Question

Chapter 01, Problem 55 todit. Solution 1.55 presents the solution to a problem that is similar to this one. Vector A has a magnitude of 128 units and points 39.0 north of west. Vector B points 67.0 east of north. Vector C points 18.0 west of south. These three vectors add to give a resultant vector that is zero. Using components, find the magnitudes of (a) vector and (b) vector c. (a) Number (b) Number

Answer 1

North West & Gr East South OF = A 0339°, OE - ASin39°, OH = B Cos67° OI = B Sim67°, OD = C. Cosicº, OG = c Sim 18° I : resultant o à B & C and is zero so Sum of them X- components & T Components are separately Rexo. Sum of x-components of Ā B & T = 0 => ST-OF+OI-OG = 0 - ACOS 39° + B Sim67" - simidº - 0 → -128 C0339° + B SIM 67° = C Sim 18° D C = (128 COS 39° +B Sin670) ->0 Sim 10º =

Sum of Y-componenent of A, B, C =0 OE tOH-OD = 0 → 128 Sin 39°+ B.C0367°– C COS 18 = 0 => C = 128 Sin 39° +BCO367° --- CACOS 18A Comparing eg. & - G122 Cos390 + B8 in 67°) 128 Sin39° + B COS67° Sim 18° cos18° => -128 C0339º. cosidº +B Sin 67Cos18°=128 Sin 39. Simpo B C0867. Simido - B&Sm67? Cos12–(@367.5mg'} = 128308339.cogien + Sin39. Siri 18° trigonometrical identities we can write- B im (67-18) = 128 Cos (39° -18°) B=128 Cos210 B = 158.34 units Masin 49029 I put the value os B in equ. ④ - C = 6-128 C03399 H158.34 Sim 679) - Sinigo = 149.759 C = 149.76 units