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.) 6.50 A -18.0 cm 18.0 cm х х 13.0 A 18.0 cm 39.0 A Top wire magnitude direction N/m º counterclockwise from the +x-axis Bottom left wire magnitude N/m direction º counterclockwise from the +x-axis Bottom right wire magnitude direction N/m º counterclockwise from the +x-axis

Answer 1

2 = 6.50 A 12= 13.0A 13= 39.0A d 2 d 3 d = 18.0cm=0.18m Top wire: i) F22 3,2 x B12 IL CÂx Mo Iz (cosson - sino f atid h2² Mo 3, IzL (cosso ĵ t sinzo in end F z = 40x10x 6.50X13:0 (cosso åt sino in 2T (0.18) (4.694x10.9 + 8.1317109 Î ) Fis: 3,2 x 1,LC) Mols (Cossi + Sin389) 2nd - MolyP3L ( 38j + Sin 38 ] 2 Td fis L 4TH X 6.50x 39 0 [casso i sin 30 î] 27 (0.18) : (- 14.083x105 + 24-343x01) (-9.39 410s it 32.5x105 f)

Magnitude: = V(29.39%+(2253 vio? z 33-85x105 Nm. = 106012 Direction: 188 - Tamil counter 32.5 9:39 clockwise from +x-anis Bottom left wire: Bay F1 = = I l x Bzi 2 IZL(-Å) x Mo II (Cossoi-Sin 307) zind 「かい Mo3,32 (-Sin 30 î - Corso f ) 2Tid 45x10x 6.50813-0 (-Sin 30 î. Cos 30 f ) 2T (0-187 (-4.6940105 ņ - 801311105 ) ㄷ 7/23 = I₂ L x B 23 = I2L(-E) x Mo33 (ů) 2nd Moly I3 zod 471xo'x 63.0x39.0 (î) 2000.15) que la v -5 56.33 X 10 (1)

3 F23 (51064X103 – 801318165 ☺) Magnitude: F 52.28x16 Nim L Direction : 0 = 368 - Tan/ 8-131 51.64 02-351° counter clockwire from tX aris Bottom right wire: B31 F317 2 3 Islx B = 93L (-(c) x 1-7 332 Modi (cosao i + Sin 30 j) 2nd ( Sim 38 î – Cos zo Ĵ ) Mo I, I3 2Td 4T HÓ'x 6.50x390 (Sin38 î– Cos 30 ſ) 271 C0-18) = (14.083465 - 24.393x1655) A 32 = Jix B32 = 33 LEÅ) x Holz (W ㄷ 2nd Mo 1273 (-1)= 411X10 x 13:0x3900 € zad sharorlar 211 (0.18) 56.33 X 105 s

J e Film : (-42541669 -243720*5) Magnitude: F3 -5 48.79810 Nlm L Oz Direction: Oz=180 + Tañ / 24.39 42.25 F3 03=2169 counter clockwise from te-axis.