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 Forces: 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

(a) find the value of x.

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Answer

To solve for the value of x, we start by applying the principles of static equilibrium. The sum of vertical forces acting on the rod must equal zero, and the sum of moments around any point must also equal zero.

Calculate the total weight of the rod:
The weight, W, of the rod can be calculated using the formula:

$W = mg = 50 ext{ kg} \times 9.81 ext{ m/s}^2 = 490.5 ext{ N}$
Set up the equilibrium equations:
Let R_C be the reaction at C and R_D be the reaction at D. Given that R_D = 2R_C, the equilibrium of vertical forces gives us:

$R_C + R_D = W$

Substituting R_D: $R_C + 2R_C = 490.5 ext{ N} \ o 3R_C = 490.5 ext{ N} \ o R_C = \frac{490.5 ext{ N}}{3} = 163.5 ext{ N}$

Therefore, R_D = 2R_C = 327 ext{ N}.
Calculate moments around point C:
Taking moments about point C:

$R_D \times 1.8 = W \times 1.0$

Substituting for R_D and W: $327 \times 1.8 = 490.5 \times (0.2 + x)$

Simplifying gives us: $587.6 = 490.5 \times (0.2 + x)$
Solve for x:
Dividing both sides by 490.5: $\frac{587.6}{490.5} = 0.2 + x$ $1.197 = 0.2 + x \to x = 1.197 - 0.2 \to x \approx 0.997 ext{ m}$ Thus, the value of x is approximately 0.997 m.

Step 2

(b) find the value of m.

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Answer

In this part, we have moved support D to point E, where EB = 0.4 metres and we need to find the mass m placed at point B.

Identify new equilibrium conditions:
The reactions must still satisfy the equilibrium conditions. Denote R_E as the reaction at E. Given that R_E = 4R_C:

$R_E = 4R_C = 4 \cdot 163.5 = 654 ext{ N}.$
Set up the vertical forces equation:
Considering the vertical forces:

$R_E + R_C = W + mg$
Moment about point E:
Taking moments about point E gives:

$R_C \times 1.6 = W \times 1.6 \to 163.5 \times 1.6 = 490.5 \times 0.4$
Simplify: $261.6 = 196.2 + 0.4m$
Solve for m:
Rearranging gives: $0.4m = 261.6 - 196.2 \to 0.4m = 65.4 \to m = \frac{65.4}{0.4} = 163.5 ext{ kg}.$ Therefore, the value of m is 163.5 kg.

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

(a) find the value of x.

(b) find the value of m.

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