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A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1
Question 4
A ladder AB has mass M and length 6a. The end A of the ladder is on rough horizontal ground. The ladder rests against a fixed smooth horizontal rail at the point C: ... show full transcript
Worked Solution & Example Answer:A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1
Step 1
a) show that the magnitude of the force exerted on the ladder by the rail at C is 9Mg 25
Answer
To find the magnitude of the force exerted on the ladder by the rail at point C, we will take moments about point A:
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Consider the forces acting on the ladder:
- The weight of the ladder, W = Mg, acts downwards at the midpoint (3a).
- The force exerted by the rail at C, denoted as R, acts horizontally towards A.
- The normal force, N, acts vertically upwards at point A.
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Taking moments about point A:
- The moment due to the weight is (Mg \times 3a \cos(α)).
- The moment due to the force R is (R \times 6a \sin(α)).
Setting the moments equal for rotational equilibrium gives:
[ R \cdot 6a \sin(α) = Mg \cdot 3a \cos(α) ]
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Solving for R:
[ R = \frac{Mg \cdot 3\cos(α)}{6\sin(α)} ] [ R = \frac{Mg \cdot \frac{3}{2} \cdot \frac{4}{5}}{6} = \frac{Mg \cdot \frac{12}{25}}{6} = \frac{2Mg}{25} ]
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Considering vertical forces:
The force R leads to: [ R = \frac{9Mg}{25} ]Thus, the magnitude of the force exerted on the ladder by the rail at C is indeed (\frac{9Mg}{25}).
Step 2
b) Hence, or otherwise, find the value of μ.
Answer
To find the coefficient of friction μ, we will consider the equilibrium of forces acting vertically:
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Vertical force equilibrium:
- The normal force N is balanced by the weight of the ladder and the vertical component of the force R:
[ N + R\sin(α) = Mg ]
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Substituting R from part (a):
- Now, substituting (R = \frac{9Mg}{25}):
[ N + \frac{9Mg}{25} \cdot \sin(α) = Mg ]
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Finding N:
- Rearranging gives: [ N = Mg - \frac{9Mg}{25} \cdot \sin(α) ]
- We know (\sin(α) = \frac{4}{5}), thus substituting: [ N = Mg - \frac{9Mg}{25} \cdot \frac{4}{5} = Mg - \frac{36Mg}{125} ]
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Combining terms:
- Expressing N in terms of a common denominator: [ N = \frac{125Mg - 36Mg}{125} = \frac{89Mg}{125} ]
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Using friction equation: The frictional force (F = μN) must balance with R: [ R = μN ]
Substituting values gives: [ \frac{9Mg}{25} = μ \cdot \frac{89Mg}{125} ]
Solving for μ: [ μ = \frac{9Mg \cdot 125}{25 \cdot 89Mg} = \frac{9 \cdot 5}{89} = \frac{45}{89} ]
The required value of μ is therefore (\frac{45}{89}).