Problem 2.97

PROBLEM 2.97 For the semicircular ring of Problem 2.91, determine the magnitude and direction of the resultant of the forces exerted by the cables at B knowing that the tensions in cables BD and BE are 220 N and 250 N, respectively.

SOLUTION For the solutions to Problems 2.91 and 2.92, we have TBD = − (120 N ) i + (140 N ) j + (120 N ) k TBE = − (120 N ) i + (150 N ) j − (160 N ) k

Then: R B = TBD + TBE

= − ( 240 N ) i + ( 290 N ) j − ( 40 N ) k and

R = 378.55 N

cosθ x = −

RB = 379 N

240 = −0.6340 378.55

θ x = 129.3° cosθ y =

290 = −0.7661 378.55

θ y = 40.0° cosθ z = −

40 = −0.1057 378.55

θ z = 96.1°

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PROBLEM 2.98 To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AB is 920 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION Have TAB = ( 920 lb )( sin 50° cos 40°i − cos 50° j + sin 50° sin 40° j) TAC = TAC ( − cos 45° sin 25°i − sin 45° j + cos 45° cos 25° j)

(a) R A = TAB + TAC

( RA ) x ∴

( RA ) x

= ΣFx = 0:

=0

( 920 lb ) sin 50° cos 40° − TAC cos 45° sin 25° = 0

or TAC = 1806.60 lb

TAC = 1807 lb

(b)

( RA ) y

= ΣFy : − ( 920 lb ) cos 50° − (1806.60 lb ) sin 45°

( RA ) y

( RA ) z

= ΣFz :

= −1868.82 lb

( 920 lb ) sin 50° sin 40° + (1806.60 lb ) cos 45° cos 25° ( RA ) z

= 1610.78 lb

∴ RA = − (1868.82 lb ) j + (1610.78 lb ) k Then: RA = 2467.2 lb

RA = 2.47 kips

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PROBLEM 2.98 CONTINUED and cosθ x = cosθ y = cosθ z =

0 =0 2467.2

θ x = 90.0°

−1868.82 = −0.7560 2467.2

θ y = 139.2°

1610.78 = 0.65288 2467.2

θ z = 49.2°

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PROBLEM 2.99 To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AC is 850 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, determine (a) the tension in AB, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION Have TAB = TAB ( sin 50° cos 40°i − cos 50° j + sin 50° sin 40° j) TAC = ( 850 lb )( − cos 45° sin 25°i − sin 45° j + cos 45° cos 25° j) (a)

( RA ) x ∴

( RA ) x

=0

= ΣFx = 0: TAB sin 50° cos 40° − ( 850 lb ) cos 45° sin 25° = 0 TAB = 432.86 lb

TAB = 433 lb

(b)

( RA ) y

= ΣFy : − ( 432.86 lb ) cos 50° − ( 850 lb ) sin 45°

( RA ) y

( RA ) z

= ΣFz :

= −879.28 lb

( 432.86 lb ) sin 50° sin 40° + (850 lb ) cos 45° cos 25° ( RA ) z

= 757.87 lb

∴ R A = − ( 879.28 lb ) j + ( 757.87 lb ) k RA = 1160.82 lb

cosθ x = cosθ y =

RA = 1.161 kips

0 =0 1160.82

θ x = 90.0°

−879.28 = −0.75746 1160.82

θ y = 139.2°

757.87 = 0.65287 1160.82

θ z = 49.2°

cosθ z =

104

PROBLEM 2.100 For the plate of Problem 2.89, determine the tension in cables AB and AD knowing that the tension if cable AC is 27 lb and that the resultant of the forces exerted by the three cables at A must be vertical.

SOLUTION With:

JJJG AC = ( 45 in.) i − ( 48 in.) j + ( 36 in.) k

( 45 in.)2 + ( −48 in.)2 + ( 36 in.)2

AC =

TAC = TAC λ AC = TAC

= 75 in.

JJJG AC 27 lb ( 45 in.) i − ( 48 in.) j + ( 36 in.) k  = AC 75 in. 

TAC = (16.2 lb ) i − (17.28 lb ) j + (12.96 ) k and

JJJG AB = − ( 32 in.) i − ( 48 in.) j + ( 36 in.) k

( −32 in.)2 + ( −48 in.)2 + ( 36 in.)2

AB =

TAB = TABλ AB = TAB

= 68 in.

JJJG AB T = AB ( −32 in.) i − ( 48 in.) j + ( 36 in.) k  AB 68 in. 

TAB = TAB ( −0.4706i − 0.7059 j + 0.5294k ) and

JJJG AD = ( 25 in.) i − ( 48 in.) j − ( 36 in.) k

AD =

( 25 in.)2 + ( −48 in.)2 + ( 36 in.)2

TAD = TADλ AD = TAD

= 65 in.

JJJG AD T = AD ( 25 in.) i − ( 48 in.) j − ( 36 in.) k  AD 65 in.

TAD = TAD ( 0.3846i − 0.7385 j − 0.5538k )

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PROBLEM 2.100 CONTINUED Now R = TAB + TAD + TAD = TAB ( −0.4706i − 0.7059 j + 0.5294k ) + (16.2 lb ) i − (17.28 lb ) j + (12.96 ) k 

+ TAD ( 0.3846i − 0.7385 j − 0.5538k ) Since R must be vertical, the i and k components of this sum must be zero. Hence:

−0.4706TAB + 0.3846TAD + 16.2 lb = 0

(1)

0.5294TAB − 0.5538TAD + 12.96 lb = 0

(2)

Solving (1) and (2), we obtain: TAB = 244.79 lb,

TAD = 257.41 lb TAB = 245 lb TAD = 257 lb

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