3. three boats exert forces on a mooring hook as shown. find the resultant force on the hook.

Explanation:

Step1: Resolve forces into x - components

For the 150 N force, $F_{1x}=0$; for the 500 N force, $F_{2x}=- 500\cos40^{\circ}$; for the 420 N force, $F_{3x}=420\cos60^{\circ}$. $F_{x}=0 - 500\cos40^{\circ}+420\cos60^{\circ}$ $F_{x}=-500\times0.766 + 420\times0.5=-383+210=-173$ N

Step2: Resolve forces into y - components

For the 150 N force, $F_{1y}=150$; for the 500 N force, $F_{2y}=500\sin40^{\circ}$; for the 420 N force, $F_{3y}=420\sin60^{\circ}$. $F_{y}=150 + 500\sin40^{\circ}+420\sin60^{\circ}$ $F_{y}=150+500\times0.643+420\times0.866=150 + 321.5+363.72 = 835.22$ N

Step3: Calculate the magnitude of the resultant force

$R=\sqrt{F_{x}^{2}+F_{y}^{2}}$ $R=\sqrt{(-173)^{2}+835.22^{2}}=\sqrt{29929+697584.4484}=\sqrt{727513.4484}\approx853$ N

Step4: Calculate the direction of the resultant force

$\theta=\arctan\left(\frac{F_{y}}{F_{x}}\right)$ $\theta=\arctan\left(\frac{835.22}{-173}\right)\approx101.6^{\circ}$ counter - clockwise from the negative x - axis

Answer:

The magnitude of the resultant force is approximately 853 N and its direction is approximately $101.6^{\circ}$ counter - clockwise from the negative x - axis.