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

Let the reaction at Q be denoted as $X$ . According to the problem, the reaction at P is $2X$ .

Using the equilibrium condition for vertical forces, we have:

$2X + X = 4.5 \text{ kg} \times 9.81 \text{ m/s}^2$

This simplifies to:

$3X = 44.145\ ext{ N}$

Therefore, solving for $X$ gives:

$X = \frac{44.145}{3} = 14.715\ ext{ N}$

Rounding to two significant figures, the magnitude of the reaction at Q is approximately 15 N.

Step 2

b) the distance AG

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Answer

To find the distance AG, we first need to calculate the position of the center of mass (G). The moments about point A can be expressed as:

$4.5g \times AG = 2X \times 0.8 + X \times 2.4$

Substituting the values found:

$4.5 \times 9.81 \times AG = 2 \times 15 \times 0.8 + 15 \times 2.4$

This leads to:

$AG = \frac{4.5 \times 9.81}{4.5g} = 1.33\ ext{ m}$

Thus, the distance AG is approximately 1.33 m.

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