Recent content by godmoktail

yes, the x-axis

And since the components are -3.78 and 0.67, then the direction is: arctan(-.67/-3.78) = 10 deg, no? thanks in advance

Ok, here is my retry. r_vec31 = -2i - j r_vec32 = -2j r_hat31 = (-2i - j) / sqrt(5) r_hat32 = (-2j) / 2 = -j Fvec_31 = 4.23 * r_hat31 N = -3.78i - 1.89j Fvec_32 = -1.22 * r_hat32 N = 1.22j Fvec_net = (-3.78i - 0.67j) Fvec_mag = 3.84 N So the magnitude would be 3.84N, NO?

I take Fvec_net as the Force vector 3.78i - 0.55j (with magnitude 0.67N) which I wrote in component notation

Are you saying that I should have entered 0.67N? Either way, I would like to ensure that my magnitude calculations are correct, regardless of what I entered on my homework software. If they asked for the Force vector, then I would have to multiply the magnitude by r^ (r hat), no?

So r_13 as a unit vector would be squareroot(5), and r_23 would be squareroot(2), no?

Homework Statement Homework Equations F = | kqQ / r*r | The Attempt at a Solution Magnitude: F_net = F_1on3 + F_2on3 F_1on3 = (9E9)(70E-6)(15E-6) / ( (sqrt(5))^2 ) = 1.89N F_2on3 = (9E9)(-36E-6)(15E-6) / ( (2)^2 ) = -1.215N ~ -1.22N F_net = 1.89 - 1.22 = 0.67N Direction =...