Question

There are two unitless vectors:

Answer 1

Answer:

$200.38^0$

Explanation:

Given the forces:

F1 = 8.92 i + 17.37 j

F2 = 8.31 i - 10.97 j

If a third vector is added to them such that they add up to to the null vector as F1 + F2 + F3 = 0

then to get F3:

F3 = -F2-F1

F3 = -(8.31 i - 10.97 j)-(8.92 i + 17.37 j )

F3 = -8.31 i + 10.97 j-8.92 i - 17.37 j

F3 = -8.31i-8.92i+10.97j-17.37j

F3 = -17.23i-6.4j

from the vector:

x = -17.23 and y = -6.4

angle of the third vector with respect to the +x-axis is expressed as:

$\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-6.4}{-17.23}\\\theta = tan^{-1}0.3715\\\theta = 20.38^0$

Hence the angle the vector makes with the x axis will be $\theta = 180+20.38 = 200.38^0$