A force of \(\text{5}\) \(\text{N}\) to the right is applied to a crate. A second force of \(\text{2}\) \(\text{N}\) to the left is also applied to the crate. Calculate algebraically the resultant of the forces applied to the crate.

Step 1: Draw a sketch

A simple sketch will help us understand the problem.

Step 2: Decide which method to use to calculate the resultant

Remember that force is a vector. Since the forces act along a straight line (i.e. the \(x\)-direction), wecan use the algebraic technique of vector addition.

Step 3: Choose a positive direction

Choose the positive direction to be to the right. This means that the negative direction is to the left.

Rewriting the problem using the choice of a positive direction gives us a force of \(\text{5}\) \(\text{N}\) in the positive \(x\)-direction and force of \(\text{2}\) \(\text{N}\) in the negative \(x\)-direction being applied to the crate.

Step 4: Now define our vectors algebraically

\begin{align*} \vec{F}_{1}& = \text{5}\text{ N} \\ \vec{F}_{2}& = -\text{2}\text{ N} \end{align*}

Step 5: Add the vectors

Thus, the resultant force is:

\begin{align*} \vec{F}_{1}+\vec{F}_{2}& = (5)+(-2) \\ & = \text{3}\text{ N} \end{align*}

Step 6: Quote the resultant

Remember that in this case a positive force means to the right: \(\text{3}\) \(\text{N}\) to the right.

A force of \(\text{40}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{30}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the magnitude of the resultant force.

Step 1: Draw a rough sketch

As before, the rough sketch looks as follows:

Step 2: Determine the length of the resultant

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}\\ (40)^{2} + (30)^{2} &= R^{2}\\ R &= \text{50}\text{ N} \end{align*}

Step 3: Quote the resultant

The magnitude of the resultant force is then \(\text{50}\) \(\text{N}\).

A force of \(\text{40}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{30}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the magnitude of the resultant force.

Step 1: Magnitude

We determined the magnitude of the resultant vector in the previous worked example to be \(\text{50}\) \(\text{N}\). The sketch of the situation is:

Step 2: Determine the direction of the resultant

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{30}}{\text{40}} \\ \alpha &= \tan^{-1}(\text{0,75}) \\ \alpha &= \text{36,87}\text{°} \end{align*}

Step 3: Quote the resultant

The resultant force is then \(\text{50}\) \(\text{N}\) at \(\text{36.9}\)\(\text{°}\) to the positive \(x\)-axis.