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Exercises

Algebraic addition of vectors

Exercise 1.5

A force of \(\text{17}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{23}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the resultant force.

We draw a rough sketch:

4497a86fa0a5ef9c87688735b0bab797.png

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.

A force of \(\text{23,7}\) \(\text{N}\) in the negative \(x\)-direction acts simultaneously to a force of \(\text{9}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the resultant force.

We draw a rough sketch:

6aaf376eb7efe1c1e0ddac18aedbaa59.png

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.

Four forces act simultaneously at a point, find the resultant if the forces are:

  • \(\vec{F}_{1}\) = \(\text{2,3}\) \(\text{N}\) in the positive \(x\)-direction

  • \(\vec{F}_{2}\) = \(\text{4}\) \(\text{N}\) in the positive \(y\)-direction

  • \(\vec{F}_{3}\) = \(\text{3,3}\) \(\text{N}\) in the negative \(y\)-direction

  • \(\vec{F}_{4}\) = \(\text{2,1}\) \(\text{N}\) in the negative \(y\)-direction

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.

The following forces act simultaneously on a pole, if the pole suddenly snaps in which direction will it be pushed:

  • \(\vec{F}_{1}\) = \(\text{2,3}\) \(\text{N}\) in the negative \(x\)-direction

  • \(\vec{F}_{2}\) = \(\text{11,7}\) \(\text{N}\) in the negative \(y\)-direction

  • \(\vec{F}_{3}\) = \(\text{6,9}\) \(\text{N}\) in the negative \(y\)-direction

  • \(\vec{F}_{4}\) = \(\text{1,9}\) \(\text{N}\) in the negative \(y\)-direction

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.

d295941e03368da87032b3556b6a5475.png

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.