Resultant Force on a Submarine

A number of tugboats are trying to manoeuvre a submarine in the harbour but they are not working as a team. Each tugboat is exerting a different force on the submarine.

Given the following force vectors, determine the resultant force:

• \(\vec{F}_{1}\) = \(\text{3.4}\) \(\text{kN}\) in the positive \(x\)-direction
• \(\vec{F}_{2}\) = \(\text{4 000}\) \(\text{N}\) in the positive \(y\)-direction
• \(\vec{F}_{3}\) = \(\text{300}\) \(\text{N}\) in the negative \(y\)-direction
• \(\vec{F}_{4}\) = \(\text{7}\) \(\text{kN}\) in the negative \(y\)-direction

Step 1: Convert to consistent S.I. units

To use the graphical method of finding the resultant we need to work in the same units. Strictly speaking in this problem all the vectors are in newtons but they have different factors which will affect the choice of scale. These need to taken into account and the simplest approach is to convert them all to a consistent unit and factor. We could use kN or N, the choice does not matter. We will choose kN. Remember that k represents a factor of \(\times 10^{3}\).

\(\vec{F}_{1}\) and \(\vec{F}_{4}\) do not require any adjustment because they are both in kN. To convert N to kN we use: \begin{align*} \text{kN} &= \times 10^{3}\\ \frac{\text{N}}{\text{kN}} &= \frac{1}{\times 10^{3}}\\ \text{N} &= \times 10^{-3}\ \text{kN} \end{align*}

To convert the magnitude of \(\vec{F}_{2}\) to kN: \begin{align*} F_2 &= \text{4 000}\text{ N}\\ F_2 &= \text{4 000} \times \text{10}^{-\text{3}}\text{ kN}\\ F_2 &= \text{4}\text{ kN} \end{align*} Therefore \(\vec{F}_{2}\) = \(\text{4}\) \(\text{kN}\) in the positive \(y\)-direction.

To convert the magnitude of \(\vec{F}_{3}\) to kN: \begin{align*} F_3 &= \text{300}\text{ N}\\ F_3 &= \text{300} \times \text{10}^{-\text{3}}\text{ kN}\\ F_3 &= \text{0,3}\text{ kN} \end{align*} Therefore \(\vec{F}_{3}\) = \(\text{0.3}\) \(\text{kN}\) in the negative \(y\)-direction. So:

• \(\vec{F}_{1}\) = \(\text{3.4}\) \(\text{kN}\) in the positive \(x\)-direction
• \(\vec{F}_{2}\) = \(\text{4}\) \(\text{kN}\) in the positive \(y\)-direction
• \(\vec{F}_{3}\) = \(\text{0.3}\) \(\text{kN}\) in the negative \(y\)-direction
• \(\vec{F}_{4}\) = \(\text{7}\) \(\text{kN}\) in the negative \(y\)-direction

Step 2: Choose a scale and draw axes

The vectors we have do have very big magnitudes so we need to choose a scale that will allow us to draw them in a reasonable space, we can use \(\text{1}\) \(\text{kN}\) : \(\text{1}\) \(\text{cm}\) as our scale for the drawings.

Step : Determine \(\vec{R}_{x}\)

There is only one vector in the \(x\)-direction, \(\vec{F}_{1}\), therefore \(\vec{R}_{x}\) = \(\vec{F}_{1}\).

Step 3: Determine \(\vec{R}_{y}\)

Then we determine the resultant of all the vectors that are parallel to the \(y\)-axis. There are three vectors \(\vec{F}_{2}\), \(\vec{F}_{3}\) and \(\vec{F}_{4}\) that we need to add. We do this using the tail-to-head method for co-linear vectors.

The single vector, \(\vec{R}_{y}\), that would give us the same outcome is:

Step 4: Draw axes

Then we draw axes that the diagram should fit on. We need our axes to extend just further than the vectors aligned with each axis. Our axes need to start at the origin and go beyond \(\text{3.4}\) \(\text{kN}\) in the positive \(x\)-direction and further than \(\text{3.3}\) \(\text{kN}\) in the negative \(y\)-direction. Our scale choice of \(\text{1}\) \(\text{kN}\) : \(\text{1}\) \(\text{cm}\) means that our axes actually need to extend \(\text{3.4}\) \(\text{cm}\) in the positive \(x\)-direction and further than \(\text{3.3}\) \(\text{cm}\) in the negative \(y\)-direction

Step 5: Draw \(\vec{R}_x\)

The length of \(\vec{R}_x\) is \(\text{3.4}\) \(\text{kN}\) so the arrow we need to draw must be \(\text{3.4}\) \(\text{cm}\) long. The arrow must point in the positive \(x\)-direction.

Step 6: Draw \(\vec{R}_y\)

The length of \(\vec{R}_y\) is \(\text{3.3}\) \(\text{kN}\) so the arrow we need to draw must be \(\text{3.3}\) \(\text{cm}\) long. The arrow must point in the negative \(y\)-direction. The important fact to note is that we are implementing the head-to-tail method so the vector must start at the end (head) of \(\vec{R}_{x}\).

Step 7: Draw the resultant vector, \(\vec{R}\)

The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a vector from the tail of \(\vec{R}_{x}\) to the head of \(\vec{R}_{y}\).

Step 8: Measure the resultant, \(\vec{R}\)

We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to the actual result. In the last diagram the resultant, \(\vec{R}\) is \(\text{4.7}\) \(\text{cm}\) long therefore the magnitude of the vector is \(\text{4.7}\) \(\text{kN}\).

The direction of the resultant we need to measure from the diagram using a protractor. The angle that the vector makes with the \(x\)-axis is \(\text{44}\)\(\text{°}\).

Step 9: Quote the final answer

\(\vec{R}\) is \(\text{4.7}\) \(\text{kN}\) at \(-\text{44}\)\(\text{°}\) from the positive \(x\)-direction.