A force of \(20 \mathrm{N}\) is directed at an angle of \(60^{\circ}\) above the \(x\) -axis. A second force of \(20 \mathrm{N}\) is directed at an angle of \(60^{\circ}\) below the \(x\) -axis. What is the vector sum of these two forces?

To calculate the components of force vectors, we can use the following formulas: - For horizontal component: \(F_x = F \cdot \cos(\theta)\) - For vertical component: \(F_y = F \cdot \sin(\theta)\) For the first force, \(F_1 = 20 N\) and the angle above the x-axis is \(\theta_1 = 60^{\circ}\): \(F_{1x} = 20 \cdot \cos(60^{\circ}) = 20 \cdot \frac{1}{2} = 10 N\) \(F_{1y} = 20 \cdot \sin(60^{\circ}) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} N\) For the second force, \(F_2 = 20 N\) and the angle below the x-axis is \(\theta_2 = 60^{\circ}\): \(F_{2x} = 20 \cdot \cos(60^{\circ}) = 20 \cdot \frac{1}{2} = 10 N\) \(F_{2y} = -20 \cdot \sin(60^{\circ}) = -20 \cdot \frac{\sqrt{3}}{2} = -10\sqrt{3} N\) (The negative sign indicates that the force points downward)

Now we add the components of each force vector to determine the sum: \(F_{x} = F_{1x} + F_{2x} = 10 N + 10 N = 20 N\) \(F_{y} = F_{1y} + F_{2y} = 10\sqrt{3} N - 10\sqrt{3} N = 0 N\)

Since \(F_y = 0 N\), the resultant vector lies entirely in the horizontal direction and has a magnitude of: \(F = \sqrt{F_{x}^{2} + F_{y}^{2}} = \sqrt{(20 N)^{2} + (0 N)^{2}} = 20 N\) Because the force is acting horizontally, the angle \(\theta = 0^{\circ}\) relative to the x-axis.

The vector sum of the two forces is \(20 N\) directed horizontally along the x-axis with an angle of \(0^{\circ}\).