# The magnitude of the resultant of two forces each having magnitude F_0 and an angle between them is 120 degrees, is xF. Then x is 1 2 3 sqrt(2)

## Question

The magnitude of the resultant of two forces each having magnitude $F_{0}$ and an angle between them is $120^{\circ}$, is $\mathrm{xF}$. Then $\mathrm{x}$ is

- 1
- 2
- 3
- $\sqrt{2}$

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## Answer

**:**

To solve for $\mathrm{x}$, we need to determine the magnitude of the resultant force when two forces of magnitude $F_{0}$ act at an angle of $120^{\circ}$ to each other.

The formula for the resultant of two forces with magnitudes $F_{0}$ forming an angle $\theta$ is given by:

$$ F_{\text {resultant }}^2 = F_{0}^2 + F_{0}^2 + 2 F_{0}^2 \cos \theta $$

In our specific case, $\theta = 120^{\circ}$, so let’s plug in these values:

$$ F_{\text {resultant }}^2 = 2 F_{0}^2 (1 + \cos 120^{\circ}) $$

Recall that $\cos 120^{\circ} = -\frac{1}{2}$, hence:

$$ F_{\text {resultant }}^2 = 2 F_{0}^2 \left(1 - \frac{1}{2}\right) $$

This simplifies to:

$$ F_{\text {resultant }}^2 = 2 F_{0}^2 \left(\frac{1}{2}\right) = F_{0}^2 $$

Thus, the magnitude of the resultant force is:

$$ F_{\text{resultant}} = \sqrt{F_{0}^2} = F_{0} $$

Since we have $F_{\text {resultant}} = \mathrm{x} F = F_{0}$, it follows that:

$$ \mathrm{x} = 1 $$

**Therefore, the correct option is ( \boxed{1} ).**

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