Lizzie is sat securely on a wooden sledge - AQA - A-Level Maths Pure - Question 17 - 2019 - Paper 2

To derive the expression for tension $T$, we need to apply Newton's second law in horizontal and vertical directions.

1. Resolving Forces Vertically:
In the vertical direction, the forces acting are the weight of Lizzie and the sledge ($Mg$) and the vertical component of the tension ($T\sin \theta$). The normal reaction force $R$ acts upward:

$R + T\sin \theta = Mg$
=> $R = Mg - T\sin \theta$.

2. Resolving Forces Horizontally:
In the horizontal direction, the only horizontal forces are the horizontal component of the tension ($T\cos \theta$) and the frictional force ($\mu R$). Using Newton's second law for motion in the horizontal direction:

$T\cos \theta - \mu R = Ma$

3. Replacing R:
Substituting for $R$ from the vertical resolution:

$T\cos \theta - \mu (Mg - T\sin \theta) = Ma$

4. Rearranging the Equation:
This expands to:

$T\cos \theta - \mu Mg + \mu T\sin \theta = Ma$

Rearranging gives:

$T (\cos \theta + \mu \sin \theta) = Ma + \mu Mg$

5. Final Expression for T:
Dividing the entire equation by $(\cos \theta + \mu \sin \theta)$ leads to:

$T = \frac{M(a + \mu g)}{\cos \theta + \mu \sin \theta}$.