Question

2 forces in magnitude - direction form (10 pts)
(problem 5.1.21)
two teenagers are pulling on ropes attached to a tree. the angle between the ropes is 30.0°. david pulls with a force of 400.0 n and stephanie pulls with a force of 300.0 n.
(a) find the vector component form of the net force. (5 pts)
(b) find the magnitude of the resultant (net) force on the tree and the angle it makes with davids rope. (5 pts)

Explanation:

Step1: Set up coordinate system

Let David's force vector \(\vec{F}_D\) be along the \(x -\)axis. So \(\vec{F}_D=(400.0,0)\) N. For Stephanie's force \(\vec{F}_S\), using the angle \(\theta = 30^{\circ}\), the \(x -\)component is \(F_{Sx}=300.0\cos(30^{\circ})\) and the \(y -\)component is \(F_{Sy}=300.0\sin(30^{\circ})\).

$$F_{Sx}=300.0\times\frac{\sqrt{3}}{2}=150\sqrt{3}\approx259.8$$

$$F_{Sy}=300.0\times\frac{1}{2} = 150$$

So \(\vec{F}_S=(259.8,150)\) N.

Step2: Find the net force vector \(\vec{F}_{net}\)

The \(x -\)component of the net force \(F_{netx}=F_{Dx}+F_{Sx}\) and the \(y -\)component \(F_{nety}=F_{Dy}+F_{Sy}\). Since \(F_{Dy} = 0\), \(F_{netx}=400.0 + 259.8=659.8\) N and \(F_{nety}=150\) N. So \(\vec{F}_{net}=(659.8,150)\) N.

Step3: Find the magnitude of the net force

Use the formula \(|\vec{F}_{net}|=\sqrt{F_{netx}^2+F_{nety}^2}\)

$$|\vec{F}_{net}|=\sqrt{(659.8)^2+(150)^2}=\sqrt{435336.04 + 22500}=\sqrt{457836.04}\approx676.7$$

N

Step4: Find the angle \(\alpha\) of the net force with David's rope

Use the formula \(\tan\alpha=\frac{F_{nety}}{F_{netx}}\)
\(\tan\alpha=\frac{150}{659.8}\approx0.227\), then \(\alpha=\arctan(0.227)\approx12.8^{\circ}\)

Answer:

(a) The vector component form of the net force is \((659.8,150)\) N.
(b) The magnitude of the resultant force is approximately \(676.7\) N and the angle it makes with David's rope is approximately \(12.8^{\circ}\)