Question

Three forces are applied on a particle of mass m= 15 kg located at the origin, as show below. The magnitudes are given to be FA = 15 N, FB = 35 N, Fc = 22 N, and the directions are 04 = 53°, measured from the negative x-axis, 03= 32°, measured from tl positive x-axis, and the direction of Fc is along the negative y-axis. a. Calculate the magnitude the net force on the particle. b. Calculate the direction of the net force on the particle relative to the x-axis. c. Calculate the magnitude and direction of the acceleration of the particle. d. Suppose FA can be changed while Fg and Fc are left fixed. Calculate the magnitude and direction FA so that the particle is in equilibrium.

Answer 1

B FA = 15N FB = 35N Fe=22N QA = 59°08= 32° > On Resolving the vector Horizontal component (FA) re (FA na FA COS OA = '15x605 53° " (*=9,027 --- along negative x-axis. vertical component Dy = fa sino a = 15% sin 53 _ .... (FOy = 11.979 --- along positive y-cixis & On resolving the vector fo : Horizontal component _(18)x= fg.Cos 0 g = 35% cOS (32) = 29.681 N --. along positive xrari's Vertical component (fby = fp Sin OB = 35% sin (32) 3 18.541 N --, along positive yeaxis

". Net e oro along ya axis. fy (FB) x + (), -fc L = 18.547+ 11.979 - 22 Fy = 8.526 N --- along positive y-axis Net force along x-axis Fx = (8) + (Fase _ = 29.68.1+(-9.027) - 20.654 N--. along positive x-axi's a'. Resultant force (E) _FEV 2 +F42 ... f = |(20.65LÝ +(8.526) :F = 22.346 N b) Direction of Net foree from positive a- axis. ..0 = tan r fy tan / 8.526 FX 20.654 10 = 22.43° 1 - from positive x-axis. -- _magnitude q_acceleration on particle - 227346 = 15 a = 1.489 m/s2 acceleration is along the direction of Direction f OY CO,

dy In order to be in equilibrium s F A_must be equal and opposite q yesultant q angle between fo & Fc is o = 90+32 =1220 ... Fragtf+ 2 fe fc Cos(122) - FREN (35) +(229 + 2 35822860.529) FR = 29.905 N T In order to be in equilibrium. magnitude of FÅ = FR - = 29.905 N in opposite direction of FR