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### Transcript of Problem 2.132

PROBLEM 2.132Two cables tied together at C are loaded as shown. Knowing that Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTIONFy = 0: TCA Q cos 30 = 0

With (a)

Q = 60 lb TCA = ( 60 lb )( 0.866 ) TCA = 52.0 lb

(b) With

Fx = 0: P TCB Q sin 30 = 0

P = 75 lb TCB = 75 lb ( 60 lb )( 0.50 ) or TCB = 45.0 lb

152

PROBLEM 2.133Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable.

SOLUTIONHave or Then forFx = 0: TCA Q cos 30 = 0

TCA = 0.8660 Q TCA 60 lb 0.8660Q < 60 lb

or From or For

Q 69.3 lbFy = 0: TCB = P Q sin 30

TCB = 75 lb 0.50Q TCB 60 lb 75 lb 0.50Q 60 lb

or Thus, Therefore,

0.50Q 15 lb Q 30 lb 30.0 Q 69.3 lb

153

PROBLEM 2.134A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces.

SOLUTIONFree-Body Diagram of Connection

Fx = 0: With

3 3 FB FC FA = 0 5 5

FA = 8 kN, FB = 16 kN FC = 4 4 (16 kN ) (8 kN ) 5 5 FC = 6.40 kN Fy = 0: FD + 3 3 FB FA = 0 5 5

With FA and FB as above: FD = 3 3 (16 kN ) (8 kN ) 5 5 FD = 4.80 kN

154

PROBLEM 2.135A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 5 kN and FD = 6 kN, determine the magnitudes of the other two forces.

SOLUTIONFree-Body Diagram of Connection

Fy = 0: FD

3 3 FA + FB = 0 5 5 3 FA 5

or With

FB = FD +

FA = 5 kN, FD = 8 kNFB = 5 3 6 kN + ( 5 kN ) 3 5 FB = 15.00 kN Fx = 0: FC + FC = = 4 4 FB FA = 0 5 5

4 ( FB FA ) 5 4 (15 kN 5 kN ) 5 FC = 8.00 kN

155

PROBLEM 2.136Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in.

SOLUTIONFree-Body Diagram of Collar

(a)

Triangle Proportions

Fx = 0: P +

4.5 ( 50 lb ) = 0 20.5 or P = 10.98 lb

(b)

Triangle Proportions

Fx = 0: P +

15 ( 50 lb ) = 0 25 or P = 30.0 lb

156

PROBLEM 2.137Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb.

SOLUTIONFree-Body Diagram of Collar

Triangle Proportions

Hence:

Fx = 0: 48 +

50 x 2 400 + x

=0

or

= x

48 2 400 + x 50

2 = 0.92 lb 400 + x 2 x 2 = 4737.7 in 2 x

(

) = 68.6 in. x

157

PROBLEM 2.138A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTIONThe force in cable DB can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with JJJ G DB = ( 480 mm ) i ( 510 mm ) j + ( 320 mm ) k DB =F = F DB

( 480 )2 + ( 510 )2 + ( 320 )2

= 770 mm

JJJ G DB 385 N ( 480 mm ) i ( 510 mm ) j + ( 320 mm ) k = F = DB 770 mm F = ( 240 N ) i ( 255 N ) j + (160 N ) k

Fx = +240 N, Fy = 255 N, Fz = +160.0 N

158

PROBLEM 2.139A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. Determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N.

SOLUTIONThe force in each cable can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with JJJ G BD = ( 0.48 m ) i + ( 0.51 m ) j ( 0.32 m ) k BD =

( 0.48 m )2 + ( 0.51 m )2 + ( 0.32 m )2

= 0.77 m

TBD = T BD = TBD

JJJ G BD TBD ( 0.48 m ) i + ( 0.51 m ) j ( 0.32 m ) k = BD 0.77 m

TBD = TBD ( 0.6234i + 0.6623j 0.4156k )

andJJJ G BE = ( 0.27 m ) i + ( 0.40 m ) j ( 0.6 m ) k

BE =

( 0.27 m )2 + ( 0.40 m )2 + ( 0.6 m )2

= 0.770 m

TBE = T BE = TBE

JJJ G BD TBE ( 0.26 m ) i + ( 0.40 m ) j ( 0.6 m ) k = BD 0.770 m

TBE = TBE ( 0.3506i + 0.5195 j 0.7792k )

Now, because of the frictionless ring at B, TBE = TBD = 385 N and the force on the support due to the two cables isF = 385 N ( 0.6234i + 0.6623j 0.4156k 0.3506i + 0.5195j 0.7792k )

= ( 375 N ) i + ( 455 N ) j ( 460 N ) k

159

PROBLEM 2.139 CONTINUEDThe magnitude of the resultant isF = Fx2 + Fy2 + Fz2 =

( 375 N )2 + ( 455 N )2 + ( 460 N )2

= 747.83 N

or F = 748 N The direction of this force is:

x = cos 1

375 747.83 455 747.83 460 747.83

or x = 120.1 or y = 52.5 or z = 128.0

y = cos 1 z = cos 1

160

PROBLEM 2.140A steel tank is to be positioned in an excavation. Using trigonometry, determine (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R.

SOLUTIONForce Triangle

(a) For minimum P it must be perpendicular to the vertical resultant R P = ( 425 lb ) cos 30 or P = 368 lb (b) R = ( 425 lb ) sin 30 or R = 213 lb

161