For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. Fig. P2.109 and P2.110

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Answer 1

Textbook Question

Chapter 2.5, Problem 2.115P

For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

Chapter 2.5, Problem 2.115P, For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three

Fig. P2.109 and P2.110

Expert Solution & Answer

To determine

The tension in three cables shown in figure $P2.109$ , knowing that weight of the plate is $792 N$ .

Answer to Problem 2.115P

The tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Explanation of Solution

The sketch of plate supported by three cables is shown in figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 1

Free body diagram at $A$ is shown in figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 2

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 support on point $A$ which is equal to the weight of the rectangular plate.

The weight of the plate is $792 N$ .

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted at $A$ in the upward direction due to the weight of the plate.

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 $A$ in the upward direction due to the weight of the plate.

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 $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $320 mm$ for $x2$ , $0 mm$ for $y2$ and $360 mm$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(320 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AB→$ .

$AB=(−320 mm)2+(−480 mm)2+(360 mm)2= 680 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $450 mm$ for $x3$ , $0 mm$ for $y3$ and $360 mm$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(450 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AC→$ .

$AC=(450 mm)2+(−480 mm)2+(360 mm)2=750 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $250 mm$ for $x4$ , $0 mm$ for $y4$ and $−360 mm$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(250 mm)i−(480 mm)j−(360 mm)k$

Calculate the magnitude of $AD→$ .

$AD=(250 mm)2−(480 mm)2−(360 mm)2=650 mm$

Substitute $AB→=−(320 mm)i−(480 mm)j+(360 mm)k$ for $AB→$ and $680 mm$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(320 mm)i−(480 mm)j+(360 mm)k680 mm=−817i−1217j+1917k$

Substitute $−817i−1217j+1917k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−817i−1217j+1917k)TAB$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AC→$ and $7.40 mm$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(250 mm)i−(480 mm)j−(360 mm)k750 mm=0.6i−0.64j+0.48k$

Substitute $0.6i−0.64j+0.48k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(0.6i−0.64j+0.48k)TAC$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AD→$ and $650 mm$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(250 mm)i−(480 mm)j−(360 mm)k650 mm=513i−9.613j−7.213k$

Substitute $513i−9.613j−7.213k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(513i−9.613j−7.213k)TAD$

Substitute $(−817i−1217j+1917k)TAB$ for $TAB$ , $(0.6i−0.64j+0.48k)TAC$ for $TAC$ , $(513i−9.613j−7.213k)TAD$ , $792j$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−817i−1217j+917k)TAB+(0.6i−0.64j+0.48k)TAC+(513i−9.613j−7.213k)TAD+792j=0(−817TAB+0.6TAC+513TAD)i+(−1217TAB−0.64TAC+−9.613TAD+792)j+(+917TAB+0.48TAC−7.213TAD)k=0$

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

$−817TAB+0.6TAC+513TAD=0$ (XII)

$−1217TAB−0.64TAC+−9.613TAD+792=0$ (XIII)

$+917TAB+0.48TAC−7.213TAD= 0$ (XIV)

Multiply equation (XII) with 12 and equation (XIV) with $−9$ and add to get new equation.

$−9617TAB+7.2TAC+6013TAD+9617TAB+5.12TAC+76.813TAD−6336=0$

$12.32TAC+136.813TAD−6336=0$ (XV)

Multiply equation (XII) with $9$ and equation (XIV) with $8$ and add to get new equation.

$−7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC−57.613TAD= 05.4TAC+4513TAD+3.84TAC−57.613TAD= 0$
$9.24TAC−12.613TAD=0$ (XVI)

Multiply equation (XV) with $12.6$ and (XVI) with $136.8$ to get $TAC$ .

$155.232TAC+1723.6813TAD−79833.6 N+1264.032TAC−1723.6813TAD=0$

$(155.232+1264.032)TAC=79833.6 NTAC=56.25 N$

Substitute $56.2 N$ for $TAC$ in equation (XVI) to get $TAD$ .

