A non-uniform rod AB, of mass m and length 5d, rests horizontally in equilibrium on two supports at C and D, where AC = DB = d, as shown in Figure 1. The centre of mass of the rod is at the point G. A particle of mass $\frac{5}{2} m$ is placed on the rod at B and the rod is on the point of tipping about D. (a) Show that $GD = \frac{5}{2} d$. The particle is moved from B to the mid-point of the rod and the rod remains in equilibrium. (b) Find the magnitude of the normal reaction between the support at D and the rod.

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A non-uniform rod AB, of mass m and length 5d, rests horizontally in equilibrium on two supports at C and D, where AC = DB = d, as shown in Figure 1 - Edexcel - A-Level Maths Mechanics - Question 4 - 2012 - Paper 1

Question 4

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A non-uniform rod AB, of mass m and length 5d, rests horizontally in equilibrium on two supports at C and D, where AC = DB = d, as shown in Figure 1. The centre of m... show full transcript

Worked Solution & Example Answer:A non-uniform rod AB, of mass m and length 5d, rests horizontally in equilibrium on two supports at C and D, where AC = DB = d, as shown in Figure 1 - Edexcel - A-Level Maths Mechanics - Question 4 - 2012 - Paper 1

Step 1

Show that $GD = \frac{5}{2} d$

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Answer

To find $GD$ , we can use the principle of moments. The moments about point D are given by:

$mg \times GD = \frac{5}{2}mg \times d$

This simplifies to:

$GD = \frac{5}{2}d$

Thus, we have shown that $GD = \frac{5}{2} d$ .

Step 2

Find the magnitude of the normal reaction between the support at D and the rod.

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Answer

When the particle is moved to the mid-point of the rod, the moments about point D change. The equation becomes:

$mg \times \frac{5d}{2} + \frac{5}{2} mg \times \frac{3d}{2} = Y \times 3d$

Solving this leads to:

$Y = \frac{17}{12} mg$

Thus, the magnitude of the normal reaction at D is $Y = \frac{17}{12} mg$ .

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