Example on Graphical Methods

In the previous lesson, you saw some examples on finding the magnitude of the resultant of vectors in two-dimensions. The following additional worked example shows this further.

Example: Finding the Resultant in Two Dimensions Graphically

Question

Given the following three force vectors, determine the resultant force:

\(\vec{F}_{1}\) = \(\text{2.3}\) \(\text{N}\) 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

Step 1: Determine \(\vec{R}_{x}\)

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

Step 2: 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 effect is:

Step 3: Choose a scale and draw axes

We choose a scale \(\text{1}\) \(\text{N}\) : \(\text{1}\) \(\text{cm}\) for the drawing.

Then we draw axes that the diagram should fit in. 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 \(\text{2.3}\) \(\text{N}\) in the positive \(x\)-direction and further than \(\text{1.4}\) \(\text{N}\) in the negative \(y\)-direction. Our scale choice of \(\text{1}\) \(\text{N}\) = \(\text{1}\) \(\text{cm}\) means that our axes actually need to extend \(\text{2.3}\) \(\text{cm}\) in the positive \(x\)-direction and further than \(\text{1.4}\) \(\text{cm}\) in the negative \(y\)-direction

Step 4: Draw \(\vec{R}_x\)

The magnitude of \(\vec{R}_x\) is \(\text{2.3}\) \(\text{N}\) so the arrow we need to draw must be \(\text{2.3}\) \(\text{cm}\) long. The arrow must point in the positive \(x\)-direction.

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

The magnitude of \(\vec{R}_y\) is \(\text{1.4}\) \(\text{N}\) so the arrow we need to draw must be \(\text{1.4}\) \(\text{cm}\) 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 6: 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 7: 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 magnitude of the vector. In the last diagram the resultant, \(\vec{R}\) is \(\text{2.7}\) \(\text{cm}\) long therefore the magnitude of the vector is \(\text{2.7}\) \(\text{N}\).

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 \(\text{31}\) degrees.

Step 8: Quote the final answer

\(\vec{R}\) is \(\text{2.7}\) \(\text{N}\) at \(-\text{31}\)\(\text{°}\) from the positive \(x\)-direction.