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SOLUTION
The parallelogram law of addition and the triangular rule are shown in Figs. a and b , respectively.
Applying the law of consines to Fig. b,
Ans.
This yields
Thus, the direction of angle of measured counterclockwise from the positive axis, is
60° 95.19° 60° 155° Ans.
F
sin 700
sin 45°
497.01 N 497 N
700 2 450 2 2(700)(450) cos 45°
If and , determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
60° 450 N
x
y
700 N
F
15
Ans: FR = 497 N f = 155 °
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
The parallelogram law of addition and the triangular rule are shown in Figs. a and b , respectively.
Applying the law of cosines to Fig. b,
Ans.
Applying the law of sines to Fig. b , and using this result, yields
u = 45.2° Ans.
sin (90° + u) 700
sin 105°
= 959.78 N = 960 N
F = 2500 2 + 700 2 - 2(500)(700) cos 105°
If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.
x
y
700 N
F
u 15
Ans: F = 960 N u = 45.2°
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SOLUTION
Parallelogram Law: The parallelogram law of addition is shown in Fig. a.
Trigonometry: Using the law of sines (Fig. b ), we have
Ans.
366 N Ans.
sin 45°
sin 75°
448 N
sin 60°
sin 75°
The vertical force acts downward at on the two-membered frame. Determine the magnitudes of the two components of F directed along the axes of and. Set 500 N.
F
F
C
B
A
Ans: FAB = 448 N FAC = 366 N
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Solve Prob. 2-4 with F = 350 lb.
SOLUTION
Parallelogram Law: The parallelogram law of addition is shown in Fig. a.
Trigonometry: Using the law of sines (Fig. b ), we have
Ans.
F (^) AC = 256 lb Ans.
F AC
sin 45°
sin 75°
FAB = 314 lb
FAB
sin 60°
sin 75°
F
C
B
A
30
45
Ans: FAB = 314 lb FAC = 256 lb
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Ans: ( F 1 )v = 2.93 kN ( F 1 ) u = 2.07 kN
Solution
Parallelogram Law. The parallelogram law of addition is shown in Fig. a , Trigonometry. Applying the sines law by referring to Fig. b.
( F 1 )v sin 45°
sin 105°
; ( F 1 )v = 2.928 kN = 2.93 kN Ans.
( F 1 ) u sin 30°
sin 105°
; ( F 1 ) u = 2.071 kN = 2.07 kN Ans.
Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components.
u
v
75
30
30
F 1 4 kN
F 2 6 kN
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Ans: ( F 2 ) u = 6.00 kN ( F 2 )v = 3.11 kN
Solution
Parallelogram Law. The parallelogram law of addition is shown in Fig. a , Trigonometry. Applying the sines law of referring to Fig. b ,
( F 2 ) u sin 75°
sin 75°
; ( F 2 ) u = 6.00 kN Ans.
( F 2 )v sin 30°
sin 75°
; ( F 2 )v = 3.106 kN = 3.11 kN Ans.
Resolve the force F 2 into components acting along the u and v axes and determine the magnitudes of the components.
u
v
75
30
30
F 1 4 kN
F 2 6 kN
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Ans: FR = 980 lb f = 19.4°
Solution
Parallelogram Law. The parallelogram law of addition is shown in Fig. a , Trigonometry. Applying the law of cosines by referring to Fig. b ,
FR = 28002 + 5002 - 2(800)(500) cos 95° = 979.66 lb = 980 lb Ans.
Using this result to apply the sines law, Fig. b ,
sin u 500
sin 95°
; u = 30.56°
Thus, the direction f of F R measured counterclockwise from the positive x axis is
f = 50 ° - 30.56° = 19.44° = 19.4° Ans.
Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
y
x
500 lb
800 lb
35
40
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SOLUTION
Parallelogram Law: The parallelogram law of addition is shown in Fig. a.
Trigonometry: Using law of cosines (Fig. b ), we have
Ans.
The angle can be determined using law of sines (Fig. b ).
