We draw a rough sketch:

Now we determine the length of the resultant.s

We note that the triangle formed by the two force vectors and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (17)^{2} + (23)^{2} &= R^{2}\\ R &= \text{28,6}\text{ N} \end{align*}

To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}}{\text{adjacent side}} \\ \tan\alpha &= \frac{\text{23}}{\text{17}} \\ \alpha &= \tan^{-1}(\text{1,353}) \\ \alpha &= \text{53,53}\text{°} \end{align*}

The resultant force is then \(\text{28,6}\) \(\text{N}\) at \(\text{53,53}\)\(\text{°}\) to the positive \(x\)-axis.

We draw a rough sketch:

Now we determine the length of the resultant.

We note that the triangle formed by the two force vectors and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (\text{23,7})^{2} + (9)^{2} &= R^{2}\\ R &= \text{25,4}\text{ N} \end{align*}

To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}}{\text{adjacent side}} \\ \tan\alpha &= \frac{\text{9}}{\text{23,7}} \\ \alpha &= \tan^{-1}(\text{0,3797}) \\ \alpha &= \text{20,79}\text{°} \end{align*}

The resultant force is then \(\text{25,4}\) \(\text{N}\) at \(\text{20,79}\)\(\text{°}\) to the \(x\)-axis. However we actually give this as \(\text{159,21}\)\(\text{°}\) (i.e. \(\text{180}\)\(\text{°}\) minus \(\text{20,79}\)\(\text{°}\)) to keep with the convention we have defined of giving vector directions.

We note that we have more than two vectors so we must first find the resultant in the \(x\)-direction and the resultant in the \(y\)-direction.

In the \(x\)-direction we only have one vector and so this is the resultant.

In the \(y\)-direction we have three vectors. We can add these algebraically to find \(\vec{R}_{y}\):

\begin{align*} \vec{F}_{1} & = \text{+4}\text{ N} \\ \vec{F}_{2} & = -\text{3,3}\text{ N} \\ \vec{F}_{3} & = -\text{2,1}\text{ N} \end{align*}

Thus, the resultant force is:

\begin{align*} \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} & = (4) + (-\text{3,3}) + (-\text{2,1}) \\ & = -\text{1,4} \end{align*}

Now we use \(\vec{R}_{x}\) and \(\vec{R}_{y}\) to find the resultant.

We note that the triangle formed by \(\vec{R}_{x}\), \(\vec{R}_{y}\) and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (\text{2,3})^{2} + (-\text{1,4})^{2} &= R^{2}\\ R &= \text{2,69}\text{ N} \end{align*}

To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}}{\text{adjacent side}} \\ \tan\alpha &= \frac{\text{1,4}}{\text{2,3}} \\ \alpha &= \tan^{-1}(\text{0,6087}) \\ \alpha &= \text{31,33}\text{°} \end{align*}

The resultant force is then \(\text{2,69}\) \(\text{N}\) at \(\text{31,33}\)\(\text{°}\) to the \(x\)-axis. However we actually give this as \(\text{328,67}\)\(\text{°}\) to keep with the convention we have defined of giving vector directions.

To determine the answer we need to find the magnitude and direction of the resultant. This is then the direction in which the pole will be pushed.

We note that we have more than two vectors so we must first find the resultant in the \(x\)-direction and the resultant in the \(y\)-direction.

In the \(x\)-direction we only have one vector and so this is the resultant.

In the \(y\)-direction we have three vectors. We can add these algebraically to find \(\vec{R}_{y}\):

\begin{align*} \vec{F}_{1} & = -\text{11,7}\text{ N} \\ \vec{F}_{2} & = -\text{6,9}\text{ N} \\ \vec{F}_{3} & = -\text{1,9}\text{ N} \end{align*}

Thus, the resultant force is:

\begin{align*} \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} & = (-\text{11,7}) + (-\text{6,9}) + (-\text{1,9}) \\ & = -\text{20,5} \end{align*}

Now we use \(\vec{R}_{x}\) and \(\vec{R}_{y}\) to find the resultant.

We note that the triangle formed by \(\vec{R}_{x}\), \(\vec{R}_{y}\) and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (-\text{2,3})^{2} + (-\text{20,5})^{2} &= R^{2}\\ R &= \text{20,63}\text{ N} \end{align*}

A rough sketch will help to determine the direction.

To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}}{\text{adjacent side}} \\ \tan\alpha &= \frac{\text{2,3}}{\text{20,5}} \\ \alpha &= \tan^{-1}(\text{0,112}) \\ \alpha &= \text{6,40}\text{°} \end{align*}

The resultant force acts in a direction of \(\text{6,40}\)\(\text{°}\) to the \(x\)-axis. However we actually give this as \(\text{186,4}\)\(\text{°}\) to keep with the convention we have defined of giving vector directions.