A fixed rough plane is inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 2

For particle A of mass 2 kg, the equation of motion can be expressed as:

$T - F - 2g \sin(30°) = 2a$

Here, ( F ) is the frictional force acting on A, and thus we have:

$F = \mu R = \frac{1}{\sqrt{3}} R$

Resolving perpendicular to the plane gives us:

$R = 2g \cos(30°)$

Substituting the expression for R into the frictional force gives:

$F = \frac{1}{\sqrt{3}} (2g \cos(30°))$

Next, substituting F into the equation for A:

$T - \frac{2g \cos(30°)}{\sqrt{3}} - g = 2a$

We now have two equations:

1. From B: ( T = 4g - 4a )
2. From A: ( T = \frac{1}{\sqrt{3}}(2g \cos(30°)) + 2g + 2a )

Setting these equal allows us to solve for ( T ) and ( a ). Upon simplifying, we find:

$2T - 4g + 4a - \frac{2g \cos(30°)}{\sqrt{3}} = 0$

After rearranging the equations and substituting for acceleration, we arrive at the final expression for tension, leading to:

$T = \frac{8g}{3}$

Substituting ( g = 9.81 : ext{m/s}^2 ):

$T = \frac{8 \times 9.81}{3} = 26.1 ext{ N}$