Question

There are two forces F1 = 10 N and F2 = 20 N and the angle between them is 60° and the value of the Resultant (F2-F1) = 10√3. Find the angle alpha between the resultant and the vector.​

Answer 1

Answer:

Explanation:

   f_1 = 10\ N.

   f_2 = 20\ N.

   Angle between \vec f_1 and \vec f_2, \theta=60^\circ.

Let \vec f_1  be along the positive x axis direction, then \vec f_2 is along the direction 60^\circ with respect to the positive x axis direction.

Assuming,

\hat i,\ \hat j are the unit vector along the positive x and y axis direction.

In unit vector notation, \vec f_1 and \vec f_2 are given as,

\vec f_1 = f_1\ \hat i=10\ \hat i\ N.\\\vec f_2 = f_2\cos(60^\circ)\ \hat i+f_2\sin(60^\circ)\ \hat j\\=20\cos(60^\circ)\ \hat i+20\sin(60^\circ)\ \hat j\\=(10\ \hat i\ +\ 17.32\ \hat j)\ N.

Therefore,

\vec f_2-\vec f_1=(10\hat i+17.32\hat j)-(10\hat i)=17.32\hat j\ N.

The resulting vector, \vec f_2-\vec f_1 is along the positive y axis direction, therefore its direction with respect to positive x axis is 90^\circ, if \alpha is the angle along the direction of \vec f_2-\vec f_1, then \alpha = 90^\circ.

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Answer 2

Answer:as for angle between resultant and F1 ....it is tan inverse of (root 3 )/5

Explanation: We know that *tanA (assume A to be alpha)= asinA/b +acosA

here a is F1 and b is F2

So tanA =5√3 /(20+5)

tanA = √3/5

So A is tan inverse of √3/5

Thus angle between two vectors is 19.08°