Question

Find the magnitude (in N/m) and direction (in degrees counterclockwise from the +x-axis) of the force that each wire experiences in the figure by using vector addition. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) 5.50 A 118.0 cm - 18.0 cm 11.0 18.0 cm 27.5 A Top wire magnitude direction 6.72e-5 N/m o counterclockwise from the +x-axis Bottom left wire magnitude 3.61e-4 direction N/m o counterclockwise from the +x-axis Bottom right wire magnitude 1.68e-4 direction N/m o counterclockwise from the +x-axis

Answer 1

'Answert f = Mo I IZ 217 face (URX10 TIM/A) (5.50A) (27.5A) => 1.68X10 Nlm fag -- (untio ma (s.50A) (1) =>6.72x105 N/m 21 (0.18m) foc = (4x16 cm) (14) (27-5A) =>3.361X10 Nm Qa (0.18m) 2x (0.18m) For top wire fax=(fAB - fAc) sin so fax = -5.oux ios Nim fay=(fabrfac) (s.30%. fay= 2.036x1öNAM 4 FACE ALAB FA=& fax't tay 2 2 Fo 1-5.0yXTOS = 2.097 XióY Nim A = 1807 tañ' (Fay )=>180t tan 2.036 Xiay OA = 103.90° For bottom left wire fox = fec- for Cs Go => 3.095 xjö Nm Soy E-fae singo' => 5.82 Xiös nim. FB = /FBy + fby2 =) 3.08X10 N/m 08 = 360'+tañl chapters) -> 34941° right fcx = fac & 6o - foc -9.52X154 N/m fey =- fac singo' => -1.45x1őY N/m fc=2.918104 N/m Oc = 180° + tan'a (home) -> 209.98 for bottom wire t