$9.24(56.25 N)−12.613TAD=0TAD= 536.25 N$

Substitute $56.25 N$ for $TAC$ and $536.25 N$ for $TAD$ in equation (XII) to get $TAB$ .

$−817TAB+0.6(56.25 N)+513( 536.25 N )=0$

$TAB= 510 N$

Therefore, the tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

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Answer 2

Textbook Question

Chapter 2.5, Problem 2.115P

For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

Chapter 2.5, Problem 2.115P, For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three

Fig. P2.109 and P2.110

Expert Solution & Answer

To determine

The tension in three cables shown in figure $P2.109$ , knowing that weight of the plate is $792 N$ .

Answer to Problem 2.115P

The tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Explanation of Solution

The sketch of plate supported by three cables is shown in figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 1

Free body diagram at $A$ is shown in figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 2

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 support on point $A$ which is equal to the weight of the rectangular plate.

The weight of the plate is $792 N$ .

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted at $A$ in the upward direction due to the weight of the plate.

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 $A$ in the upward direction due to the weight of the plate.

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 $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $320 mm$ for $x2$ , $0 mm$ for $y2$ and $360 mm$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(320 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AB→$ .

$AB=(−320 mm)2+(−480 mm)2+(360 mm)2= 680 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $450 mm$ for $x3$ , $0 mm$ for $y3$ and $360 mm$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(450 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AC→$ .

$AC=(450 mm)2+(−480 mm)2+(360 mm)2=750 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $250 mm$ for $x4$ , $0 mm$ for $y4$ and $−360 mm$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(250 mm)i−(480 mm)j−(360 mm)k$

Calculate the magnitude of $AD→$ .

$AD=(250 mm)2−(480 mm)2−(360 mm)2=650 mm$

Substitute $AB→=−(320 mm)i−(480 mm)j+(360 mm)k$ for $AB→$ and $680 mm$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(320 mm)i−(480 mm)j+(360 mm)k680 mm=−817i−1217j+1917k$

Substitute $−817i−1217j+1917k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−817i−1217j+1917k)TAB$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AC→$ and $7.40 mm$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(250 mm)i−(480 mm)j−(360 mm)k750 mm=0.6i−0.64j+0.48k$

Substitute $0.6i−0.64j+0.48k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(0.6i−0.64j+0.48k)TAC$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AD→$ and $650 mm$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(250 mm)i−(480 mm)j−(360 mm)k650 mm=513i−9.613j−7.213k$

Substitute $513i−9.613j−7.213k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(513i−9.613j−7.213k)TAD$

Substitute $(−817i−1217j+1917k)TAB$ for $TAB$ , $(0.6i−0.64j+0.48k)TAC$ for $TAC$ , $(513i−9.613j−7.213k)TAD$ , $792j$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−817i−1217j+917k)TAB+(0.6i−0.64j+0.48k)TAC+(513i−9.613j−7.213k)TAD+792j=0(−817TAB+0.6TAC+513TAD)i+(−1217TAB−0.64TAC+−9.613TAD+792)j+(+917TAB+0.48TAC−7.213TAD)k=0$

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

$−817TAB+0.6TAC+513TAD=0$ (XII)

$−1217TAB−0.64TAC+−9.613TAD+792=0$ (XIII)

$+917TAB+0.48TAC−7.213TAD= 0$ (XIV)

Multiply equation (XII) with 12 and equation (XIV) with $−9$ and add to get new equation.

$−9617TAB+7.2TAC+6013TAD+9617TAB+5.12TAC+76.813TAD−6336=0$

$12.32TAC+136.813TAD−6336=0$ (XV)

Multiply equation (XII) with $9$ and equation (XIV) with $8$ and add to get new equation.

$−7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC−57.613TAD= 05.4TAC+4513TAD+3.84TAC−57.613TAD= 0$
$9.24TAC−12.613TAD=0$ (XVI)

Multiply equation (XV) with $12.6$ and (XVI) with $136.8$ to get $TAC$ .

