Mecánica Vectorial Para Ingenieros

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)

R = 8.4 kN

We measure:

α = 19°
R = 8.4 kN

1

19°

PROBLEM 2.2
The cable stays AB and AD help support pole AC. Knowing that the
tension is 500 N in AB and 160N in 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:

α = 51.3°, β = 59°

(a)

(b)

We measure:

R = 575 N, α = 67°
R = 575 N

2

67°

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

SOLUTION
(a)

(b)

R = 37 lb, α = 76°

We measure:

R = 37 lb

3

76°

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
magnitudeand direction of their 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

4

86.5°

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 theresultant R of the
forces exerted by the rods on the lever is to be vertical, (b) the
corresponding magnitude of R.

SOLUTION

Graphically, by the triangle law
F2 ≅ 108 N

We measure:

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)5

R = 75.0 N

PROBLEM 2.6
Two control 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 ofthe component along b-b′ ?

SOLUTION

Using the triangle rule and the Law of Sines
sin β
sin 40°
=
35 lb
50 lb

(a)

sin β = 0.44995

β = 26.74°
α + β + 40° = 180°

Then:

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) Usingtrigonometry, determine the angle α knowing that the
component along b-b′ is 30 lb. (b) What is 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
at A. 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° −...