By Russell Charles Hibbeler

Teacher recommendations handbook (ISM) for Engineering Mechanics Statics, thirteenth version (c2013) by way of Russell Charles Hibbeler. (ch 01-08)

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**Additional info for Engineering Mechanics Statics - Intructor Solutions manual (ch 01-08)**

**Sample text**

Two forces act on the screw eye. If F1 = 400 N and F2 = 600 N, determine the angle u(0° … u … 180°) between them, so that the resultant force has a magnitude of FR = 800 N. F1 u SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. Applying law of cosines to Fig. 5° Ans. F2 2–22. Two forces F1 and F2 act on the screw eye. If their lines of action are at an angle u apart and the magnitude of each force is F1 = F2 = F, determine the magnitude of the resultant force FR and the angle between FR and F1.

B), we have F FAB 500 = sin 60° sin 75° FAB = 448 N C Ans. FAC 500 = sin 45° sin 75° FAC = 366 N 30Њ Ans. *2–8. Solve Prob. 2-7 with F = 350 lb. B 45Њ SOLUTION A Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have F FAB 350 = sin 60° sin 75° FAB = 314 lb C Ans. FAC 350 = sin 45° sin 75° FAC = 256 lb 30Њ Ans. 2–9. Resolve F1 into components along the u and v axes and determine the magnitudes of these components. v F1 F2 SOLUTION 150 N 30 30 Sine law: 105 F1v 250 = sin 30° sin 105° F1v = 129 N Ans.

U F A 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, FBC = 28502 + 6502 - 2(850)(650) cos 30° Ans. 64 lb = 434 lb Using this result and applying the sine law to Fig. 5° Ans. f 45Њ C 2–15. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal. FA u A SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig.