Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9781259639272
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 4.3, Problem 4.107P

Chapter 4.3, Problem 4.107P, PROBLEM 4.107 Solve Problem 4.106 for a = 1.5 m. PROBLEM 4.106 The 6-m pole ABC is acted upon by a

PROBLEM 4.107

Solve Problem 4.106 for a = 1.5 m.

PROBLEM 4.106 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a =3 m, determine the tension in each cable and the reaction at A.

Expert Solution & Answer
Check Mark
To determine

The tension in each of the cable and the reaction at A.

Answer to Problem 4.107P

The tension in cable BD is 525N_ , tension in cable BE is 105N_. Also the reaction at (105N)i^+(840N)j^+(140N)k^_.

Explanation of Solution

Refer figure1 shown below. The figure shows that the pole ABC is acted upon by a certain force. The pole is balanced by cables BD and BE about a ball and socket joint A. The forces along the cables can be resolved into its x and y components. P is the force acting on the pole, and T is the tension along the cables.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.107P

The sum of moments along line AC is zero at equilibrium.

MAC=0 (I)

Here, MA is the sum of moments along line AC.

Write the equation to find the position vector of point B.

rB=yj^ (II)

Here, rB is the position vector of point B, y is y component of the position vector.

Write the equation to find the position vector of point C.

rC=yj^ (III)

Here, rC is the position vector of point C, y is the y component of the position vector.

Write the equation to find the position vector CF.

CF=xi^+yj^+zk^ (IV)

Here, CF is the position vector CF , x is the x component of vector, y is the y component of vector, z is the z component of vector

Write the equation to find the magnitude of vector CF.

CF=x2+y2+z2 (V)

Here, CF is the magnitude of vector CF , x is the x component of vector, y is the y component of vector, z is the z component of vector

Write the equation to find the position vector BD.

BD=xi^+yj^+zk^ (VI)

Here, BD is the position vector, x is the x component of vector, y is the y component of vector, z is the z component of vector

Write the equation to find the magnitude of vector BD.

BD=x2+y2+z2 (VII)

Here, BD is the magnitude of vector BD , x is the x component of vector, y is the y component of vector, z is the z component of vector

Write the equation to find the position vector BE.

BE=xi^+yj^+zk^ (VIII)

Here, BE is the position vector , x is the x component of vector, y is the y component of vector, z is the z component of vector

Write the equation to find the magnitude of vector BE.

BE=x2+y2+z2 (IX)

Here, BE is the magnitude of vector BE , x is the x component of vector, y is the y component of vector, z is the z component of vector

Write the equation to find the force acting along line CF.

P=PCFCF (X)

Here, P is the force vector along the line CF, P is the magnitude of force, CF is the position vector along line CF, and CF is the magnitude of the position vector.

Write the equation to find the tension in cable BD.

TBD=TBDBDBD (XI)

Here, TBD is the tension vector along the cable BD, TBD is the magnitude of tension, BD is the position vector corresponding to cable, BD is the magnitude of position vector.

Write the equation to find the tension in cable BE.

TBE=TBEBEBE (XII)

Here, TBE is the tension vector along the cable BE, TBE is the magnitude of tension, BE is the position vector corresponding to cable, BE is the magnitude of position vector.

Write the equation to find the sum of moments along the point A.

MA=rB×TBD+rB×TBE+rC×P (XIII)

Here, rB is the position vector of point B, TBD is the tension in cable BD, TBE is the tension in cable BE, rC is the position vector of point C, P is the force at point C.

Since the sum of moments at point A is zero, rewrite equation (XI).

rB×TBD+rB×TBE+rC×P=0 (XIV)

Substitute for rB , rC , TBD , TBE , P and take the determinant of equation (XII).

|i^j^k^030122|TBDBD+|i^j^k^030122|TBEBE+|i^j^k^0603124|PCE=0 (XV)

Find the determinant value of equation (XIII) and equate the coefficients of unit vectors i^ , j^ , and k^.

