Universidad Autónoma de Guadalajara Campus

TabascoFACULTAD DE INGENIERÍA PETROLERA

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2.43. Two cables are tied together at C and are loaded as shown. Knowingthat a 5 20°, determine the tension (a) in cable AC, (b) incable BC.

Law of sines:

a)

b)

TAC= 2128.99NTBC= 1735.49N

TACsen70 °= TBC

sen50 °= 1962Nsen60 °

TAC= 1962 Nsen60 °sen70°= 2128.99N

TBC= 1962Nsen60 °sen50°= 1735.49N

2.45 Two cables are tied together at C and are loaded as shown. Determinethe tension (a) in cable AC, (b) in cable BC.

a)

b)

TAC =439.69N

TBC = 326.35N

TACsen60 °= TBC

sen 40°= 500Nsen80 °

TAC= 500 Nsen80 °sen60°= 439.69NTBC= 500Nsen80 °sen60°= 326.35N

2.46 Two cables are tied together at C and are loaded as shown. Knowingthat P 5 500 N and a 5 60°, determine the tension in (a) incable AC, (b) in cable BC.

a)

b)

TAC= 305NTBC= 514N

TACsen35 °= TBC

sen75 °= 500Nsen70 °

TAC= 500Nsen70 °sen35°= 305N

TBC= 500Nsen70 °sen75°= 514N

2.47 Two cables are tied together at C and are loaded as shown. Determinethe tension (a) in cable AC, (b) in cable BC.

W = mg

W = (200Kg)(p.8m/s2)

W = 1962N

a)

b)

TAC= 586NTBC= 2190N

TACsen15 °= TBC

sen105 °= 1962 Nsen60 °

TAC= 1962Nsen60 °sen15°= 586N

TBC= 51962Nsen60 ° sen105°= 2190N

2.48 Knowing that a 5 20°, determine the tension (a) in cable AC, (b) inrope BC.

a)

b)

TAC= 1244NTBC= 115.4N

TACsen110°= TBCsen5 °= 1200 Nsen65 °

TAC= 1200Nsen65 °sen110°= 1244N

TBC= 1200Nsen65 °sen5°= 115.4N

2.49 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P= 500 lb and Q=650 lb, determine the magnitudes of the forces exerted on the rods A and B.

Resolving the forces into x and y directions:R=P+Q+F A+FB=0

Substituting Components:R=−(500lb ) j+[ (650 lb )cos (50° ) ] i−[ (650 lb )sin (50 ° ) ] j+FBi−(FAcos 50° ) i+(F A sin50 ° ) j=0

In the y direction (one unknown force)−500 lb−(650 lb )sin 50 °+FA sin 50 °=0 F A=

(500 lb )+ (650lb ) sin 50°sin 50°

¿1302.70 lb

In the x-direction:(650 lb )cos 50°+Fb−F Acos50 °=0 Fb=F A cos50 °−(650 lb )cos50 ° (1302.70 lb )cos50 °− (650lb ) cos50 ° ¿419.55 lb

2.50 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA =750 lb and FB =400 lb, determine the magnitudes of P and Q.

Resolving the forces into x and y directions:R=P+Q+F A+FB=0

Substituting Components:

R=−Pj+Qcos (50 ° ) i−Qcos (50 ° ) j−[ (750 lb )cos50 ° ] i+ [ (750 lb ) sin50 ° ] j+(400lb ) i

In the x direction (one unknown force)Qcos50°−[ (750lb ) cos50 ° ]+400 lb=0 Q=

(750 lb )cos50 °−400 lbcos 50°

¿127.710 lb

In the y-direction:−P−Q sin 50 °+(750lb ) sin 50 °=0 P=−Qsin 50 °+(750 lb) sin 50 ° ¿−(127.710lb ) sin 50°+ (750 lb ) sin50 ° ¿476.70 lb

2.51 A 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.

∑ F x=0 :35F B−FC−

35FA=0

F A=8kN FB=16 kN FC=

45

(16kN )−45(8kN )

