Dynamics and inclined planes

In a nutshell

To solve problems where there is motion on an inclined plane, you can use the same principles already seen in static equilibrium problems. However any motion in a particular plane can be considered by using $F=ma$ in a particular direction. You can still use $\sum F=0$ for a plane where there is no motion as well as $F=\mu R$ .

Equations

DESCRIPTION	EQUATION
When an object accelerates in a particular plane, use Newton's second law	$F = m \times a$
When an object does not accelerate in a particular plane, the resultant force is zero.	$\sum F = 0$
Friction	$F=\mu R$

Variable definitions

QUANTITY NAME	SYMBOL	UNIT NAME	UNIT
$Force \ of \ friction$	$F$	$Newtons$	$N$
$Reaction \ force$	$R$	$Newtons$	$N$
$Weight$	$W$	$Newtons$	$N$
$Acceleration$	$a$	$Meters \ per \ second\ squared$	$m.s^{-2}$
$Coefficient \ of \ friction$	$\mu$	$no \ units$	$-$

Dynamics and inclined planes

Procedure

1.	Draw a force diagram.
2.	Resolve the forces into different components; refer to the inclined plane.
3.	For the plane where there is no acceleration, use $\sum F =0$ to form equations.
4.	For the plane where there is acceleration, use $F=ma$ to form equations.
5.	Solve the equations.

Example 1

A fox of mass $m$ slides from rest on a slide inclined at $\theta = 52^{\circ}$ to the horizontal. The fox accelerates down the slide at $a=2.4 \ m.s^{-2}$ . Calculate the coefficient of friction $\mu .$

First draw a force diagram:

Maths; Application of forces; KS5 Year 13; Dynamics and inclined planes

Resolve the forces into components, parallel and perpendicular to the plane. You should add labels to your diagram as follows:

Maths; Application of forces; KS5 Year 13; Dynamics and inclined planes

Resolving perpendicular to the plane gives:

$R=W \cos(52) \\ R=mg \cos(52)$ 

Resolving parallel to the plane gives:

$\begin{aligned}F &=ma \\P = W \sin(52) - F &= ma \\mg \sin(52) - \mu \times mg \cos(52) &= m \times 2.4 \\\end{aligned}$ 

Solve for $\mu$ to give:

$\begin{aligned}\mu &= \dfrac{mg \sin(52)-2.4m}{mg \cos(52)} \\\\\mu&= \underline{0.882 \ (3 \ s.f.)}\end{aligned}$ 