substitute $3059i−5059j+959k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(3059i−5059j+959k)TAC$

Substitute $(−20 ft)i−(100 ft)j−(74 ft)k$ for $AD→$ and $126 ft$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(−20 ft)i−(100 ft)j−(74 ft)k126 ft=−1063i−5063j−3763k$

Substitute $−1063i−5063j−3763k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(−1063i−5063j−3763k)TAD$

Substitute $(−421i−2021j+521k)TAB$ for $TAB$ , $(3059i−5059j+959k)TAC$ for $TAC$ , $(−1063i−5063j−3763k)TAD$ , $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−421i−2021j+521k)TAB+(3059i−5059j+959k)TAC+(−1063i−5063j−3763k)TAD+Pj=0(−421TAB+3059TAC−1063TAD)i+(−2021TAB−5059TAC−5063TAD+P)j+(+521TAB+959TAC−3763TAD)k=0$

Since total force is zero. Equate force along each direction as zero.

$−421TAB+3059TAC−1063TAD=0$ (XII)

$−2021TAB−5059TAC−5063TAD+P=0$ (XIII)

$+521TAB+959TAC−3763TAD= 0$ (XIV)

Substitute $590 lb$ for $TAC$ in equation (XII) and (XIV) to modify the equation.

$−421TAB+3059(590 lb)−1063TAD=0$

$−421TAB+300 lb−1063TAD=0$ (XV)

$+521TAB+959(590 lb)−3763TAD= 0$

$+521TAB+90 lb−3763TAD= 0$ (XVI)

Multiply equation (XV) by $5$ and XVI by $−10$ and add to get $TAC$ .

$−2021TAB+1500 lb−5063TAD+2021TAB+360 lb−14863TAD= 01860 lb−19863TAD= 0TAD= 591.82 lb$

Substitute $591.82 lb$ for $TAD$ in equation (XVI) to get $TAB$ .

$+521TAB+90 lb−3763(591.82 lb)= 0TAB= 1081.82 lb$

Substitute $1081.82 lb$ for $TAB$ , $590 lb$ for $TAC$ and $591.82 lb$ for $TAD$ in equation (XIII) to get $P$ .

$−2021(1081.82 lb)−5059(590 lb)−5063(591.82 lb)+P=0P= 2000 lb$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $2000 lb_$ .

Answer 14

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A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at

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Answer to Problem 2.112P

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