VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 4.3, Problem 4.105P

Chapter 4.3, Problem 4.105P, PROBLEM 4.105 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each

PROBLEM 4.105

A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A.

Expert Solution & Answer
Check Mark
To determine

The tension in each cable and the reaction at the ball and socket joint at A.

Answer to Problem 4.105P

The tension in cable BD is 1100lb_ , cable BE is 1100lb_ , and the reaction in joint A is (1200lb)i^(560lb)j^_.

Explanation of Solution

The free body diagram corresponding to the system of cable is shown in figure 1 plotted below.

A down ward force of magnitude 840lb acts at point C. The resultant tension experienced in cables are to be found in the problem.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 4.3, Problem 4.105P

Write the equation to find the position vector BD.

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

Here, BD is the position vector connecting point B and point D, x is the x coordinate, y is the y coordinate, and z is the z coordinate.

Write the equation to find the position vector BE.

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

Here, BE is the position vector connecting point B and point E, x is the x coordinate , y is the y coordinate, and z is the z coordinate.

Write the equation to find the magnitude of BD.

BD=x2+y2+z2 (III)

Here, BD is the magnitude of the vector, x is the x coordinate, y is the y coordinate, and z is the z coordinate.

Write the equation to find the magnitude of BE.

BE=x2+y2+z2 (IV)

Here, BE is the magnitude of the vector, x is the x coordinate, y is the y coordinate, z is the z coordinate.

TBD is the tension in the cable BD.

Write the equation to find the tension along cable BD.

TBD=TBDBDBD (V)

Here, TBD is the tension vector in the direction of unit vector, TBD is the magnitude of the tension, BD is the position vector, and BD is the magnitude of the position vector.

Write the equation to find the tension along cable BE.

TBE=TBEBEBE (VI)

Here, TBE is the tension vector along the direction of unit vector, TBE is the magnitude of the tension, BE is the position vector connecting point B and E, BE is the magnitude of the tension.

Write the equation to find the moments at point B due to force at D.

MBD=rB×TBD (VII)

Here, MBD is the moment at B due to force at D, rBD is the perpendicular distance between the line of action between forces at B and D, TBD is the tension in the wire.

Write the equation to find the moment at point B due to force at point E.

MBE=rB×TBE (VIII)

Here, MBE is the moment of force at point B due to force at point E, rB is the perpendicular distance between the line of action of force at points B and E, TBE is the tension at cable BE.

Write the equation to find the moment of force at point B due to force at point C.

MBC=rC×TBC (IX)

Here, MBC is the moment of force at point B due to force at point C, rC is the perpendicular distance between the point B and C, TBC is the tension at cable BC.

Write the equation to find the sum of moment of force at point B.

MB=MBD+MBE+MBC (X)

Here, MB is the sum of moments at point B, MBD is the moments about cable BD, MBE is the moment about cable BE, and MBC is the moment about cable BE.

Substitute equation (VII), (VIII), and (IX) in equation (X) to find the MB.

MB=rB×TBD+rB×TBE+rC×TBC (XI)

Substitute BDBD for TBD , BEBE for TBE , and 840 for TCB in equation (XI) to get MB.

MB=rB×TBD(BDBD)+rB×TBE(BEBE)+rC×TBC(840) (XII)

Since sum of all external force acting on a body is zero, equate the equation for moments to zero.

rB×TBD(BDBD)+rB×TBE(BEBE)+rC×TBC(840)=0 (XIII)

Write the equation to find the sum of x components of force.

Fx=0AxxBD(TBD)xBE(TBE)=0 (XIV)

Write the equation to find the sum of y components of force.

Fy=0Ay+yBD(TBD)+yBE(TBE)TCB=0 (XV)

Write the equation to find the sum of z component of forces.

Fz=0Az+zBD(TBD)zBE(TBE)=0 (XVI)

Conclusion:

Substitute 6ft for x , 7ft for y , and 6ft for z in equation (I) to get BD.

BD=(6ft)i^+(7ft)j^+(6ft)k^

Substitute 6ft for x , 7ft for y , and 6ft for z in equation (II) to get BE.

BE=(6ft)i^+(7ft)j^(6ft)k^

Substitute 6ft for x , 7ft for y , and 6ft for z in equation (III) to get BD.

BD=(6ft)2+(7ft)2+(6ft)2=11ft

Substitute 6ft for x , 7ft for y , and 6ft for z in equation (IV) to get BE.

BE=(6ft)2+(7ft)2(6ft)2=11ft

Substitute 11ft for BD , (6ft)i^+(7ft)j^+(6ft)k^ for BD , (6ft)i^+(7ft)j^+(6ft)k^ for BE , 6i^ for rB , 6i^ for rB , and 10i^ for rc in equation (XIII) to get the value of tensions.

6i^×TBD11(6i^+7j^+6k^)+6i^×TBE11(6i^+7j^6k^)+10i^×(840j^)=04211TBDk^3611TBDj^+4211TBEk^+3611TBEj^8400k^=0

Equate the coefficients of unit vectors i^ , j^ , and k^ to zero.

Equate the coefficient of unit vector i^.

3611TBD+3611TBE=0TBE=TBD

Equate the coefficients of unit vector k^.

4211TBD+4211TBE8400=02(4211TBD)=8400TBD=TBE=1100lb

Substitute 6 for x , 11 for BD , 1100lb for TBD , 11 for BE , and 1100lb for TBE in equation (XIV) to get Ax.

Ax611(1100lb)611(1100lb)=0Ax=0

Substitute 7 for y , 11 for BD , 1100lb for TBD , 11 for BE , 840lb for TCB and 1100lb for TBE, in equation (XV) to get Ay.

Ay+711(1100lb)+711(1100lb)840lb=0Ay=560lb

Substitute 6 for x , 11 for BD , 1100lb for TBD , 11 for BE , and 1100lb for TBE in equation (XVI) to get Az.

Az+611(1100lb)611(1100lb)=0Az=0

Therefore, the tension in cable BD is 1,100lb_ , cable BE is 1,100lb_ , and the reaction in joint A is (1200lb)i^(560lb)j^_.

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Chapter 4 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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