Resultant Using Vectors

Determine, by resolving into components, the resultant of the following four forces acting at a point:

• \(\vec{F}_1\)=\(\text{3.5}\) \(\text{N}\) at \(\text{45}\)\(\text{°}\) to the positive \(x\)-axis.
• \(\vec{F}_2\)=\(\text{2.7}\) \(\text{N}\) at \(\text{63}\)\(\text{°}\) to the positive \(x\)-axis.
• \(\vec{F}_3\)=\(\text{1.3}\) \(\text{N}\) at \(\text{127}\)\(\text{°}\) to the positive \(x\)-axis.
• \(\vec{F}_4\)=\(\text{2.5}\) \(\text{N}\) at \(\text{245}\)\(\text{°}\) to the positive \(x\)-axis.

Step 1: Sketch the problem

Draw all of the vectors on the Cartesian plane. This does not have to be precisely accurate because we are solving algebraically but vectors need to be drawn in the correct quadrant and with the correct relative positioning to each other.

We are going to record the various components in a table to help us manage keep track of the calculation. For each vector we need to determine the components in the \(x\)- and \(y\)-directions.

Step 2: Determine components of \(\vec{F}_1\)

\begin{align*} \sin(\theta) &= \frac{F_{1y}}{F_1} \\ \sin( \text{45}\text{°}) &= \frac{F_{1y}}{\text{3,5}} \\ F_{1y} &= \left(\sin (\text{45}\text{°})\right)\left( \text{3,5}\right) \\ &= \text{2,47}\text{ N} \end{align*}

Secondly we find the magnitude of the horizontal component, \({F}_{\mathrm{1x}}\):

\begin{align*} \cos(\theta) &= \frac{F_{1x}}{F_1} \\ \cos( \text{45}\text{°}) &= \frac{F_{1x}}{\text{3,5}} \\ F_{1x} &= \left(\cos (\text{45}\text{°})\right)\left( \text{3,5}\right) \\ &= \text{2,47}\text{ N} \end{align*}

Step 3: Determine components of \(\vec{F}_2\)

\begin{align*} \sin(\theta) &= \frac{F_{2y}}{F_2} \\ \sin(\text{63}\text{°}) &= \frac{F_{2y}}{\text{2,7}} \\ F_{2y} &= \left(\sin(\text{63}\text{°})\right)\left( \text{2,7}\right) \\ &= \text{2,41}\text{ N} \end{align*}

Secondly we find the magnitude of the horizontal component, \({F}_{\mathrm{2x}}\):

\begin{align*} \cos\theta &= \frac{F_{2x}}{F_2} \\ \cos( \text{63}\text{°}) &= \frac{F_{2x}}{\text{2,7}} \\ F_{2x} &= \left(\cos(\text{63}\text{°})\right)\left( \text{2,7}\right) \\ &= \text{1,23}\text{ N} \end{align*}

Step 4: Determine components of \(\vec{F}_3\)

\begin{align*} \sin(\theta) &= \frac{F_{3y}}{F_3} \\ \sin( \text{127}\text{°}) &= \frac{F_{3y}}{\text{1,3}} \\ F_{3y} &= \left(\sin \text{127}\text{°}\right)\left( \text{1,3}\right) \\ &= \text{1,04}\text{ N} \end{align*}

Secondly we find the magnitude of the horizontal component, \({F}_{\mathrm{3x}}\):

\begin{align*} \cos(\theta) &= \frac{F_{3x}}{F_3} \\ \cos (\text{127}\text{°}) &= \frac{F_{3x}}{\text{1,3}} \\ F_{3x} &= \left(\cos \text{127}\text{°}\right)\left( \text{1,3}\right) \\ &= -\text{0,78}\text{ N} \end{align*}

Step 5: Determine components of \(\vec{F}_4\)

\begin{align*} \sin(\theta) &= \frac{F_{4y}}{F_4} \\ \sin( \text{245}\text{°}) &= \frac{F_{4y}}{\text{2,5}} \\ F_{4y} &= \left(\sin( \text{245}\text{°})\right)\left( \text{2,5}\right) \\ &= -\text{2,27}\text{ N} \end{align*}

Secondly we find the magnitude of the horizontal component, \({F}_{\mathrm{4x}}\):

\begin{align*} \cos(\theta) &= \frac{F_{4x}}{F_4} \\ \cos( \text{245}\text{°}) &= \frac{F_{4x}}{\text{2,5}} \\ F_{4x} &= \left(\cos( \text{245}\text{°})\right)\left( \text{2,5}\right) \\ &= -\text{1,06}\text{ N} \end{align*}

Step 6: Determine components of resultant

Sum the various component columns to determine the components of the resultant. Remember that if the component was negative don't leave out the negative sign in the summation.

Now that we have the components of the resultant, we can use the Theorem of Pythagoras to determine the magnitude of the resultant, R.

\begin{align*} R^2 &= (R_y)^2 + (R_x)^2 \\ & = (\text{1,86})^2 + (\text{3,65})^2 \\ &= \text{16,78}\\ R &= \text{4,10}\text{ N} \end{align*}

We can also determine the angle with the positive \(x\)-axis.

\begin{align*} \tan(\alpha) &= \frac{\text{1,86}}{\text{3,65}}\\ \alpha &= \tan^{-1}(\frac{\text{3,65}}{\text{1,86}})\\ \alpha &= \text{27,00}\text{°} \end{align*}

Step 7: Quote the final answer

The resultant has a magnitude of \(\text{4.10}\) \(\text{N}\) at and angle of \(\text{27.00}\)\(\text{°}\) to the positive \(x\)-direction.