Thus, the direction of F R measured from the x axis is
f = 33.16° - 30° = 3.16° Ans.
f
u = 33.16°
sin u = 0.
sin u 6
sin 100°
u
= 10.80 kN = 10.8 kN
FR = 28 2 + 6 2 - 2(8)(6) cos 100°
The plate is subjected to the two forces at A and B as shown. If , determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.
u = 60°
A
B
F (^) A 8 kN
F (^) B 6 kN
40
u
Ans: FR = 10.8 kN f = 3.16°
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SOLUTION
Ans.
Ans.
sin 40°
Fb sin 60°
; Fb = 26.9 lb
sin 40°
Fa sin 80°
; Fa = 30.6 lb
The force acting on the gear tooth is Resolve this force into two components acting along the lines aa and bb.
F = 20 lb.
80
60 a
a
b
b
F
Ans: Fa = 30.6 lb Fb = 26.9 lb
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The component of force F acting along line aa is required to be 30 lb. Determine the magnitude of F and its component along line bb.
SOLUTION
Ans.
Ans.
sin 80°
Fb sin 60°
; Fb = 26.4 lb
sin 80°
F
sin 40°
; F = 19.6 lb
80
60 a
a
b
b
F
Ans: F = 19.6 lb Fb = 26.4 lb
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Force F acts on the frame such that its component acting along member AB is 650 lb, directed from B towards A. Determine the required angle and the component acting along member BC. Set and u = 30°.
F = 850 lb
f (0° … f … 45°)
SOLUTION
The parallelogram law of addition and the triangular rule are shown in Figs. a and b , respectively.
Applying the law of cosines to Fig. b,
Ans.
Using this result and applying the sine law to Fig. b , yields
Ans.
sin (45° + f) 850
sin 30°
f = 33.5°
= 433.64 lb = 434 lb
F (^) BC = 28502 + 6502 - 2(850)(650) cos 30°
A
B
C
F
45
u
f
Ans: FBC = 434 lb f = 33.5°
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SOLUTION
Ans.
Ans.
sin 1.47°
sin u
; u = 2.37°
FR = 2 (30.85) 2 + (50) 2 - 2(30.85)(50) cos 1.47° = 19.18 = 19.2 N
sin 73.13°
sin (70° - u¿)
; u¿^ = 1.47°
F¿^ = 2 (20)^2 + (30) 2 - 2(20)(30) cos 73.13° = 30.85 N
Determine the magnitude and direction of the resultant of the three forces by first finding the resultant F ¿ = F 1 + F 2 and then forming F R = F ¿ + F 3.
F R = F 1 + F 2 + F 3
y
x
F 2 20 N
F 1 30 N
20
3
5 (^4) F 3 50 N
Ans: FR = 19.2 N u = 2.37° c
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Determine the design angle for strut AB so that the 400-lb horizontal force has a component of 500 lb directed from A towards C. What is the component of force acting along member AB? Take f = 40°.
u (0° … u … 90°)
SOLUTION
Parallelogram Law: The parallelogram law of addition is shown in Fig. a.
Trigonometry: Using law of sines (Fig. b ), we have
Ans.
Thus,
Using law of sines (Fig. b )
FAB = 621 lb Ans.
FAB
sin 86.54°
sin 40°
c = 180° - 40° - 53.46° = 86.54°
u = 53.46° = 53.5°
sin u = 0.
sin u 500
sin 40° 400
A
C
B
400 lb u
f
Ans: u = 53.5° FAB = 621 lb
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SOLUTION
Parallelogram Law: The parallelogram law of addition is shown in Fig. a.
Trigonometry: Using law of cosines (Fig. b ), we have
The angle can be determined using law of sines (Fig. b ).
f = 38.3° Ans.
sin f = 0.
sin f 400
sin 30°
f
FAC = 2400 2 + 600 2 - 2(400)(600) cos 30° = 322.97 lb
Determine the design angle between struts AB and AC so that the 400-lb horizontal force has a component of 600 lb which acts up to the left, in the same direction as from B towards A. Take u = 30°.
f (0° … f … 90°) (^) A
C
B
400 lb u
f
Ans: f = 38.3°
