shows three ropes tied together in a knot. One of your friends pulls on a rope with 3.0 units of force and another pulls on a second rope with 5.0 units of force. How hard and in what direction must you pull on the third rope to keep the knot from moving?

Consider that the knot in the ropes as particle in static equilibrium. The below figure shows the three ropes tied together in a knot.

From the above figure the magnitude of the force $F_1$is 3 units and the magnitude of the force$F_2$ is 5 units. Let the magnitude of the third vector be $F_3$.

Angle made by ${\overrightarrow{F}}_1$with the positive x-axis,${\theta }_1=0^{°}$

Angle made by ${\overrightarrow{F}}_2$with the positive x-axis, ${\theta }_2={120}^{°}$

Angle made by ${\overrightarrow{F}}_3$with the negative x-axis,$\theta$

Now express the vectors using the unit vectors.

$F_1=3i\text{and}{\overrightarrow{F}}_2=-5\sin {30}^{°}i+5\cos {30}^{°}j$Since,${\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=0$

Rearrange the equation in terms of vector${\overrightarrow{F}}_3$

$\begin{array}{l}{\overrightarrow{F}}_3=-\left({\overrightarrow{F}}_1+{\overrightarrow{F}}_2\right)\\ {\overrightarrow{F}}_1=\left(3\right)\cos 0^{°}\widehat{i}+\left(3\right)\sin 0^{°}\widehat{j}\\ {\overrightarrow{F}}_2=-5\sin {30}^{°}\widehat{i}+5\cos {30}^{°}\widehat{j}\\ =-2.5\widehat{i}+4.33\widehat{j}\end{array}$

Therefore, the vector,$\overrightarrow{F_3}=-[3\widehat{i}-2.5\widehat{i}+4.33\widehat{j}]=-0.5\widehat{i}-4.33\widehat{j}$

Magnitude of the third force,

$F_3=\sqrt{{(-0.5)}^2+{(-4.33)}^2}$$=4.36$units

Therefore, the required force you must pull on the third rope to keep the knot from moving is 4.36 units

Direction of ${\overrightarrow{F}}_3$is,

$\theta ={\tan }^{-1}\left(\frac{-4.36}{-0.5}\right)={83.4}^{°}\ln 3^{\text{rd}}$quadrant

Hence, makes an angle of ${83.4}^{°}$below the negative x-axis.