Equate the coefficients of unit vector i^ to zero.

2TBD+2TBE+2413P=0 (XVI)

Here, TBD is the magnitude of tension in cable BD, TBE is the magnitude of tension in cable BE, P is the magnitude of force.

Equate the coefficient of unit vector k^ to zero.

TBDTBE+1813P=0 (XVII)

Add equations (XIV) to twice of equation (XV) to get the value of tensions in cables.

2TBD+2TBE+2413P+2(TBDTBF+1813P)=4TBD+6013P=0 (XVIII)

The sum of all external forces is zero. Therefore sum of tensions in cables, force at P and S is zero.

TBD+TBE+P+A=0 (XIX)

Here, TBD is the tension in cable BD, TBE is the tension in cable BE, P is the force at point P, and A is the magnitude of reaction force at point A.

Conclusion:

Substitute 6m for y in equation (II) to get rB.

rB=3j^

Substitute 6m for y in equation (III) to get rC.

rC=6j^

Substitute 1.5m for x , 6m for y , and 2m for z in equation (IV) to get CF.

CF=1.5i^6j^+2k^

Substitute 1.5m for x , 6m for y , and 2m for z in equation (V) to get CF.

CF=(1.5m)2+(6m)2+(2m)2=6.5m

Substitute 1.5m for x , 3m for y , and 3m for z in equation (VI) to get BD.

BD=1.5i^3j^3k^

Substitute 1.5m for x , 3m for y , and 3m for z in equation (VII) to get BD.

BD=(1.5m)2+(3m)2+(3m)2=4.5m

Substitute 1.5m for x , 3m for y , and 3m for z in equation (VIII) to get BE

BE=1.5i^3j^+3k^

Substitute 1.5m for x , 3m for y , and 3m for z in equation (IX) to get BE

BE=(1.5m)2+(3m)2+(3m)2=4.5m

Substitute (1.5i^6j^+2k^) for CF and 6.5m for CF in equation (X) to get P.

P=P7(1.5i^6j^+2k^)=P13(3i^12j^+4k^)

Substitute (1.5i^3j^3k^) for BD and 4.5m for BD in equation (XI) to get TBD.

TBD=TBD4.5(1.5i^3j^3k^)=TBD3(i^2j^2k^)

Substitute (1.5i^2j^+2k^) for BE and 4.5m for BE in equation (XII) to get TBE.

TBE=TBE3(i^2j^+2k^)

Substitute P13(3i^12j^+4k^) for P , TBD3(i^2j^2k^) for TBD , and TBE3(i^2j^+2k^) for TBE , 3j^ for rB , 6j^ for rC in equation (XIII) to get MA in determinant form.

|i^j^k^030122|TBD3+|i^j^k^030122|TBE3+|i^j^k^0603124|P13=0

Solve equation (XVIII) to get TBD.

TBD=1513P

Substitute 1513P for TBD in equation (XVII) to get TBE.

TBE=313P

Substitute 455N for P in above equation to get TBD.

TBD=1513(455N)=525N

Substitute 455N for P in the above equation to find TBE.

TBE=313(455N)=105N

Substitute 5253 for TBD , 1053 for TBE , 45513 for P , and AX for A in equation (XIX) to get Ax.

5253+105345513(3)+Ax=0Ax=105N

Similarly substitute 5253 for TBD , 1053 for TBE , 45513 for P , and Ay in equation (XIX) to get Ay.

Ay=5253(2)1053(2)45513(12)+Ay=0Ay=840N

Substitute 5253 for TBD , 1053 for TBE , 45513 for P , and Az for A in equation (XIX) to get Az.

5253(2)+1053(2)+45513(4)+Az=0Az=140N

The vector form of the reaction force is (105N)i^+(840N)j^+(140N)k^

Therefore, the tension in cable BD is 525N_ , tension in cable BE is 105N_. Also the reaction at (105N)i^+(840N)j^+(140N)k^_.

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Chapter 4 Solutions

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

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