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 Further Forces & Newtons Laws: 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 2

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 2

Step 1

find the value of x.

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Answer

To solve for x, we start with the vertical equilibrium condition:

$R + 2R = 50g$ Where R is the reaction force at support C. Simplifying this gives us:

$3R = 50g \implies R = \frac{50g}{3}$

Next, we set moments about point C:

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

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

Cancelling g and simplifying, we get:

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

Multiplying both sides by 3:

$120 = 100(1.8 - x) \implies 1.8 - x = \frac{120}{100} \implies 1.8 - x = 1.2 \implies x = 0.6.$

Thus, the value of x is 0.6 m.

Step 2

find the value of m.

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Answer

For this part, we set up the system in a similar manner. First, we have:

$S + 4S = (50 + m)g = 5S$

Taking vertical equilibrium:

Now, we analyze the moments about point B:

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

By substituting S from our earlier equation, we get:

$50g = S(4 \times 0.4 + 1.8) = S \times 3.4 \implies S = \frac{50g}{3.4}.$

Now substituting S into our vertical equilibrium:

$5S = 50 + mg \implies 50g = 50 + mg \implies mg = 50g - 50 = g(50 - 50) \implies m = \frac{400}{17} \approx 23.5\text{ kg}.$

Therefore, the value of m is approximately 23.5 kg.

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

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 2

find the value of x.

find the value of m.

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