Question

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?

Short answer

The vector sum of the two forces is 20 N directed horizontally along the x-axis with an angle of 0 degrees.

Step-by-step solution

Calculate horizontal and vertical components for each force

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)

Add the corresponding components together to find the final vector components

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

Calculate the magnitude and angle of the resultant vector

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.

Final Answer

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