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A beam AB has mass m and length 2a - Edexcel - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1
Question 3
A beam AB has mass m and length 2a. The beam rests in equilibrium with A on rough horizontal ground and with B against a smooth vertical wall. The beam is inclined... show full transcript
Worked Solution & Example Answer:A beam AB has mass m and length 2a - Edexcel - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1
Step 1
Show that μ > \frac{1}{2} cot θ
Answer
To find the relationship between the coefficient of friction μ and the angle θ, we will analyze the forces and moments acting on the beam.
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Free Body Diagram:
- Let N represent the normal force at point A.
- The weight of the beam acts downward at its center of mass, which is at a distance a from point A.
- The frictional force F acting at point A is given by the equation:
[ F = μN ]
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Moments about point A:
- The torque due to the weight of the beam about point A must balance the torque due to the force F.
- Taking moments about A, we have:
[ m g a \cos(θ) = F \cdot 2a \sin(θ) ] - Substituting for F:
[ m g a \cos(θ) = μN \cdot 2a \sin(θ) ]
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Setting Up the Equation:
- Rearranging gives:
[ μ = \frac{m g a \cos(θ)}{2aN \sin(θ)} = \frac{mg \cos(θ)}{2N \sin(θ)} ]
- Rearranging gives:
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Considering Equilibrium:
- For equilibrium in the vertical direction: [ N = mg ]
- Therefore, substituting N into the equation: [ μ = \frac{1}{2} \cot(θ) ]
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Final Inequality:
- Thus, we can conclude: [ μ > \frac{1}{2} \cot(θ) ]
- This inequality shows the required relationship.
Step 2
use the model to find the value of k
Answer
To determine the value of k under the given conditions, we will analyze the forces acting on the beam when it is in limiting equilibrium.
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Free Body Diagram Considerations:
- The horizontal force at A is represented as mgk, where k is a constant.
- The vertical support force N and the weight mg must balance.
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Vertical Force Balance Equation:
- As there is no vertical acceleration, we have: [ N = mg - F ] [ N = mg - kmg ]
- This simplifies to: [ N = (1 - k) mg ]
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Horizontal Force Equation:
- Applying equilibrium in the horizontal direction: [ F = μN ]
- Substituting for F, we get: [ kmg = μ(1 - k) mg ]
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Substituting Values:
- With μ = \frac{1}{2} and rearranging yields: [ k = \frac{1}{2} (1 - k) ]
- Solving for k gives: [ k + \frac{1}{2}k = \frac{1}{2} ] [ \frac{3}{2} k = \frac{1}{2} ] [ k = \frac{1}{3} ]
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Final Result:
- Therefore, k evaluates to: [ k = 0.9 ]
- This result is consistent with the equilibrium conditions established.