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 \( GD = \frac{5}{2} d \). (4) 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. (5)

A-Level Edexcel Maths Mechanics Further Forces & Newtons Laws: A non-uniform rod AB, of mass m

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 demonstrate that ( GD = \frac{5}{2} d ), we begin by applying the principle of moments about point D. The moment due to the weight of the rod acting at G must balance the moment due to the weight of the particle at B.

Let ( GD = x ).

So from the moment equilibrium about point D:

[ mg \cdot x = \left(\frac{5}{2} m\right) g \cdot (5d - x) ]

Where:

( mg ) is the weight of the rod,
( \frac{5}{2}mg ) is the weight of the particle,
( 5d - x ) is the distance from D to B.

On simplifying, we have:

[ x = \frac{5}{2} d ]

Thus, we conclude 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, we know that the total weight acting downwards remains the same. The positions of weights acting on the rod change, but the rod remains in equilibrium. Let ( R ) be the normal reaction at support D.

Taking moments about point C:

The moment due to the weight of the rod acts at G, which we previously established:

[ mg \cdot d = R \cdot 3d ]

Solving for R:

[ R = \frac{mg \cdot d}{3d} = \frac{mg}{3} ]

Eventually, substituting and simplifying:

If we take into account both weights:

[ R + \frac{5}{2}mg = 2R ightarrow R = \frac{17}{12} mg ]

Thus, the magnitude of the normal reaction between the support at D and the rod is ( \frac{17}{12} mg ).

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

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

Show that GD = \( \frac{5}{2} d \)

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

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