Q 3 continoun
sing
the Tur forcos, vec(F)_(AC) and vec(F)_(BC), ore seperoted by 120\deg .
The net force magnitude
Sinco cos(180\deg )=-(1)/(2).
F_(C)=\sqrt({:((2.5\times 10^(-5))/(L))^(2)+((3.75\times 10^(-5))/(L))^(2)-(2.5\times 10^(-5))/(L)*(3.75.10^(-5))/(L)))
Simplify:
F_(c)=(1)/(L)\sqrt((2.5\times 10^(-5))^(2)+(3.75\times 10^(-5))^(2)-(2.5\times 10^(-5))(3.75\times 10^(-5)))
F_(C)=(1)/(10.9375\times 10^(-10))
F_(C)=(3.306\times 10^(-5))/(L)= mognitude
Direction! the net forco points slightly tomerd
wire B, bosed on the realtive magnitudes of
EAs end inBC.