A small ring of mass 0.25 kg is threaded on a fixed rough horizontal rod. The ring is pulled upwards by a light string which makes an angle 40° with the horizontal, as shown in Figure 3. The string and the rod are in the same vertical plane. The tension in the string is 1.2 N and the coefficient of friction between the ring and the rod is μ. Given that the ring is in limiting equilibrium, find (a) the normal reaction between the ring and the rod, (b) the value of μ.

A-Level Edexcel Maths Mechanics Moments: A small ring of mass 0.25 kg is

A small ring of mass 0.25 kg is threaded on a fixed rough horizontal rod - Edexcel - A-Level Maths Mechanics - Question 5 - 2007 - Paper 1

Question 5

A small ring of mass 0.25 kg is threaded on a fixed rough horizontal rod. The ring is pulled upwards by a light string which makes an angle 40° with the horizontal, ... show full transcript

Worked Solution & Example Answer:A small ring of mass 0.25 kg is threaded on a fixed rough horizontal rod - Edexcel - A-Level Maths Mechanics - Question 5 - 2007 - Paper 1

Step 1

(a) the normal reaction between the ring and the rod

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Answer

To find the normal reaction, we start by analyzing the forces acting on the ring. The forces involved include:

Weight of the ring: This force acts downwards and is calculated as: $W = mg = 0.25 ext{ kg} imes 9.81 ext{ m/s}^2 = 2.4525 ext{ N}$
Tension in the string: The vertical component of this force can be found using trigonometry: $T_y = 1.2 ext{ N} imes ext{sin}(40°) \ ext{T}_y \ = 1.2 imes 0.6428 \ = 0.77136 ext{ N}$

Setting up the equilibrium equation in the vertical direction, we have:

$R + T_y = W \ R + 0.77136 = 2.4525 \ R = 2.4525 - 0.77136 \ R = 1.68114 ext{ N}$

Thus, the normal reaction between the ring and the rod is approximately 1.68 N.

Step 2

(b) the value of μ

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Answer

To find the value of the coefficient of friction μ, we can use the relationship between the frictional force and the normal reaction:

The horizontal component of tension $T_x$ can be expressed as: $T_x = T imes ext{cos}(40°) \ T_x = 1.2 ext{ N} imes 0.7660 \ T_x \approx 0.9192 ext{ N}$
The frictional force is equal to the horizontal component of tension under limiting equilibrium: $F = ext{μ} imes R$

Rearranging gives: $μ = rac{F}{R} = rac{T_x}{R} \ μ = rac{0.9192 ext{ N}}{1.68114 ext{ N}} \ μ ext{ approximately equals } 0.5477.$

Therefore, the value of the coefficient of friction μ is approximately 0.55.

A small ring of mass 0.25 kg is threaded on a fixed rough horizontal rod - Edexcel - A-Level Maths Mechanics - Question 5 - 2007 - Paper 1

Worked Solution & Example Answer:A small ring of mass 0.25 kg is threaded on a fixed rough horizontal rod - Edexcel - A-Level Maths Mechanics - Question 5 - 2007 - Paper 1

(a) the normal reaction between the ring and the rod

(b) the value of μ

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