Vector Mechanics for Engineers: Statics, 11th Edition
Vector Mechanics for Engineers: Statics, 11th Edition
11th Edition
ISBN: 9780077687304
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek
Publisher: McGraw-Hill Education
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Chapter 2.5, Problem 2.107P

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N.

Chapter 2.5, Problem 2.107P, Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0,

Fig.P2.107and P2.108

Expert Solution & Answer
Check Mark
To determine

The value of P shown in figure P2.107 in the textbook for which tension in the cable shown AD is 305N.

Answer to Problem 2.107P

The force P at the point A shown in figure P2.107 is 960N_.

Explanation of Solution

The sketch of the cables connected at A, where the forces P and Q applied is shown in below figure1.

Vector Mechanics for Engineers: Statics, 11th Edition, Chapter 2.5, Problem 2.107P , additional homework tip  1

Free body diagram at A is shown in figure 2.

Vector Mechanics for Engineers: Statics, 11th Edition, Chapter 2.5, Problem 2.107P , 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 the force exerted in the x direction and Q is the magnitude of force exerted in the y direction.

The force Q exerted at point A is zero and the tension in cable AD is 305N.

Let TAB, TAC, and P are the tension vector in cable AB, AC, AD and force exerted in the x direction respectively and Q is the force at A along the y direction.

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 x direction.

P=Pj (X)

Here, P is the vector notation of weight of the body and P is the magnitude of weight.

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

Q=Qj

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 total force.

TAB+TAC+TAD+P+Q=0

Since Q=0N, write the above equation in simplified form

TAB+TAC+TAD+P=0 (XI)

Conclusion:

Substitute 960mm for x1 , 240mm for y1 , 0mm for z1 , 0mm for x2 , 0mm for y2 and 380mm for z2 in equation (I) to get AB.

AB=(960mm)i(240mm)j+(380mm)k

Calculate the magnitude of AB.

AB=(960mm)2+(240mm)2+(320mm)2=1060mm

Substitute 960mm for x1 , 240mm for y1 , 0mm for z1 , 0mm for x3 , 0mm for y3 and 320mm for z3 in equation (II) to get AC.

AC=(960mm)i(240mm)j(320mm)k

Calculate the magnitude of AC.

AC=(960mm)2+(240mm)2+(320mm)2=1040mm

Substitute 960mm for x1 , 240mm for y1 , 0mm for z1 ,0mm for x4 , 960mm for y4 and 0mm for z4 in equation (III) to get AD.

AD=(960mm)i+(720mm)j(220mm)k

Calculate the magnitude of AD.

AD=(960mm)2+(720mm)2+(220mm)2=1220mm

Substitute (960mm)i(240mm)j+(380mm)k for AB and 1060mm for AB in equation (V) to get λAB.

λAB=(960mm)i(240mm)j+(380mm)k1060mm=4853i1253j+1953k

Substitute 4853i1253j+1953k for λAB in equation (IV) to get TAB.

TAB=(4853i1253j+1953k)TAB

Substitute (960mm)i(240mm)j(320mm)k for AC and 7.40mm for AC in equation
(VII) to get λAC.

λAC=(960mm)i(240mm)j(320mm)k1040mm=1213i313j413k

Substitute 1213i313j413k for λAC in equation (VI) to get TAC.

TAC=(1213i313j413k)TAC

Substitute (960mm)i+(720mm)j(220mm)k for AD and 6.50mm for AD in equation (IX) to get λAD.

λAD=(960mm)i+(720mm)j(220mm)k1220mm

Substitute (960mm)i+(720mm)j(220mm)k1220mm for λAD and 305N for TAD in equation (VIII) to get TAD.

TAD=((960mm)i+(720mm)j(220mm)k1220mm)(305N)=(240N)i+(180N)j(55N)k

Substitute (4853i1253j+1953k)TAB for TAB , (1213i313j413k)TAC for TAC , (240N)i+(180N)j(55N)k and Pi for P in the in equation (XI) to get force P exerted at point A.

(4853i1253j+1953k)TAB+(1213i313j413k)TAC+(240N)i+(180N)j(55N)k+Pj=0(4853TAB1213TAC240N+P)i+(1253TAB313TAC+180N)j+(+1953TAB413TAC55N)k=0

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

4853TAB1213TAC240N+P=0 (XII)

1253TAB313TAC+180N=0 (XIII)

1953TAB413TAC55N=0 (XIV)

Multiply equation (XIII) by 43 to modify the equation.

48159TAB413TAC+180N=0 (XV)

Add above equation with equation (XIV) to get TAB.

48159TAB+413TAC240N+(1953TAB413TAC55N)=048159TAB240N+(1953TAB55N)=0TAB=446.71N

Substitute 446.71N for TAB in equation (XIV) to get TAC.

1953(446.71N)413TAC55N=0TAC=341.71N

Substitute 341.71N for TAC, 446.71N for TAB in equation (XII) to get P.

4853(446.71N)1213(341.71N)240N+P=0P=960N

Therefore, the force P exerted at A is 960Ni_.

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Chapter 2 Solutions

Vector Mechanics for Engineers: Statics, 11th Edition

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