2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-1. If 0 = 60° and F = 450 N, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. y F x 15° 700 N 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, FR= V7002 + 4502 - 2(700)(450) cos 45° = 497.01 N = 497 N Ans. This yields FR 60°-150=45' F=450N $82 260° x ? = 95.19° 150 0 700 sin x sin 45° 497.01 Thus, the direction of angle $ of FR measured counterclockwise from the positive x axis, is 700N (a) ?= a + 60° = 95.19° + 60° = 155° Ans. 700N 4500 F=450N x FR (b) Ans: FR = 497 N f =155° 22
@ 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-2. 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. y F u x 15º 1 700 N 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, F = 25002 + 7002 - 2(500)(700) cos 105° = 959.78 N = 960 N Ans. y 90'+15"=1050 7F FR=500N Applying the law of sines to Fig. b, and using this result, yields 90°0 ---- sin (90° + u) sin 105° 700 959.78 u = 45.2° Ans. - x 150 700N (a) 700 N 171050 F FR=500N 90-0 (b) Ans: F = 960 N 11 = 45.2° 23
@ 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-3. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis. SOLUTION FR = 2(250)2 + (375)2 - 2(250)(375) cos 75° = 393.2 = 393 lb y F1 = 250 lb +30% + Ans. 393.2 250 sin 75° sin u u = 37.89° f= 360° - 45° + 37.89° = 353º 24 Ans. x - - a 45 F 2 = 375 lb y Fs=25046 x FR 75 150 F. = 375 46 Ans: FR = 393 lb f = 353°