$155.232TAC+1723.6813TAD−79833.6 N+1264.032TAC−1723.6813TAD=0$

$(155.232+1264.032)TAC=79833.6 NTAC=56.25 N$

Substitute $56.2 N$ for $TAC$ in equation (XVI) to get $TAD$ .

$9.24(56.25 N)−12.613TAD=0TAD= 536.25 N$

Substitute $56.25 N$ for $TAC$ and $536.25 N$ for $TAD$ in equation (XII) to get $TAB$ .

$−817TAB+0.6(56.25 N)+513( 536.25 N )=0$

$TAB= 510 N$

Therefore, the tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

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Answer 3

Textbook Question

Chapter 2.5, Problem 2.115P

For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

Chapter 2.5, Problem 2.115P, For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three

Fig. P2.109 and P2.110

Expert Solution & Answer

To determine

The tension in three cables shown in figure $P2.109$ , knowing that weight of the plate is $792 N$ .

Answer to Problem 2.115P

The tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Explanation of Solution

The sketch of plate supported by three cables is shown in figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 1

Free body diagram at $A$ is shown in figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 2

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 support on point $A$ which is equal to the weight of the rectangular plate.

The weight of the plate is $792 N$ .

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted at $A$ in the upward direction due to the weight of the plate.

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 $A$ in the upward direction due to the weight of the plate.

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 $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $320 mm$ for $x2$ , $0 mm$ for $y2$ and $360 mm$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(320 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AB→$ .

$AB=(−320 mm)2+(−480 mm)2+(360 mm)2= 680 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $450 mm$ for $x3$ , $0 mm$ for $y3$ and $360 mm$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(450 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AC→$ .

$AC=(450 mm)2+(−480 mm)2+(360 mm)2=750 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $250 mm$ for $x4$ , $0 mm$ for $y4$ and $−360 mm$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(250 mm)i−(480 mm)j−(360 mm)k$

Calculate the magnitude of $AD→$ .

$AD=(250 mm)2−(480 mm)2−(360 mm)2=650 mm$

Substitute $AB→=−(320 mm)i−(480 mm)j+(360 mm)k$ for $AB→$ and $680 mm$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(320 mm)i−(480 mm)j+(360 mm)k680 mm=−817i−1217j+1917k$

Substitute $−817i−1217j+1917k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−817i−1217j+1917k)TAB$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AC→$ and $7.40 mm$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(250 mm)i−(480 mm)j−(360 mm)k750 mm=0.6i−0.64j+0.48k$

Substitute $0.6i−0.64j+0.48k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(0.6i−0.64j+0.48k)TAC$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AD→$ and $650 mm$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(250 mm)i−(480 mm)j−(360 mm)k650 mm=513i−9.613j−7.213k$

Substitute $513i−9.613j−7.213k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(513i−9.613j−7.213k)TAD$

Substitute $(−817i−1217j+1917k)TAB$ for $TAB$ , $(0.6i−0.64j+0.48k)TAC$ for $TAC$ , $(513i−9.613j−7.213k)TAD$ , $792j$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−817i−1217j+917k)TAB+(0.6i−0.64j+0.48k)TAC+(513i−9.613j−7.213k)TAD+792j=0(−817TAB+0.6TAC+513TAD)i+(−1217TAB−0.64TAC+−9.613TAD+792)j+(+917TAB+0.48TAC−7.213TAD)k=0$

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

$−817TAB+0.6TAC+513TAD=0$ (XII)

$−1217TAB−0.64TAC+−9.613TAD+792=0$ (XIII)

$+917TAB+0.48TAC−7.213TAD= 0$ (XIV)

Multiply equation (XII) with 12 and equation (XIV) with $−9$ and add to get new equation.

