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

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Lizzie is sat securely on a wooden sledge. The combined mass of Lizzie and the sledge is $M$ kilograms. The sledge is being pulled forward in a straight line along... show full transcript

Worked Solution & Example Answer:Lizzie is sat securely on a wooden sledge - AQA - A-Level Maths: Pure - Question 17 - 2019 - Paper 2

Step 1

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

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Answer

To find the tension TT in the rope, we must resolve the forces acting on the sledge in both vertical and horizontal directions.

  1. Vertical Forces: The vertical component of the tension must balance the weight of Lizzie and the sledge and the frictional force.

    The equation for vertical forces can be written as: TsinθMgμR=0(1)T \sin \theta - Mg - \mu R = 0 \quad (1)

    Here, RR is the normal reaction force.

  2. Horizontal Forces: The horizontal component of the tension must provide the net force causing the acceleration of the system. The net force equation is:

    TcosθFf=Ma(2)T \cos \theta - F_f = Ma \quad (2)

    where Ff=μRF_f = \mu R.

  3. Substituting and Combining Equations: From the vertical forces equation (1), we can express RR in terms of TT:

    R=TsinθMgμ(3)R = \frac{T \sin \theta - Mg}{\mu} \quad (3)

    Substituting equation (3) into equation (2):

    Tcosθμ(TsinθMgμ)=MaT \cos \theta - \mu \left( \frac{T \sin \theta - Mg}{\mu} \right) = Ma

    This leads to: Tcosθ(TsinθMg)=MaT \cos \theta - (T \sin \theta - Mg) = Ma

    Rearranging gives us:

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

    Therefore, we find:

    T=M(a+μg)cosθ+μsinθT = \frac{M(a + \mu g)}{\cos \theta + \mu \sin \theta}

Step 2

Explain why her value for \( \mu \) may be incorrect.

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Answer

Lizzie's determination of the coefficient of friction μ\mu may be incorrect because the sledge is at rest in the scenario described. Since it is at this state, the friction could be at its limiting value, which means:

  • The calculated friction might not represent the static friction under normal conditions.
  • Thus, if the sledge starts to move, the kinetic friction could differ from the static friction value used in calculating μ\mu. This discrepancy can skew the result, leading to an inaccurate representation of the friction coefficient under moving conditions.

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