Resolving forces

In a nutshell

Resolving forces turns forces acting at different directions into a set of components that are parallel and perpendicular to the surface an object is on. This is done through the use of free-body diagrams, a way of representing forces on a body, and trigonometry. This method can be used to analyse forces and motion on a slope.

Definitions

KEY WORD	DEFINITION
Resolving a force	Splitting a force into two perpendicular components
Normal contact force	The force a surface exerts on an object resting on it

Equations

Description	equation
Force parallel to a slope	$F_x = mg\sin\theta$
Force perpendicular to a slope	$F_y = mg\cos\theta$

Constants

Constant	Symbol	Value
$gravitational\ acceleration$	$g$	$9.81\,ms^{-2}$

Variable definitions

Quantity name	Symbol	Derived unit	SI unit
$parallel\ force$	$F_\parallel$	$N$	$kg\,m\,s^{-2}$
$perpendicular\ force$	$F_\perp$	$N$	$kg\,m\,s^{-2}$
$mass$	$m$	$kg$	$kg$
$angle$	$\theta$	$\degree$	$rad$

Free-body diagrams

A free-body diagram is a way of representing the forces acting on an object in a simplified way. Forces are drawn as arrows, where the length of the arrow represents the magnitude of the force. The forces usually point from the centre of mass of the object.

Example

This is the free-body diagram of an object sitting on a surface. A force is being applied to the left.

Physics; Motion; KS5 Year 12; Resolving forces

$F$ is the force being applied, $W$ is the weight of the object and $N$ is called the normal contact force. The normal contact force is the reaction force to the weight force, as a result of the surface pushing against the object.

Note: If the centre of mass is not known, it can be assumed to be the centre of the object.

Object on a slope

Stationary object

When an object is on a slope, the forces are resolved in the directions perpendicular and parallel to the slope itself.

Example

The free-body diagram of an object on a frictionless slope looks like this.

Physics; Motion; KS5 Year 12; Resolving forces

While the weight force $W$ still acts straight down, the normal contact force $N$ acts perpendicular to the surface. The weight force can be resolved into components parallel and perpendicular to the slope using the following equations.

$F_\parallel = mg\sin\theta$

$F_\perp = mg\cos\theta$

Since the object is not moving perpendicular to the slope, the perpendicular component of the weight force has to be equal to the normal contact force.

$F_\perp = mg\cos\theta = N$

Moving object

When another force is applied to the object, the motion of the object can be calculated by adding up all the parallel and perpendicular components and finding the resultant.

Example

An object with a mass of $2\,kg$ is on a frictionless slope at an angle of $30\degree$ . There is a force of magnitude $10\,N$ acting on the object up the slope. Calculate the object's acceleration.

First, draw a free-body diagram. The free-body diagram looks like this.

Physics; Motion; KS5 Year 12; Resolving forces

To find the resultant force, the parallel and perpendicular components should be summed. First, write down the equation for the parallel component of the weight force.

 $F_\parallel = mg\sin\theta$

Next, substitute in the values.

 $F_{down} = 2\times 9.81 \times \sin30$ 

$F_{down} = 9.81N$ 

The component perpendicular to the slope does not need to be calculated, as it and the normal contact force cancel out and do not contribute to the object's movement. If the $10N$ force were not parallel to the slope, then the perpendicular components of those forces would need to be calculated.

Next, calculate the resultant parallel force. The parallel component of weight acts in the opposite direction to the applied force.

$F_T = 10 - 9.81$ 

$F_T = 0.19N$ 

Finally, use this force to calculate the object's acceleration. First, write down the equation.

$F = ma$ 

Next, rearrange the equation for acceleration.

$a = \dfrac{F}{m}$ 

Finally, substitute in the values.

$a = \dfrac{0.19}{2}$ 

$a =0.095\,ms^{-2}$ 

The object is accelerating at $\underline{0.095\,ms^{-2}}$ up the slope.