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 Newtons Second Law: 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 determine the distance x, we use the conditions of equilibrium for moments about a point. Let R_C be the reaction at support C and R_D be the reaction at support D.

Since the rod is balanced, we have: $R_D = 2 R_C$

The total weight of the rod acts through its center of mass, which is at the midpoint (1m from A). Therefore, taking moments about point C:

$R_D imes (2 - 0.2) = 50 imes 1$

Substituting for R_D: $2 R_C (1.8) = 50$ $R_C = \frac{50}{3.6} = 13.89 ext{ N}$

Then, substituting back to find R_D: $R_D = 2(13.89) = 27.78 ext{ N}$

Now to find x:

Using the equation of vertical forces: $R_C + R_D = 50 \\ 13.89 + 27.78 = 50$

Thus: $x = 2 - 0.2 - 1 = 0.8 ext{ m}$

Step 2

find the value of m.

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Answer

With the support now at E, we apply similar principles. Let R_E be the reaction at support E, and we know: $R_E = 4R_C$

The total length changes to 2 - 0.4 = 1.6 m from E to B. Taking moments about point C:

The equations for forces give us: $R_E + R_C = 50 + mg$

For moments about C with respect to the new position: $R_E (1.6) = 50 (1.4) + m g (1.4)$

Substituting R_E: $4R_C (1.6) = 70 + 1.4mg$

Since we know R_C: $4(13.89)(1.6) = 70 + 1.4mg$

Calculating R_E: $88.13 = 70 + 1.4mg \\ 1.4mg = 18.13 \\ m = \frac{18.13}{1.4g} \\ m = 1.83 ext{ kg (approximately)}$

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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