Finding the Resultant Force

Finding the Resultant Force

The easiest way to determine a resultant force is to draw a free body diagram. Remember that we use the length of the arrow to indicate the vector's magnitude and the direction of the arrow to show which direction it acts in.

After we have done this, we have a diagram of vectors and we simply find the sum of the vectors to get the resultant force.

(a) Force diagram of 2 forces acting on a box. (b) Free body diagram of the box.

For example, two people push on a box from opposite sides with forces of \(\text{4}\) \(\text{N}\) and \(\text{6}\) \(\text{N}\) respectively as shown in the (a) part of the figure above. The free body diagram in the (b) part of the figure above shows the object represented by a dot and the two forces are represented by arrows with their tails on the dot.

As you can see, the arrows point in opposite directions and have different lengths. The resultant force is \(\text{2}\) \(\text{N}\) to the left. This result can be obtained algebraically too, since the two forces act along the same line. First, as in motion in one direction, choose a frame of reference. Secondly, add the two vectors taking their directions into account.

For the example, assume that the positive direction is to the right, then:\begin{align*} F_R & = (\text{4}) + (-\text{6}) \\ & = -\text{2} \\ & = -\text{2}\text{ N}\ \text{to the left.} \end{align*}

Remember that a negative answer means that the force acts in the opposite direction to the one that you chose to be positive. You can choose the positive direction to be any way you want, but once you have chosen it you must use it consistently for that problem.

As you work with more force diagrams in which the forces exactly balance, you may notice that you get a zero answer (e.g. \(\text{0}\) \(\text{N}\)). This simply means that the forces are balanced and the resultant is zero.

Once a force diagram has been drawn the techniques of vector addition introduced in another tutorial can be used. Depending on the situation you might choose to use a graphical technique such as the tail-to-head method or the parallelogram method, or else an algebraic approach to determine the resultant. Since force is a vector quantity all of these methods apply.

A good strategy is:

• resolve all forces into components parallel to the \(x\)- and \(y\)-directions;
• calculate the resultant in each direction, \(\vec{R}_x\) and \(\vec{R}_y\), using co-linear vectors; and
• use \(\vec{R}_x\) and \(\vec{R}_y\) to calculate the resultant, \(\vec{R}\).

Example: Finding the Resultant Force

Question

A car (experiencing a gravitational force of magnitude \(\text{12 000}\) \(\text{N}\) and a normal force of the same magnitude) applies a force of \(\text{2 000}\) \(\text{N}\) on a trailer (experiencing a gravitational force of magnitude \(\text{2 500}\) \(\text{N}\) and normal force of the same magnitude). A constant frictional force of magnitude \(\text{200}\) \(\text{N}\) is acting on the trailer, and a constant frictional force of magnitude \(\text{300}\) \(\text{N}\) is acting on the car.

1. Draw a force diagram of all the forces acting on the car.

2. Draw a free body diagram of all the forces acting on the trailer.

3. Use the force diagram to determine the resultant force on the trailer.

Step 1: Draw the force diagram for the car.

The question asks us to draw all the forces on the car. This means that we must include horizontal and vertical forces.

Step 2: Determine the resultant force on the trailer.

To find the resultant force we need to add all the horizontal forces together. We do not add vertical forces as the movement of the car and trailer will be in a horizontal direction, and not up or down.\({F}_{R}=\text{2 000}+\left(-\text{200}\right)=\text{1 800} \text{N to the right}\).

Optional: Simulation of Forces and Motion

Another Example

A donkey (experiencing a gravitational force of \(\text{2 500}\) \(\text{N}\)) is trying to pull a cart (force due to gravity of \(\text{800}\) \(\text{N}\)) with a force of \(\text{400}\) \(\text{N}\). The rope between the donkey and the cart makes an angle of \(\text{30}\)\(\text{°}\) with the cart. The cart does not move.

Draw a free body diagram of all the forces acting on the donkey.

Draw a force diagram of all the forces acting on the cart.

Where:

\begin{align*} \vec{F}_{f} & \text{ frictional force} \\ \vec{F}_{2} & \text{ force of cart on donkey} \\ \vec{F}_{g} & \text{ gravitational force} \\ \vec{N} & \text{ normal force} \end{align*}

Find the magnitude and direction of the frictional force preventing the cart from moving.

The cart is not moving so this is static friction only. We note that the frictional force acts in the \(x\)-direction only and will be equal to the \(x\) component of the force applied by the donkey. The direction will be in the opposite direction that the donkey is pulling. The frictional force is:

\begin{align*} \vec{F}_{f} & = F_{A}\cos \theta \\ & = \text{400} \cos(\text{30}) \\ & = \text{346,41}\text{ N} \text{ in the opposite direction to the applied force} \end{align*}