$−9617TAB+7.2TAC+6013TAD+9617TAB+5.12TAC+76.813TAD−6336=0$

$12.32TAC+136.813TAD−6336=0$ (XV)

Multiply equation (XII) with $9$ and equation (XIV) with $8$ and add to get new equation.

$−7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC−57.613TAD= 05.4TAC+4513TAD+3.84TAC−57.613TAD= 0$
$9.24TAC−12.613TAD=0$ (XVI)

Multiply equation (XV) with $12.6$ and (XVI) with $136.8$ to get $TAC$ .

$155.232TAC+1723.6813TAD−79833.6 N+1264.032TAC−1723.6813TAD=0$

$(155.232+1264.032)TAC=79833.6 NTAC=56.25 N$

Substitute $56.2 N$ for $TAC$ in equation (XVI) to get $TAD$ .

$9.24(56.25 N)−12.613TAD=0TAD= 536.25 N$

Substitute $56.25 N$ for $TAC$ and $536.25 N$ for $TAD$ in equation (XII) to get $TAB$ .

$−817TAB+0.6(56.25 N)+513( 536.25 N )=0$

$TAB= 510 N$

Therefore, the tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

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Answer 4

Textbook Question

Chapter 2.5, Problem 2.115P

For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

Chapter 2.5, Problem 2.115P, For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three

Fig. P2.109 and P2.110

Expert Solution & Answer

To determine

The tension in three cables shown in figure $P2.109$ , knowing that weight of the plate is $792 N$ .

Answer to Problem 2.115P

The tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Explanation of Solution

The sketch of plate supported by three cables is shown in figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 1

Free body diagram at $A$ is shown in figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 2

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 support on point $A$ which is equal to the weight of the rectangular plate.

The weight of the plate is $792 N$ .

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted at $A$ in the upward direction due to the weight of the plate.

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 $A$ in the upward direction due to the weight of the plate.

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 $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $320 mm$ for $x2$ , $0 mm$ for $y2$ and $360 mm$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(320 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AB→$ .

$AB=(−320 mm)2+(−480 mm)2+(360 mm)2= 680 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $450 mm$ for $x3$ , $0 mm$ for $y3$ and $360 mm$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(450 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AC→$ .

$AC=(450 mm)2+(−480 mm)2+(360 mm)2=750 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $250 mm$ for $x4$ , $0 mm$ for $y4$ and $−360 mm$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(250 mm)i−(480 mm)j−(360 mm)k$

Calculate the magnitude of $AD→$ .

$AD=(250 mm)2−(480 mm)2−(360 mm)2=650 mm$

Substitute $AB→=−(320 mm)i−(480 mm)j+(360 mm)k$ for $AB→$ and $680 mm$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(320 mm)i−(480 mm)j+(360 mm)k680 mm=−817i−1217j+1917k$

Substitute $−817i−1217j+1917k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−817i−1217j+1917k)TAB$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AC→$ and $7.40 mm$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(250 mm)i−(480 mm)j−(360 mm)k750 mm=0.6i−0.64j+0.48k$

Substitute $0.6i−0.64j+0.48k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(0.6i−0.64j+0.48k)TAC$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AD→$ and $650 mm$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(250 mm)i−(480 mm)j−(360 mm)k650 mm=513i−9.613j−7.213k$

Substitute $513i−9.613j−7.213k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(513i−9.613j−7.213k)TAD$

Substitute $(−817i−1217j+1917k)TAB$ for $TAB$ , $(0.6i−0.64j+0.48k)TAC$ for $TAC$ , $(513i−9.613j−7.213k)TAD$ , $792j$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−817i−1217j+917k)TAB+(0.6i−0.64j+0.48k)TAC+(513i−9.613j−7.213k)TAD+792j=0(−817TAB+0.6TAC+513TAD)i+(−1217TAB−0.64TAC+−9.613TAD+792)j+(+917TAB+0.48TAC−7.213TAD)k=0$

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

$−817TAB+0.6TAC+513TAD=0$ (XII)

$−1217TAB−0.64TAC+−9.613TAD+792=0$ (XIII)

$+917TAB+0.48TAC−7.213TAD= 0$ (XIV)

Multiply equation (XII) with 12 and equation (XIV) with $−9$ and add to get new equation.

