Two forces F1 and F2 are used to pull a car, which met an accident. The angle between the two force is θ. Find the value of θ for the resultant force is equal to \(\sqrt{(F_1^2 + F_2^2)}\)
θ = 90
Understanding Resultant Force Calculation
When two forces, say F1 and F2, act on an object, their combined effect is represented by a single force called the resultant force, often denoted by R. The magnitude of this resultant force depends on the magnitudes of the individual forces (F1 and F2) and the angle (θ) between them.
Formula for Resultant Force
The general formula to calculate the magnitude of the resultant force (R) of two forces acting at an angle θ is derived using the parallelogram law of vector addition:
$$ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta} $$
Applying the Given Condition
In this specific problem, we are given a condition for the resultant force:
$$ R = \sqrt{(F_1^2 + F_2^2)} $$
Step-by-Step Solution
To find the angle θ, we need to equate the general formula for the resultant force with the given condition:
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Set the two expressions for R equal:
$$ \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta} = \sqrt{(F_1^2 + F_2^2)} $$
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To simplify, square both sides of the equation:
$$ F_1^2 + F_2^2 + 2F_1F_2 \cos\theta = F_1^2 + F_2^2 $$
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Subtract (F12 + F22) from both sides:
$$ 2F_1F_2 \cos\theta = 0 $$
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Assuming that the forces F1 and F2 are non-zero (as they are used to pull a car), we can divide by 2F1F2:
$$ \cos\theta = \frac{0}{2F_1F_2} $$
$$ \cos\theta = 0 $$
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Now, we need to find the angle θ whose cosine is 0. From trigonometry, we know that the cosine function is zero at 90 degrees (or $\frac{\pi}{2}$ radians).
$$ \theta = 90^\circ $$
Conclusion
Therefore, the angle θ between the two forces F1 and F2 for their resultant force to be equal to \(\sqrt{(F_1^2 + F_2^2)}\) is 90°.
Options Analysis
- θ = 0°: If θ = 0°, cos(0) = 1, and R = \(\sqrt{F_1^2 + F_2^2 + 2F_1F_2}\) = \(F_1 + F_2\).
- θ = 45°: If θ = 45°, cos(45) = \(\frac{1}{\sqrt{2}}\), and R = \(\sqrt{F_1^2 + F_2^2 + \sqrt{2}F_1F_2}\).
- θ = 90°: If θ = 90°, cos(90) = 0, and R = \(\sqrt{F_1^2 + F_2^2 + 0}\) = \(\sqrt{F_1^2 + F_2^2}\). This matches the given condition.
- θ = 125°: If θ = 125°, cos(125) is negative, resulting in R < \(\sqrt{F_1^2 + F_2^2}\).
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