8/14/2019 Problem 2.38

1/6

PROBLEM 2.38

Knowing that 50 , = determine the resultant of the three forces

shown.

SOLUTION

The resultant forceR has thex- andy-components:

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

7.6264 lbxR =

and

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

289.59 lbyR =

Further:

290tan

7.6 =

1 290tan 88.57.6

= =

Thus: 290 lb=R 88.5

38

8/14/2019 Problem 2.38

2/6

PROBLEM 2.39

Determine (a) the required value of if the resultant of the three forces

shown is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

For an arbitrary angle , we have:

( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cosx xR F = = + +

(a) So, forR to be vertical:

( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cos 0x xR F = = + + =

Expanding,

( )cos 3 cos cos35 sin sin 35 0 + =

Then:

13

cos35

tan sin35

=

or

11 3

cos35tan 40.265

sin35

= = 40.3 =

(b) Now:

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

252 lbR R= =

39

8/14/2019 Problem 2.38

3/6

PROBLEM 2.40

For the beam of Problem 2.37, determine (a) the required tension in cable

BCif 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 cableBCis 145 lb, determine

the resultant of the three forces exerted at pointB of beamAB.

SOLUTION

We have:

( ) ( )84 12 3156 lb 100 lb116 13 5

x x BCR F T= = +

or 0.724 84 lbx BCR T= +

and

( ) ( )80 5 4

156 lb 100 lb116 13 5

y y BCR F T= =

0.6897 140 lby BCR T=

(a) So, forR 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

8/14/2019 Problem 2.38

4/6

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 tensionsexerted at point A of the boom must be directed along AB,

(b) the corresponding magnitude of the resultant.

SOLUTION

Choosex-axis along barAB.

Then

(a) Require

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

or 7.2909 kNAET =

7.29 kNAET =

(b) xR F=

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

9.03 kN=

9.03 kNR =

41

8/14/2019 Problem 2.38

5/6

PROBLEM 2.42

For the block of Problems 2.35 and 2.36, determine (a) the required value

of 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 , = determine the resultant of the

three forces shown.

Problem 2.36: Knowing that 65 , = 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 = = + + (1)

( ) ( )400 N sin 600 N cosy yR F = = (2)

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

Thus600

tan 1.5400

= =

56.3 =

(b) Substituting for in Equation (1):

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

1021.1 NxR =

1021 NxR R= =

42

8/14/2019 Problem 2.38

6/6

PROBLEM 2.43

Two cables are tied together at Cand 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 2

16 in. 12 in. 20 in.AC = + =

( ) ( )2 2

20 in. 21 in. 29 in.BC = + =

Then, from the Free Body Diagram of point C:

16 210: 0

20 29x AC BCF T T = + =

or29 4

21 5BC ACT T=

and12 20

0: 600 lb 020 29

y AC BCF T T = + =

or12 20 29 4

600 lb 020 29 21 5

AC ACT T

+ =

Hence: 440.56 lbACT =

(a) 441 lbACT =

(b) 487 lbBCT =

43