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 Further Forces & Newtons Laws: 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

96%

114 rated

Answer

To find the normal reaction, we consider the forces acting on the ring. The vertical forces include the weight of the ring ( W = mg = 0.25 imes 9.81 = 2.4525 ext{ N} ) and the vertical component of the tension in the string.

The tension in the string is given as 1.2 N, and its vertical component can be calculated as:

$T_y = 1.2 imes ext{sin}(40°) \approx 1.2 imes 0.6428 \approx 0.7714 ext{ N}$

In equilibrium, the sum of the vertical forces must equal zero:

$R + T_y = W \Rightarrow R + 0.7714 = 2.4525 \Rightarrow R = 2.4525 - 0.7714 \approx 1.6811 ext{ N}$

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

Step 2

b) the value of μ

99%

104 rated

Answer

To find the coefficient of friction μ, we start by analyzing the horizontal forces acting on the ring. The frictional force (F) can be expressed as:

$F = μR$

The horizontal component of the tension in the string can be found as:

$T_x = 1.2 imes ext{cos}(40°) \approx 1.2 imes 0.7660 \approx 0.9192 ext{ N}$

At limiting equilibrium, the frictional force equals the horizontal component of the tension:

$μR = T_x \Rightarrow μ = \frac{T_x}{R}$

Substituting the previously calculated values:

$μ = \frac{0.9192}{1.6811} \approx 0.5464$

Therefore, the value of 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

Join the A-Level students using SimpleStudy...