PROBLEM 2-44 (page 41, 13th edition) . The magnitude of the resultant force acting on the bracket is 400 N. Determine the magnitude of F . Take Ø = 30°. Disregard the u axis. y u F2 = 650 N 1 4 5 3 F1 45° 1¢ x 45° F3= 500 N Figure: 02_P044-045-046 Copyright @2013 Pearson Education, publishing as Prentice Hall
2-44 (page 41) 13th edition COMPLETION OF PROBLEM · Determine F, so that FD = 400N : R F. =! F · Cartesian Vector Method (suppressing units) F= F i + Fp j F = ! F = "650 $5( +F cos 30° + 500 cos 450 Rx X 3# " % + F sin 30° ( 500 sin 450 Ry y $ =1 #5 & F = ! F = 650 2 'Ry Fp = VFR + F L'Rx 400= (!390+F1cos30°+500cos45°)2 +(520+F 1 sin30° ! 500 sin45°)2 · Solution Solve using Solver on a TI-83 Plus calculator: or F = ! 417N - F = 314N The negative sign indicates for that answer that F, must act in the opposite direction to that shown in the diagram.
PROBLEM 2-51 (page 41, 13th edition) PROBLEM 2-40 (page 40, 12th edition) · Determine the magnitude and direction measured counterclockwise from the positive x-axis of the resultant force of the three forces acting on the ring A. Take F = 500N and ! = 20°. y F 600 N 3 5 4 400 N 1 30° 1 A x A PROB02_039-040.jpg Copyright @ 2010 Pearson Prentice Hall, Inc.