$−9617TAB+7.2TAC+6013TAD+9617TAB+5.12TAC+76.813TAD−6336=0$

$12.32TAC+136.813TAD−6336=0$ (XV)

Multiply equation (XII) with $9$ and equation (XIV) with $8$ and add to get new equation.

$−7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC−57.613TAD= 05.4TAC+4513TAD+3.84TAC−57.613TAD= 0$
$9.24TAC−12.613TAD=0$ (XVI)

Multiply equation (XV) with $12.6$ and (XVI) with $136.8$ to get $TAC$ .

$155.232TAC+1723.6813TAD−79833.6 N+1264.032TAC−1723.6813TAD=0$

$(155.232+1264.032)TAC=79833.6 NTAC=56.25 N$

Substitute $56.2 N$ for $TAC$ in equation (XVI) to get $TAD$ .

$9.24(56.25 N)−12.613TAD=0TAD= 536.25 N$

Substitute $56.25 N$ for $TAC$ and $536.25 N$ for $TAD$ in equation (XII) to get $TAB$ .

$−817TAB+0.6(56.25 N)+513( 536.25 N )=0$

$TAB= 510 N$

Therefore, the tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The tension in three cables shown in figure $P2.109$ , knowing that weight of the plate is $792 N$ .

Answer to Problem 2.115P

The tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Explanation of Solution

The sketch of plate supported by three cables is shown in figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 1

Free body diagram at $A$ is shown in figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 2

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 support on point $A$ which is equal to the weight of the rectangular plate.

The weight of the plate is $792 N$ .

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted at $A$ in the upward direction due to the weight of the plate.

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 $A$ in the upward direction due to the weight of the plate.

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 $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $320 mm$ for $x2$ , $0 mm$ for $y2$ and $360 mm$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(320 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AB→$ .

$AB=(−320 mm)2+(−480 mm)2+(360 mm)2= 680 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $450 mm$ for $x3$ , $0 mm$ for $y3$ and $360 mm$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(450 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AC→$ .

$AC=(450 mm)2+(−480 mm)2+(360 mm)2=750 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $250 mm$ for $x4$ , $0 mm$ for $y4$ and $−360 mm$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(250 mm)i−(480 mm)j−(360 mm)k$

Calculate the magnitude of $AD→$ .

$AD=(250 mm)2−(480 mm)2−(360 mm)2=650 mm$

Substitute $AB→=−(320 mm)i−(480 mm)j+(360 mm)k$ for $AB→$ and $680 mm$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(320 mm)i−(480 mm)j+(360 mm)k680 mm=−817i−1217j+1917k$

Substitute $−817i−1217j+1917k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−817i−1217j+1917k)TAB$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AC→$ and $7.40 mm$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(250 mm)i−(480 mm)j−(360 mm)k750 mm=0.6i−0.64j+0.48k$

Substitute $0.6i−0.64j+0.48k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(0.6i−0.64j+0.48k)TAC$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AD→$ and $650 mm$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(250 mm)i−(480 mm)j−(360 mm)k650 mm=513i−9.613j−7.213k$

Substitute $513i−9.613j−7.213k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(513i−9.613j−7.213k)TAD$

Substitute $(−817i−1217j+1917k)TAB$ for $TAB$ , $(0.6i−0.64j+0.48k)TAC$ for $TAC$ , $(513i−9.613j−7.213k)TAD$ , $792j$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−817i−1217j+917k)TAB+(0.6i−0.64j+0.48k)TAC+(513i−9.613j−7.213k)TAD+792j=0(−817TAB+0.6TAC+513TAD)i+(−1217TAB−0.64TAC+−9.613TAD+792)j+(+917TAB+0.48TAC−7.213TAD)k=0$

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

$−817TAB+0.6TAC+513TAD=0$ (XII)

$−1217TAB−0.64TAC+−9.613TAD+792=0$ (XIII)

$+917TAB+0.48TAC−7.213TAD= 0$ (XIV)

Multiply equation (XII) with 12 and equation (XIV) with $−9$ and add to get new equation.

