Solucionario estatica

PROBLEM 2.1
Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION
(a)

(b)

We measure:

R = 8.4 kN

α = 19°
R = 8.4 kN

19°

1

PROBLEM 2.2
The cable stays AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 Nin AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

We measure: (a)

α = 51.3°, β = 59°

(b)

We measure:

R = 575 N, α = 67° R = 575 N

67°

2

PROBLEM 2.3
Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 15 lband Q = 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION
(a)

(b)

We measure:

R = 37 lb, α = 76° R = 37 lb

76°

3

PROBLEM 2.4
Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 45 lb and Q = 15 lb, determine graphically the magnitude and direction oftheir resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION
(a)

(b)

We measure:

R = 61.5 lb, α = 86.5° R = 61.5 lb

86.5°

4

PROBLEM 2.5
Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the left-hand rod is F1 = 120 N, determine (a) the required force F2 in the right-hand rod if the resultant R of the forcesexerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Graphically, by the triangle law We measure:
F2 ≅ 108 N R ≅ 77 N By trigonometry: Law of Sines F2 R 120 = = sin α sin 38° sin β

α = 90° − 28° = 62°, β = 180° − 62° − 38° = 80°
Then:
F2 R 120 N = = sin 62° sin 38° sin 80° or (a) F2 = 107.6 N (b) R = 75.0 N

5

PROBLEM 2.6
Twocontrol rods are attached at A to lever AB. Using trigonometry and knowing that the force in the right-hand rod is F2 = 80 N, determine (a) the required force F1 in the left-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the Law of Sines F1 R 80 = = sin α sin 38° sin β

α = 90° − 10° = 80°, β =180° − 80° − 38° = 62°
Then: F1 R 80 N = = sin 80° sin 38° sin 62° or (a) F1 = 89.2 N (b) R = 55.8 N

6

PROBLEM 2.7
The 50-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Using trigonometry, determine the angle α knowing that the component along a-a′ is 35 lb. (b) What is the corresponding value of the component along b-b′ ?

SOLUTION

Using the triangle ruleand the Law of Sines (a) sin β sin 40° = 35 lb 50 lb sin β = 0.44995

β = 26.74°
Then:

α + β + 40° = 180° α = 113.3°

(b) Using the Law of Sines:

Fbb′ 50 lb = sin α sin 40° Fbb′ = 71.5 lb

7

PROBLEM 2.8
The 50-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Using trigonometry, determine the angle α knowing that the component along b-b′ is 30 lb. (b) Whatis the corresponding value of the component along a-a′ ?

SOLUTION

Using the triangle rule and the Law of Sines (a) sin α sin 40° = 30 lb 50 lb sin α = 0.3857

α = 22.7°
(b)

α + β + 40° = 180° β = 117.31°
Faa′ 50 lb = sin β sin 40°
 sin β  Faa′ = 50 lb    sin 40°  Faa′ = 69.1 lb

8

PROBLEM 2.9
To steady a sign as it is being lowered, two cables are attached to the sign atA. Using trigonometry and knowing that α = 25°, determine (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the Law of Sines Have:

α = 180° − ( 35° + 25° )
= 120°

Then:

P R 360 N = = sin 35° sin120° sin 25° or (a) P = 489 N (b) R = 738 N

9...