A hemispherical bowl of radius r is fixed with its rim horizontal. A thin uniform rod rests in equilibrium on the rim of the bowl with one end resting on the inner surface of the bowl at A, as shown in the diagram. The rod has length 2a and weight W. The point of contact between the rod and the rim is B, and the rim has center C. The rod is in a vertical plane containing C. The rod is inclined at θ to the horizontal and the line AC is inclined at 2θ to the horizontal. The contacts at A and B are smooth. In any order, show that (i) the contact force acting on the rod at A has magnitude W tan θ, (ii) the contact force acting on the rod at B has magnitude $ \frac{W cos 2θ}{cos θ} $, (iii) $ 2r cos 2θ = a cos θ $.

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A hemispherical bowl of radius r is fixed with its rim horizontal - CIE - A-Level Further Maths - Question 4 - 2010 - Paper 1

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A hemispherical bowl of radius r is fixed with its rim horizontal. A thin uniform rod rests in equilibrium on the rim of the bowl with one end resting on the inner s... show full transcript

Worked Solution & Example Answer:A hemispherical bowl of radius r is fixed with its rim horizontal - CIE - A-Level Further Maths - Question 4 - 2010 - Paper 1

Step 1

(i) the contact force acting on the rod at A has magnitude W tan θ

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Answer

To determine the contact force at A, we analyze the forces acting on the rod.

Resolve the weight of the rod (W) into components. The perpendicular component to the surface at A will be represented using trigonometric identities.

The vertical component of the weight can be given by: $W_{Vert} = W imes ext{sin} θ$

Since rod A is smooth, the normal force ( R_A) will only balance this vertical component: $R_A = \frac{W_{Vert}}{\text{cos} θ}$ $R_A = W \text{tan} θ.$

Step 2

(ii) the contact force acting on the rod at B has magnitude W cos 2θ / cos θ

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Answer

To find the contact force at B, consider the horizontal and vertical components acting at this point:

Applying the conditions for equilibrium in the vertical direction, we can resolve for the forces.

Using the geometry and properties of the circle at B, we have:

$R_B = \frac{W cos 2θ}{sin θ}.$

Rewriting this gives: $R_B = \frac{W cos 2θ}{cos θ}.$

Step 3

(iii) 2r cos 2θ = a cos θ

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Answer

We take moments about point A to express the equilibrium condition providing:

Taking the moment about A: $\text{Moment due to } R_B = R_B \times 2a \cos θ.$
The balance of moments provides: $2r \cos 2θ = a \cos θ$ after substituting and simplifying the expressions for forces and distances.

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A hemispherical bowl of radius r is fixed with its rim horizontal - CIE - A-Level Further Maths - Question 4 - 2010 - Paper 1

Worked Solution & Example Answer:A hemispherical bowl of radius r is fixed with its rim horizontal - CIE - A-Level Further Maths - Question 4 - 2010 - Paper 1

(i) the contact force acting on the rod at A has magnitude W tan θ

(ii) the contact force acting on the rod at B has magnitude W cos 2θ / cos θ

(iii) 2r cos 2θ = a cos θ

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