$−9617TAB+7.2TAC+6013TAD+9617TAB+5.12TAC+76.813TAD−6336=0$

$12.32TAC+136.813TAD−6336=0$ (XV)

Multiply equation (XII) with $9$ and equation (XIV) with $8$ and add to get new equation.

$−7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC−57.613TAD= 05.4TAC+4513TAD+3.84TAC−57.613TAD= 0$
$9.24TAC−12.613TAD=0$ (XVI)

Multiply equation (XV) with $12.6$ and (XVI) with $136.8$ to get $TAC$ .

$155.232TAC+1723.6813TAD−79833.6 N+1264.032TAC−1723.6813TAD=0$

$(155.232+1264.032)TAC=79833.6 NTAC=56.25 N$

Substitute $56.2 N$ for $TAC$ in equation (XVI) to get $TAD$ .

$9.24(56.25 N)−12.613TAD=0TAD= 536.25 N$

Substitute $56.25 N$ for $TAC$ and $536.25 N$ for $TAD$ in equation (XII) to get $TAB$ .

$−817TAB+0.6(56.25 N)+513( 536.25 N )=0$

$TAB= 510 N$

Therefore, the tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Answer 8

Expert Solution & Answer

To determine

The tension in three cables shown in figure $P2.109$ , knowing that weight of the plate is $792 N$ .

Answer to Problem 2.115P

The tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Explanation of Solution

The sketch of plate supported by three cables is shown in figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 1

Free body diagram at $A$ is shown in figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 2

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 support on point $A$ which is equal to the weight of the rectangular plate.

The weight of the plate is $792 N$ .

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted at $A$ in the upward direction due to the weight of the plate.

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 $A$ in the upward direction due to the weight of the plate.

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 $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $320 mm$ for $x2$ , $0 mm$ for $y2$ and $360 mm$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(320 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AB→$ .

$AB=(−320 mm)2+(−480 mm)2+(360 mm)2= 680 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $450 mm$ for $x3$ , $0 mm$ for $y3$ and $360 mm$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(450 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AC→$ .

$AC=(450 mm)2+(−480 mm)2+(360 mm)2=750 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $250 mm$ for $x4$ , $0 mm$ for $y4$ and $−360 mm$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(250 mm)i−(480 mm)j−(360 mm)k$

Calculate the magnitude of $AD→$ .

$AD=(250 mm)2−(480 mm)2−(360 mm)2=650 mm$

Substitute $AB→=−(320 mm)i−(480 mm)j+(360 mm)k$ for $AB→$ and $680 mm$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(320 mm)i−(480 mm)j+(360 mm)k680 mm=−817i−1217j+1917k$

Substitute $−817i−1217j+1917k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−817i−1217j+1917k)TAB$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AC→$ and $7.40 mm$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(250 mm)i−(480 mm)j−(360 mm)k750 mm=0.6i−0.64j+0.48k$

Substitute $0.6i−0.64j+0.48k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(0.6i−0.64j+0.48k)TAC$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AD→$ and $650 mm$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(250 mm)i−(480 mm)j−(360 mm)k650 mm=513i−9.613j−7.213k$

Substitute $513i−9.613j−7.213k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(513i−9.613j−7.213k)TAD$

Substitute $(−817i−1217j+1917k)TAB$ for $TAB$ , $(0.6i−0.64j+0.48k)TAC$ for $TAC$ , $(513i−9.613j−7.213k)TAD$ , $792j$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−817i−1217j+917k)TAB+(0.6i−0.64j+0.48k)TAC+(513i−9.613j−7.213k)TAD+792j=0(−817TAB+0.6TAC+513TAD)i+(−1217TAB−0.64TAC+−9.613TAD+792)j+(+917TAB+0.48TAC−7.213TAD)k=0$

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

$−817TAB+0.6TAC+513TAD=0$ (XII)

$−1217TAB−0.64TAC+−9.613TAD+792=0$ (XIII)

$+917TAB+0.48TAC−7.213TAD= 0$ (XIV)

Multiply equation (XII) with 12 and equation (XIV) with $−9$ and add to get new equation.

