Two forces have the same module F, form an angle alpha<90° and are applied to the same point. The modulus of the resultant force is:

Hi everyone!

The answer for this one is 2 * F * cos (alpha/2)
I tried solving using the formula F = (F1² + F2² + 2 * F1 * F2 * cos alpha) ^ 1/2
But i think i got lost in my calculation
What would be a step by step solution?

The problem states that two forces, denoted by F, are applied to the same point and form an angle \alpha, where \alpha < 90^{\circ}. The question is asking for the modulus of the resultant force, denoted by c.

We can use the concept of the “law of cosines” to calculate the net force. The law of cosines states that in a triangle with sides a, b and c and an angle of \alpha between sides a and b, the following equation holds:

c^2 = a^2 + b^2 - 2ab \cos(\alpha)

In this case, a and b are the magnitudes of the two forces, and c is the magnitude of the net force. So, we can rewrite the equation as:

c^2 = F^2 + F^2 - 2F^2 \cos(\alpha)

Now we can simplify the equation and find the magnitude of the net force:

c = \sqrt{2F^2 - 2F^2 \cos(\alpha)}

We know that \cos(\frac{\alpha}{2}) = \sqrt{\frac{1+\cos(\alpha)}{2}}
so we can substitute the last expression and get:

c = 2F \cos(\frac{\alpha}{2})

So, the formula for the modulus of the resultant force is 2F \cos(\frac{\alpha}{2}).

In summary, we used the law of cosines and trigonometric identities to find the modulus of the resultant force resulting from the application of two forces at a certain angle (you are expected to memorize the common laws and identities for the IMAT exam). The magnitude of the net force is given by the formula c = 2F \cos(\frac{\alpha}{2}).