FC=6.40kN

∑ F y=0:−FD+35FB−

35FA=0

FD=35

(16kN )−35(8 kN)

FD=4.80kN

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

∑ F y=0:−FD−35FA+

35FB=0

FB=F D+35FA

FD=8kN and F A=5kN

FB=53 [6 kN+ 3

5(5kN )]

FB=15.00 kN ∑ F x=0 :−FC+

45FB−

45F A=0

FC=45 (FB−F A )

¿ 45(15kN−5kN )

FC=8.00 kN

2.53 Two 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.

∑ F y=0:T CA−Qcos30 °=0Q=60lb T CA=(60 lb )(0.866) T CA=52.0 lb

∑ F x=0 :P−T CB−Qsin30 °=0 P=75 lb T CB=75lb−(60 lb)(0.50) T CB=45.0 lb

2.63 Collar 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.

α=tan−1 20x a) α=tan−1 20

4.5=77.36 ° b)α=tan−1 20

15=53.13 °

P→

=−P i

W→

=W cosα i+W sinα j

R x→

=−P+W cosα

R y→

=W sin α −P+W cos α=0 W cos α=P

a)W cos α=P 50cos77.36 °=P 10.94 lb=P

b)W cos α=P 50cos53.13 °=P 30 lb=P

w=50 P

2.64 Collar 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.

α=tan−1 20x

P→

=−P i

W→

=W cosα i+W sinα j

R x→

=−P+W cosα

R y→

=W sin α −P+W cos α=0 W cos α=P 50cosα=48 cos α= 4850 α=cos−1 4850 α=cos−1 4850 α=16.26 °

α=tan−1 20x

=16.26

20x

=tan 16.26

x= 20tan 16.26

x=68.57 in

w=50 P=48

2.65 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that β = 20°, determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chap. 4.)

W=mg=160 kg (9.81ms2 )=1569.6 NF Magnitude Compx CompyT1 P P sin 20 P cos20T2 P P sin 20 P cos20T3 P −P cosα P sinαW 1569.6 N 0 -1569.6N

∑ F x=2P sin 20−P cos α Ec.1∑ F y=2 Pcos20+P sin α−1569.6 Ec. 2

Usando la ec 1:α=cos−1−2 P sin 20

−P=cos−12sin 20=46.83°

Usando la ec 2:Para α=+46.83°

2 Pcos 20+P sin α=1569.6P[cos20+sin 46.83]=1569.6

P= 1569.6cos20+sin 46.83

P=601.64N

Para α=−46.83 °

2 Pcos 20+P sin α=1569.6P[cos20+sin−46.83]=1569.6

P= 1569.6cos20+sin−46.83

P=1364.94N

P T3 T1, T2  α β

W

2.66 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that α = 40°, determine (a) the angle β, (b) the magnitude of the force P that must be exerted on the free end of the rope to maintain equilibrium. (See the hint for Prob. 2.65.)

W=mg=160 kg (9.81ms2 )=1569.6NF Magnitude Compx CompyT1 P P sin β P cos βT2 P P sin β P cos βT3 P −P cos40 P sin 40W 1569.6 N 0 -1569.6N

∑ F x=2P sin β−P cos40 Ec.1∑ F y=2 Pcos β+Psin 40−1569.6 Ec. 2

Usando la ec 1:β=sin−1 P cos40

2P=sin−1 cos 40

2=22.52°

Usando la ec 2:2 Pcos β+P sin 40−1569.6=0P[2cos22.52+sin 40]=1569.6

P= 1569.62cos22.52+sin 40

P=630.29N

P

T3 β T1, T2  α=40°

W

2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Prob. 2.65.)

∑ F y=2T−600

T=6002

T=300

∑ F y=2T−600

T=6002

T=300

∑ F y=3T−600

T=6003

T=200

∑ F y=3T−600

T=6003

T=200

∑ F y=4T−600

T=6004

T=150