$−9617TAB+7.2TAC+6013TAD+9617TAB+5.12TAC+76.813TAD−6336=0$

$12.32TAC+136.813TAD−6336=0$ (XV)

Multiply equation (XII) with $9$ and equation (XIV) with $8$ and add to get new equation.

$−7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC−57.613TAD= 05.4TAC+4513TAD+3.84TAC−57.613TAD= 0$
$9.24TAC−12.613TAD=0$ (XVI)

Multiply equation (XV) with $12.6$ and (XVI) with $136.8$ to get $TAC$ .

$155.232TAC+1723.6813TAD−79833.6 N+1264.032TAC−1723.6813TAD=0$

$(155.232+1264.032)TAC=79833.6 NTAC=56.25 N$

Substitute $56.2 N$ for $TAC$ in equation (XVI) to get $TAD$ .

$9.24(56.25 N)−12.613TAD=0TAD= 536.25 N$

Substitute $56.25 N$ for $TAC$ and $536.25 N$ for $TAD$ in equation (XII) to get $TAB$ .

$−817TAB+0.6(56.25 N)+513( 536.25 N )=0$

$TAB= 510 N$

Therefore, the tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The tension in three cables shown in figure $P2.109$ , knowing that weight of the plate is $792 N$ .

Answer 12

Answer to Problem 2.115P

The tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Answer 13

Explanation of Solution

The sketch of plate supported by three cables is shown in figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 1

Free body diagram at $A$ is shown in figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 2.5, Problem 2.115P , additional homework tip 2

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 support on point $A$ which is equal to the weight of the rectangular plate.

The weight of the plate is $792 N$ .

Let $TAB$ , $TAC$ , $TAD$ and $P$ are the tension vector in cable $AB$ , $AC$ , $AD$ and force exerted at $A$ in the upward direction due to the weight of the plate.

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 $A$ in the upward direction due to the weight of the plate.

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 $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $320 mm$ for $x2$ , $0 mm$ for $y2$ and $360 mm$ for $z2$ in equation (I) to get $AB→$ .

$AB→=−(320 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AB→$ .

$AB=(−320 mm)2+(−480 mm)2+(360 mm)2= 680 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $450 mm$ for $x3$ , $0 mm$ for $y3$ and $360 mm$ for $z3$ in equation (II) to get $AC→$ .

$AC→=(450 mm)i−(480 mm)j+(360 mm)k$

Calculate the magnitude of $AC→$ .

$AC=(450 mm)2+(−480 mm)2+(360 mm)2=750 mm$

Substitute $0 mm$ for $x1$ , $480 mm$ for $y1$ , $0 mm$ for $z1$ , $250 mm$ for $x4$ , $0 mm$ for $y4$ and $−360 mm$ for $z4$ in equation (III) to get $AD→$ .

$AD→=(250 mm)i−(480 mm)j−(360 mm)k$

Calculate the magnitude of $AD→$ .

$AD=(250 mm)2−(480 mm)2−(360 mm)2=650 mm$

Substitute $AB→=−(320 mm)i−(480 mm)j+(360 mm)k$ for $AB→$ and $680 mm$ for $AB$ in equation (V) to get $λAB$ .

$λAB=−(320 mm)i−(480 mm)j+(360 mm)k680 mm=−817i−1217j+1917k$

Substitute $−817i−1217j+1917k$ for $λAB$ in equation (IV) to get $TAB$ .

