A non-uniform rod AB has length 3 m and mass 4.5 kg. The rod rests in equilibrium, in a horizontal position, on two smooth supports at P and at Q, where AP = 0.8 m and QB = 0.6 m, as shown in Figure 1. The centre of mass of the rod is at G. Given that the magnitude of the reaction of the support at P on the rod is twice the magnitude of the reaction of the support at Q on the rod, find (a) the magnitude of the reaction of the support at Q on the rod, (b) the distance AG.

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A non-uniform rod AB has length 3 m and mass 4.5 kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2012 - Paper 1

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A non-uniform rod AB has length 3 m and mass 4.5 kg. The rod rests in equilibrium, in a horizontal position, on two smooth supports at P and at Q, where AP = 0.8 m a... show full transcript

Worked Solution & Example Answer:A non-uniform rod AB has length 3 m and mass 4.5 kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2012 - Paper 1

Step 1

(a) the magnitude of the reaction of the support at Q on the rod

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Answer

To find the magnitude of the reaction force at Q, denote it as $X$ . Since the reaction at P is twice that at Q, the reaction at P can be represented as $2X$ .

Using the equilibrium condition for vertical forces, we have:

$2X + X = 4.5 g$

Given $g = 9.81 \ m/s^2$ , substituting in yields:

$3X = 4.5 \times 9.81$

Now calculate:

$X = \frac{4.5 \times 9.81}{3} = 14.7 \ N \ (or \ 15 \ N)$

Step 2

(b) the distance AG

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Answer

To find the distance AG, use the formula for the centre of mass. The centre of mass (G) can be found using:

$M[G] = \frac{4.5g \cdot \left( 0.8 + 3 \times 0.6 \right)}{4.5g}$

Rearranging gives:

$AG = \frac{4}{3} \ m$

Calculating this yields:

$AG = 1.33 \ m$

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