PROBLEM 2.38 Knowing that 50 ,α = ° determine the resultant of the three forces shown.

SOLUTION

The resultant force R has the x- and y-components:

( ) ( ) ( )140 lb cos50 60 lb cos85 160 lb cos50x xR F= Σ = ° + ° − °

7.6264 lbxR = −

and

( ) ( ) ( )140 lb sin 50 60 lb sin85 160 lb sin 50y yR F= Σ = ° + ° + °

289.59 lbyR =

Further:

290tan7.6

α =

1 290tan 88.57.6

α −= = °

Thus: 290 lb=R 88.5°

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PROBLEM 2.39 Determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

For an arbitrary angle ,α we have:

( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cosx xR F α α α= Σ = + + ° −

(a) So, for R to be vertical:

( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cos 0x xR F α α α= Σ = + + ° − =

Expanding,

( )cos 3 cos cos35 sin sin 35 0α α α− + ° − ° =

Then:

13cos35

tansin 35

α° −

=°

or

1

1 3cos35tan 40.265

sin 35α − ° −

= = ° ° 40.3α = °

(b) Now:

( ) ( ) ( )140 lb sin 40.265 60 lb sin 75.265 160 lb sin 40.265y yR R F= = Σ = ° + ° + °

252 lbR R= =

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PROBLEM 2.40 For the beam of Problem 2.37, determine (a) the required tension in cable BC if the resultant of the three forces exerted at point B is to be vertical, (b) the corresponding magnitude of the resultant.

Problem 2.37: Knowing that the tension in cable BC is 145 lb, determine the resultant of the three forces exerted at point B of beam AB.

SOLUTION

We have:

( ) ( )84 12 3156 lb 100 lb116 13 5x x BCR F T= Σ = − + −

or 0.724 84 lbx BCR T= − +

and

( ) ( )80 5 4156 lb 100 lb116 13 5y y BCR F T= Σ = − −

0.6897 140 lby BCR T= −

(a) So, for R to be vertical,

0.724 84 lb 0x BCR T= − + =

116.0 lbBCT =

(b) Using

116.0 lbBCT =

( )0.6897 116.0 lb 140 lb 60 lbyR R= = − = −

60.0 lbR R= =

40

PROBLEM 2.41 Boom AB is held in the position shown by three cables. Knowing that the tensions in cables AC and AD are 4 kN and 5.2 kN, respectively, determine (a) the tension in cable AE if the resultant of the tensions exerted at point A of the boom must be directed along AB, (b) the corresponding magnitude of the resultant.

SOLUTION

Choose x-axis along bar AB.

Then

(a) Require

( ) ( )0: 4 kN cos 25 5.2 kN sin 35 sin 65 0y y AER F T= Σ = ° + ° − ° =

or 7.2909 kNAET =

7.29 kNAET =

(b) xR F= Σ

( ) ( ) ( )4 kN sin 25 5.2 kN cos35 7.2909 kN cos65= − ° − ° − °

9.03 kN= −

9.03 kNR =

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PROBLEM 2.42 For the block of Problems 2.35 and 2.36, determine (a) the required value of α of the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant.

Problem 2.35: Knowing that 35 ,α = ° determine the resultant of the three forces shown.

Problem 2.36: Knowing that 65 ,α = ° determine the resultant of the three forces shown.

SOLUTION

Selecting the x axis along ,aa′ we write

( ) ( )300 N 400 N cos 600 N sinx xR F α α= Σ = + + (1)

( ) ( )400 N sin 600 N cosy yR F α α= Σ = − (2)

(a) Setting 0yR = in Equation (2):

Thus 600tan 1.5400

α = =

56.3α = °

(b) Substituting for α in Equation (1):

( ) ( )300 N 400 N cos56.3 600 N sin 56.3xR = + ° + °

1021.1 NxR =

1021 NxR R= =

42

PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram

From the geometry, we calculate the distances:

( ) ( )2 216 in. 12 in. 20 in.AC = + =

( ) ( )2 220 in. 21 in. 29 in.BC = + =

Then, from the Free Body Diagram of point C:

16 210: 020 29x AC BCF T TΣ = − + =

or 29 421 5BC ACT T= ×

and 12 200: 600 lb 020 29y AC BCF T TΣ = + − =

or 12 20 29 4 600 lb 020 29 21 5AC ACT T + × − =

Hence: 440.56 lbACT =

(a) 441 lbACT =

(b) 487 lbBCT =

43