Question

1 forces in vector component form (10 pts)
(problem 5.1.19)
two ropes are attached to a tree, and forces of \\( \vec{f}_{1}=2.0 \hat{i}+4.0 \hat{j} \mathrm{~n} \\) and \\( \vec{f}_{2}=3.0 \hat{i}+6.0 \hat{j} \mathrm{~n} \\) are applied. the forces are coplanar (in the same plane).
(a) what is the resultant (net force) of these two force vectors, in vector component form? (5 pts)
(b) find the magnitude and direction of this net force. (5 pts)

Explanation:

Step1: Find the resultant force vector

To find the resultant force vector \(\vec{F}_{net}\), we add the corresponding components of \(\vec{F}_{1}\) and \(\vec{F}_{2}\).
If \(\vec{F}_{1}=a_{1}\hat{i}+b_{1}\hat{j}\) and \(\vec{F}_{2}=a_{2}\hat{i}+b_{2}\hat{j}\), then \(\vec{F}_{net}=(a_{1} + a_{2})\hat{i}+(b_{1}+b_{2})\hat{j}\).
Here, \(a_{1} = 2.0\), \(b_{1}=4.0\), \(a_{2}=3.0\), \(b_{2}=6.0\).
So, \(\vec{F}_{net}=(2.0 + 3.0)\hat{i}+(4.0+6.0)\hat{j}=5.0\hat{i}+10.0\hat{j}\)

Step2: Find the magnitude of the net force

The magnitude of a vector \(\vec{F}=x\hat{i}+y\hat{j}\) is given by \(|\vec{F}|=\sqrt{x^{2}+y^{2}}\).
For \(\vec{F}_{net}=5.0\hat{i}+10.0\hat{j}\), \(x = 5.0\) and \(y=10.0\)
\(|\vec{F}_{net}|=\sqrt{(5.0)^{2}+(10.0)^{2}}=\sqrt{25 + 100}=\sqrt{125}=5\sqrt{5}\approx11.2\space N\)

Step3: Find the direction of the net force

The direction \(\theta\) of a vector \(\vec{F}=x\hat{i}+y\hat{j}\) is given by \(\tan\theta=\frac{y}{x}\)
For \(\vec{F}_{net}=5.0\hat{i}+10.0\hat{j}\), \(\tan\theta=\frac{10.0}{5.0}=2\)
\(\theta=\arctan(2)\approx63.4^{\circ}\) above the \(x -\) axis (since both \(x\) and \(y\) components are positive)

Answer:

a. The resultant force vector is \(\vec{F}_{net}=5.0\hat{i}+10.0\hat{j}\space N\)
b. The magnitude of the net force is approximately \(11.2\space N\) and the direction is approximately \(63.4^{\circ}\) above the \(x -\) axis.