Resolving Forces

Simple Model

Example: Two people are pushing and pulling on a box such that they are both providing force in the same direction.

This example is already resolved into horizontal and vertical components, as the box is on a flat surface.

Let’s say the system is in equilibrium – there is no resultant force; the horizontal forces sum to 0\text{ N}, and the vertical forces sum to 0\text{ N}, also.

So, T + \text{Thrust} = F and R = W.

Say that the box weighs 5\text{ kg}. Then, from W = mg, we have W = 5 \times 9.8 = 49\text{ N}, meaning that the normal reaction force, R, = 49\text{ N}.

On a Tilted Plane

Example: A car is driving up a hill while being towed by a truck.

Again, let’s say the system is in equilibrium.

Now, we need to resolve horizontally and vertically, as we can see that W isn’t perpendicular to T, \text{Thrust} and F.

Let’s rotate the system again.

Resolving in the “new horizontal”, we have T + \text{ Thrust} = F + W\sin 30°.

Resolving in the “new vertical”, we have R = W\cos 30°.

Resolving in Vector Form

We can also resolve forces that are given in vector format. Generally, for a 2D system, we’ll have \textbf{i} indicating a horizontal component and \textbf{j} representing the vertical component, though this is not always the case.

The process here is pretty intuitive – operate in the same way you would in standard algebra.

For example, let F_1 = (2\textbf{i} + \textbf{j}), F_2 = (4\textbf{i}), F_3 = (3\textbf{j}).

Then, F_1 + F_2 - F_3 = (2 + 4)\textbf{i} + (1 - 3)\textbf{j} = 6\textbf{i} - 2\textbf{j}.