Find the angle between two forces of equal magnitudes when the magnitude of their resultant is also equal to the magnitude of either of these forces.
Understand the Problem
The question asks to determine the angle between two forces of equal magnitude, given that the magnitude of their resultant force is equal to the magnitude of either of those individual forces. We will use vector addition principles and trigonometric relationships to find the angle.
Answer
$120^{\circ}$
Answer for screen readers
The angle between the two forces is $120^{\circ}$.
Steps to Solve
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Define variables and state the given information Let the magnitude of each force be $F$. Thus $|F_1| = |F_2| = F$. The magnitude of the resultant force is also $F$, so $|R| = F$. Let the angle between the two forces be $\theta$.
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Use the law of cosines to find the magnitude of the resultant force The magnitude of the resultant force $R$ of two forces $F_1$ and $F_2$ can be found using the law of cosines: $R^2 = F_1^2 + F_2^2 + 2|F_1||F_2|cos(\theta)$
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Substitute the given values into the equation from Step 2 Since $|F_1| = |F_2| = F$ and $|R| = F$, substitute these values into the equation: $F^2 = F^2 + F^2 + 2(F)(F)cos(\theta)$
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Simplify and solve for $cos(\theta)$ $F^2 = 2F^2 + 2F^2cos(\theta)$ $-F^2 = 2F^2cos(\theta)$ $cos(\theta) = -\frac{F^2}{2F^2}$ $cos(\theta) = -\frac{1}{2}$
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Find the value of $\theta$ To find the angle $\theta$, take the inverse cosine of $-\frac{1}{2}$: $\theta = cos^{-1}(-\frac{1}{2})$ $\theta = 120^{\circ}$ or $\frac{2\pi}{3}$ radians
The angle between the two forces is $120^{\circ}$.
More Information
When two forces of equal magnitude have a resultant force of the same magnitude, they must be oriented $120^{\circ}$ apart. This geometric arrangement leads to an equilateral triangle when visualizing the forces as vectors.
Tips
A common mistake is to incorrectly apply the law of cosines or to make errors in the algebraic simplification. Another mistake is forgetting that the angle is the inverse cosine of -1/2 and not 1/2.
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