Question

f1=10N and F2=20N and angle between them is 60° find f2-f1 and tan alpha​

Answer 1

Answer:

tan alpha is root 3 f2 -f1 = 10N

Explanation:

see attached ans

Answer 2

Answer:

• $\vec f_2-\vec f_1=17.32\hat j\ N.$

• $|\vec f_2-\vec f_1|=17.32\ N.$

• $\tan\alpha = \infty$.

Explanation:

Given:

• $f_1 = 10\ N$.

• $f_2 = 20\ N$.

• 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 $\alpha = 90^\circ$.

which gives,

$\tan\alpha = \tan90^\circ = \infty$.