Lizzie is sat securely on a wooden sledge - AQA - A-Level Maths Pure - Question 17 - 2019 - Paper 2
Question 17
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 T in the rope, we must resolve the forces acting on the sledge in both vertical and horizontal directions.
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:
Tsinθ−Mg−μR=0(1)
Here, R is the normal reaction force.
Horizontal Forces:
The horizontal component of the tension must provide the net force causing the acceleration of the system. The net force equation is:
Tcosθ−Ff=Ma(2)
where Ff=μR.
Substituting and Combining Equations:
From the vertical forces equation (1), we can express R in terms of T:
R=μTsinθ−Mg(3)
Substituting equation (3) into equation (2):
Tcosθ−μ(μTsinθ−Mg)=Ma
This leads to:
Tcosθ−(Tsinθ−Mg)=Ma
Rearranging gives us:
T(cosθ+μsinθ)=Ma+Mg
Therefore, we find:
T=cosθ+μsinθM(a+μg)
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 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. This discrepancy can skew the result, leading to an inaccurate representation of the friction coefficient under moving conditions.