Question

Two forces each of magnitude 'f' and making an angle $\phi$ act on a body, then the resultant force is

• A. $f\sqrt{2(1-sin\phi)}$
• B. $f\sqrt{2(1+sin\phi)}$
• C. $2f.sin(\frac{\phi}{2})$
• D. $2f.cos(\frac{\phi}{2})$

Answer 1

Answer: (D)

$2f.cos(\frac{\phi}{2})$

Answer 2

The magnitude of the resultant force R of two forces of magnitudes $F_1$ and $F_2$ acting at an angle $\theta$ is given by the formula:

$R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta}$

In this problem, the two forces each have a magnitude $f$, so $F_1 = f$ and $F_2 = f$. The angle between them is $\phi$, so $\theta = \phi$.

Substituting these values into the formula:

$R = \sqrt{f^2 + f^2 + 2(f)(f)\cos\phi}$ $R = \sqrt{2f^2 + 2f^2\cos\phi}$ $R = \sqrt{2f^2(1 + \cos\phi)}$

Now, we use the trigonometric half-angle identity for cosine, which states that $1 + \cos\phi = 2\cos^2(\frac{\phi}{2})$.

Substitute this identity into the expression for R:

$R = \sqrt{2f^2(2\cos^2(\frac{\phi}{2}))}$ $R = \sqrt{4f^2\cos^2(\frac{\phi}{2})}$ $R = \sqrt{(2f\cos(\frac{\phi}{2}))^2}$ $R = |2f\cos(\frac{\phi}{2})|$

Since the magnitude of the resultant force is always non-negative, and assuming the angle $\phi$ is between 0 and $\pi$ (inclusive), $\frac{\phi}{2}$ will be between 0 and $\frac{\pi}{2}$ (inclusive). In this range, $\cos(\frac{\phi}{2})$ is non-negative. Thus, $|\cos(\frac{\phi}{2})| = \cos(\frac{\phi}{2})$.

So, the magnitude of the resultant force is:

$R = 2f\cos(\frac{\phi}{2})$