# Resolving Forces

## Resolving Forces Revision

**Resolving Forces**

In the Forces section. we mentioned that forces have **components**. In most systems we work with, this will generally be resolved **horizontally** and **vertically**.

You’ll need to make sure your trigonometry is up to scratch!

Make sure you are happy with the following topics before continuing.

**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°.

**Note:**

Remember, for any x,

-1 \leq \sin x \leq 1

and

-1 \leq \cos x \leq 1.

If you are resolving in any direction and a component of the force is **greater** than the force itself, something has gone wrong.

**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}.

## Resolving Forces Example Questions

**Question** **1: ** A 2\text{ kg} ball is rolled along a flat surface with a fixed horizontal force of 15\text{ N}. The ball is rolled into the mud, and stops moving. What is the frictional force acting on the ball once it has stopped moving? What is the reaction force applied?

**[2 marks]**

Take g = 9.8\text{ ms}^{-2}.

The ball stops moving – this means that the system is in equilibrium.

The friction force is equal to the thrust force, 15\text{ N}.

The reaction force is equal to the weight in this scenario. That is, R = W. Since W = mg, we have R = W = 2 \times 9.8 = 19.6\text{ N}.

**Question 2: **A car’s handbrake is applied on a 15° decline.

Given that the car has a mass of 1000\text{ kg} and assuming that friction has no effect on the system, what is the resistive force, F, applied to the car by the handbrake?

Assume g = 9.8\text{ ms}^{-2}.

**[2 marks]**

As we know the car is static, we know F will be equal to the forces acting in the opposite direction.

Resolving parallel to F, we have

\begin{aligned}F&=W\sin 15°\\[1.2em]&=mg\sin 15°\\[1.2em]&=1000 \times 9.8 \times \sin 15°\\[1.2em]&=2536.43 \text{ N (to 2dp)}\end{aligned}

**Question 3:** We have a particle being pulled from 4 different directions. It experiences forces

(\textbf{i} + 2\textbf{j}), \left( \dfrac{5}{2}\textbf{i} - \dfrac{1}{2}\textbf{j}\right) , (-2\textbf{i}, -\textbf{j}) and (-\textbf{i}).

Give the vector for the resultant force applied.

**[1 mark]**

We have (\textbf{i} + 2\textbf{j}), \left( \dfrac{5}{2}\textbf{i} - \dfrac{1}{2}\textbf{j}\right) , (-2\textbf{i}, -\textbf{j}) and (-\textbf{i}).

Then, \left( \left( 1 + \dfrac{5}{2} - 2 - 1\right) \textbf{i} + \left( 2 - \dfrac{1}{2} - 1\right) \textbf{j}\right) = \left( \dfrac{1}{2}\textbf{i} + \dfrac{1}{2}\textbf{j}\right)