A uniform rod AB has length 2 m and mass 50 kg. The rod is in equilibrium in a horizontal position, resting on two smooth supports at C and D, where AC = 0.2 metres and DB = x metres, as shown in Figure 5. Given that the magnitude of the reaction on the rod at D is twice the magnitude of the reaction on the rod at C, a) find the value of x. The support at D is now moved to the point E on the rod, where EB = 0.4 metres. A particle of mass m kg is placed on the rod at B, and the rod remains in equilibrium in a horizontal position. Given that the magnitude of the reaction on the rod at E is four times the magnitude of the reaction on the rod at C, b) find the value of m.

A-Level Edexcel Maths Mechanics Moments: A uniform rod AB has length 2 m

A uniform rod AB has length 2 m and mass 50 kg - Edexcel - A-Level Maths Mechanics - Question 8 - 2013 - Paper 1

Question 8

A uniform rod AB has length 2 m and mass 50 kg. The rod is in equilibrium in a horizontal position, resting on two smooth supports at C and D, where AC = 0.2 metres ... show full transcript

Worked Solution & Example Answer:A uniform rod AB has length 2 m and mass 50 kg - Edexcel - A-Level Maths Mechanics - Question 8 - 2013 - Paper 1

Step 1

find the value of x.

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Answer

To find the value of x, we can set up the equations based on the conditions of equilibrium.

Vertical Equilibrium: The sum of the upward forces must equal the downward forces:

$R + 2R = 50g$

Where R is the reaction force at C, leading to:

$3R = 50g$

Rearranging gives:

$R = \frac{50g}{3}$
Moments About C: Taking moments about point C helps us eliminate R:

$50g \times 0.8 = (1.8 - x) \times 2R$

Substituting R:

$50g \times 0.8 = (1.8 - x) \times \frac{100g}{3}$

Simplifying:

$40 = (1.8 - x) \times \frac{100}{3}$

Therefore:

$40 \times 3 = 100(1.8 - x)$

$120 = 180 - 100x$

Rearranging gives:

$100x = 180 - 120$

$x = 0.6$ .

Step 2

find the value of m.

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Answer

With the support moved to point E (where EB = 0.4 m), we need to set up new equilibrium equations:

Vertical Equilibrium:

$S + 4S = (50 + m)g$

Simplifying gives:

$5S = (50 + m)g$
Moments About B: Taking moments about B:

$50g \times 1 = 4S \times 0.4 + S \times 1.8$

Substituting for S:

$50g \times 1 = 4\left( \frac{50 + m}{5} \right) \times 0.4 + \left( \frac{50 + m}{5} \right) \times 1.8$

Which simplifies to:

$50 = 0.32(50 + m) + 0.36(50 + m)$

Combining terms:

$50 = (0.32 + 0.36)(50 + m)$

Thus,

$50 = 0.68(50 + m)$

Rearranging gives:

$m = \frac{50}{0.68} - 50$

Therefore:

$m = 73.53 - 50 \\ m = 23.53$ .

A uniform rod AB has length 2 m and mass 50 kg - Edexcel - A-Level Maths Mechanics - Question 8 - 2013 - Paper 1

Worked Solution & Example Answer:A uniform rod AB has length 2 m and mass 50 kg - Edexcel - A-Level Maths Mechanics - Question 8 - 2013 - Paper 1

find the value of x.

find the value of m.

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