$TAB=(−817i−1217j+1917k)TAB$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AC→$ and $7.40 mm$ for $AC$ in equation
(VII) to get $λAC$ .

$λAC=(250 mm)i−(480 mm)j−(360 mm)k750 mm=0.6i−0.64j+0.48k$

Substitute $0.6i−0.64j+0.48k$ for $λAC$ in equation (VI) to get $TAC$ .

$TAC=(0.6i−0.64j+0.48k)TAC$

Substitute $(250 mm)i−(480 mm)j−(360 mm)k$ for $AD→$ and $650 mm$ for $AD$ in equation (IX) to get $λAD$ .

$λAD=(250 mm)i−(480 mm)j−(360 mm)k650 mm=513i−9.613j−7.213k$

Substitute $513i−9.613j−7.213k$ for $λAD$ in equation (VIII) to get $TAD$ .

$TAD=(513i−9.613j−7.213k)TAD$

Substitute $(−817i−1217j+1917k)TAB$ for $TAB$ , $(0.6i−0.64j+0.48k)TAC$ for $TAC$ , $(513i−9.613j−7.213k)TAD$ , $792j$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$ .

$(−817i−1217j+917k)TAB+(0.6i−0.64j+0.48k)TAC+(513i−9.613j−7.213k)TAD+792j=0(−817TAB+0.6TAC+513TAD)i+(−1217TAB−0.64TAC+−9.613TAD+792)j+(+917TAB+0.48TAC−7.213TAD)k=0$

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

$−817TAB+0.6TAC+513TAD=0$ (XII)

$−1217TAB−0.64TAC+−9.613TAD+792=0$ (XIII)

$+917TAB+0.48TAC−7.213TAD= 0$ (XIV)

Multiply equation (XII) with 12 and equation (XIV) with $−9$ and add to get new equation.

$−9617TAB+7.2TAC+6013TAD+9617TAB+5.12TAC+76.813TAD−6336=0$

$12.32TAC+136.813TAD−6336=0$ (XV)

Multiply equation (XII) with $9$ and equation (XIV) with $8$ and add to get new equation.

$−7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC−57.613TAD= 05.4TAC+4513TAD+3.84TAC−57.613TAD= 0$
$9.24TAC−12.613TAD=0$ (XVI)

Multiply equation (XV) with $12.6$ and (XVI) with $136.8$ to get $TAC$ .

$155.232TAC+1723.6813TAD−79833.6 N+1264.032TAC−1723.6813TAD=0$

$(155.232+1264.032)TAC=79833.6 NTAC=56.25 N$

Substitute $56.2 N$ for $TAC$ in equation (XVI) to get $TAD$ .

$9.24(56.25 N)−12.613TAD=0TAD= 536.25 N$

Substitute $56.25 N$ for $TAC$ and $536.25 N$ for $TAD$ in equation (XII) to get $TAB$ .

$−817TAB+0.6(56.25 N)+513( 536.25 N )=0$

$TAB= 510 N$

Therefore, the tension in cable $AB$ is $510 N_$ , the tension in the cable $AC$ is $56.2 N_$ and the tension in the cable $AD$ is $536 N_$ .

Answer 14

Ch. 2.1 - Two forces are applied as shown to a hook....Ch. 2.1 - Two forces are applied as shown to a bracket...Ch. 2.1 - Prob. 2.3P Ch. 2.1 - Prob. 2.4P Ch. 2.1 - A stake is being pulled out of the ground by means...Ch. 2.1 - A telephone cable is clamped at A to the pole AB....Ch. 2.1 - A telephone cable is clamped at A to the pole AB....Ch. 2.1 - A disabled automobile is pulled by means of two...Ch. 2.1 - A disabled automobile is pulled by means of two...Ch. 2.1 - Two forces are applied as shown to a hook support....

For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. Fig. P2.109 and P2.110

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Answer to Problem 2.115P

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