Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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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
Check Mark
To determine

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

Answer to Problem 2.115P

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

Explanation of Solution

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

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.

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 792N.

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,yandz direction.

Write the equation of vector distance AB.

AB=(x2x1)i+(y2y1)j+(z2z1)k (I)

Here, AB is the vector distance of the cable AB, the variables x1,y1and z1 are the coordinates of point A and x2,y2and z2 are the coordinates of point B.

Write the vector distance of the cable AC.

AC=(x3x1)i+(y3y1)j+(z3z1)k (II)

Here, AC is the vector distance of the cable AC, the variables x1,y1and z1 are the coordinates of point A and x3,y3and z3 are the coordinates of point C.

Write the vector distance of the cable AD.

AD=(x4x1)i+(y4y1)j+(z4z1)k (III)

Here, AD is the vector distance of the cable AD, the variables x1,y1and z1 are the coordinates of point A and x4,y4and 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=ABAB (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=ACAC (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=ADAD (IX)

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

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

AB=(320mm)i(480mm)j+(360mm)k

Calculate the magnitude of AB.

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

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

AC=(450mm)i(480mm)j+(360mm)k

Calculate the magnitude of AC.

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

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

AD=(250mm)i(480mm)j(360mm)k

Calculate the magnitude of AD.

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

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

λAB=(320mm)i(480mm)j+(360mm)k680mm=817i1217j+1917k

Substitute 817i1217j+1917k for λAB in equation (IV) to get TAB.

TAB=(817i1217j+1917k)TAB

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

λAC=(250mm)i(480mm)j(360mm)k750mm=0.6i0.64j+0.48k

Substitute 0.6i0.64j+0.48k for λAC in equation (VI) to get TAC.

TAC=(0.6i0.64j+0.48k)TAC

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

λAD=(250mm)i(480mm)j(360mm)k650mm=513i9.613j7.213k

Substitute 513i9.613j7.213k for λAD in equation (VIII) to get TAD.

TAD=(513i9.613j7.213k)TAD

Substitute (817i1217j+1917k)TAB for TAB , (0.6i0.64j+0.48k)TAC for TAC , (513i9.613j7.213k)TAD, 792j for P in the in equation (XI) to get force P exerted at point A.

(817i1217j+917k)TAB+(0.6i0.64j+0.48k)TAC+(513i9.613j7.213k)TAD+792j=0(817TAB+0.6TAC+513TAD)i+(1217TAB0.64TAC+9.613TAD+792)j+(+917TAB+0.48TAC7.213TAD)k=0

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

817TAB+0.6TAC+513TAD=0 (XII)

1217TAB0.64TAC+9.613TAD+792=0 (XIII)

+917TAB+0.48TAC7.213TAD=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+76.813TAD6336=0

12.32TAC+136.813TAD6336=0 (XV)

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

7217TAB+5.4TAC+4513TAD+7217TAB+3.84TAC57.613TAD=05.4TAC+4513TAD+3.84TAC57.613TAD=0
9.24TAC12.613TAD=0 (XVI)

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

155.232TAC+1723.6813TAD79833.6N+1264.032TAC1723.6813TAD=0

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

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

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

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

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

TAB=510N

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

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Chapter 2 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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