PROBLEM 2.132 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.

SOLUTION

0: cos30 0y CAF T QΣ = − ° =

With 60 lbQ =

(a) ( )( )60 lb 0.866CAT =

52.0 lbCAT =

(b) 0: sin 30 0x CBF P T QΣ = − − ° =

With 75 lbP =

( )( )75 lb 60 lb 0.50CBT = −

or 45.0 lbCBT =

152

PROBLEM 2.133 Two 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.

SOLUTION

Have 0: cos30 0x CAF T QΣ = − ° =

or 0.8660 QCAT =

Then for 60 lbCAT ≤

0.8660 60 lbQ <

or 69.3 lbQ ≤

From 0: sin 30y CBF T P QΣ = = − °

or 75 lb 0.50CBT Q= −

For 60 lbCBT ≤

75 lb 0.50 60 lbQ− ≤

or 0.50 15 lbQ ≥

Thus, 30 lbQ ≥

Therefore, 30.0 69.3 lbQ≤ ≤

153

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

SOLUTION

Free-Body Diagram of Connection

3 30: 05 5x B C AF F F FΣ = − − =

With 8 kN, 16 kNA BF F= =

( ) ( )4 416 kN 8 kN5 5CF = −

6.40 kNCF =

3 30: 05 5y D B AF F F FΣ = − + − =

With AF and BF as above:

( ) ( )3 316 kN 8 kN5 5DF = −

4.80 kNDF =

154

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

SOLUTION

Free-Body Diagram of Connection

3 30: 05 5y D A BF F F FΣ = − − + =

or 35B D AF F F= +

With 5 kN, 8 kNA DF F= =

( )5 36 kN 5 kN3 5BF = +

15.00 kNBF =

4 40: 05 5x C B AF F F FΣ = − + − =

( )45C B AF F F= −

( )4 15 kN 5 kN5

= −

8.00 kNCF =

155

PROBLEM 2.136 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.

SOLUTION

Free-Body Diagram of Collar

(a) Triangle Proportions

( )4.50: 50 lb 020.5xF PΣ = − + =

or 10.98 lbP =

(b) Triangle Proportions

( )150: 50 lb 025xF PΣ = − + =

or 30.0 lbP =

156

PROBLEM 2.137 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.

SOLUTION

Free-Body Diagram of Collar

Triangle Proportions

Hence: 2

ˆ500: 48 0ˆ400

xxF

xΣ = − + =

+

or 248ˆ ˆ40050

x x= +

( )2 2ˆ ˆ0.92 lb 400x x= +

2 2ˆ 4737.7 inx =

ˆ 68.6 in.x =

157

PROBLEM 2.138 A 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.

SOLUTION

The 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

( ) ( ) ( )480 mm 510 mm 320 mmDB = − +i j k

( ) ( ) ( )2 2 2480 510 320 770 mmDB = + + =

( ) ( ) ( )385 N 480 mm 510 mm 320 mm770 mmDB

DBF FDB

= = = − + F i j kλ

( ) ( ) ( )240 N 255 N 160 N= − +F i j k

240 N, 255 N, 160.0 Nx y zF F F= + = − = +

158

PROBLEM 2.139 A 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.

SOLUTION

The 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

( ) ( ) ( )0.48 m 0.51 m 0.32 mBD = − + −i j k

( ) ( ) ( )2 2 20.48 m 0.51 m 0.32 m 0.77 mBD = − + + − =

( ) ( ) ( )0.48 m 0.51 m 0.32 m0.77 m

BDBD BD BD

BD TT TBD

= = = − + − T i j kλ

( )0.6234 0.6623 0.4156BD BDT= − + −T i j k

and

( ) ( ) ( )0.27 m 0.40 m 0.6 mBE = − + −i j k

( ) ( ) ( )2 2 20.27 m 0.40 m 0.6 m 0.770 mBE = − + + − =

( ) ( ) ( )0.26 m 0.40 m 0.6 m0.770 m

BEBE BE BE

BD TT TBD

= = = − + − T i j kλ

( )0.3506 0.5195 0.7792BE BET= − + −T i j k

Now, because of the frictionless ring at B, 385 NBE BDT T= = and the force on the support due to the two cables is

( )385 N 0.6234 0.6623 0.4156 0.3506 0.5195 0.7792= − + − − + −F i j k i j k

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

159

PROBLEM 2.139 CONTINUED

The magnitude of the resultant is

( ) ( ) ( )2 2 22 2 2 375 N 455 N 460 N 747.83 Nx y zF F F F= + + = − + + − =

or 748 NF =

The direction of this force is:

1 375cos747.83xθ

− −= or 120.1xθ = °

1 455cos747.83yθ

−= or 52.5yθ = °

1 460cos747.83zθ

− −= or 128.0zθ = °

160

PROBLEM 2.140 A 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.

SOLUTION

Force Triangle

(a) For minimum P it must be perpendicular to the vertical resultant R

( ) 425 lb cos30P∴ = °

or 368 lb=P

(b) ( )425 lb sin 30R = °

or 213 lbR =

161