A fixed rough plane is inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 1

For particle A, the equation is:

$T - F - 2g \sin(30°) = 2a$

Substituting ( \sin(30°) = \frac{1}{2} ):

$T - F - g = 2a$

Resolving forces perpendicular to the plane gives:

$R = 2g \cos(30°)$

Calculating ( \cos(30°) = \frac{\sqrt{3}}{2} ), we find:

$R = 2g \cdot \frac{\sqrt{3}}{2} = g\sqrt{3}$

The frictional force ( F ) is given by:

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

Substituting ( F = g ) into the equation for A:

$T - g - g = 2a$

This simplifies to:

$T - 2g = 2a \\ T = 2a + 2g$

Substituting ( T = 4g - 4a ) from B's equation into A's gives:

$2a + 2g = 4g - 4a \\ 6a = 2g \\ a = \frac{g}{3}$

Now substituting back to find T:

$T = 4g - 4\left(\frac{g}{3}\right) = 4g - \frac{4g}{3} = \frac{12g}{3} - \frac{4g}{3} = \frac{8g}{3}$

For( g = 26 ):

$T = \frac{8 \times 26}{3} = \frac{208}{3} \approx 69.33 ext{ N}$