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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
Find the normal reaction between the ring and the rod
Answer
To find the normal reaction, consider the forces acting on the ring. We have:
- The weight of the ring, which can be calculated using the formula: ( W = mg = 0.25 , \text{kg} \times 9.81 , \text{m/s}^2 \approx 2.45 , \text{N} ).
- The tension in the string, given as 1.2 N, acting at an angle of 40°.
Using the vertical component of the forces:
[ R + 1.2 \sin(40°) = 0.25g ]
Substituting the values:
[ R + 1.2 \sin(40°) = 2.45 ]
Solving for R:
[ R = 2.45 - 1.2 \sin(40°) \approx 1.7 , \text{N} ]
Thus, the normal reaction ( R ) between the ring and the rod is approximately 1.7 N.
Step 2
Find the value of μ
Answer
To find the coefficient of friction ( μ ), we consider the horizontal forces acting on the ring. The frictional force ( F ) is given by:
[ F = μR ]
The horizontal component of the tension can be calculated as follows:
[ F = 1.2 \cos(40°) \approx 1.2 \times 0.766 \approx 0.919 , \text{N} ]
Equating the frictional force to the horizontal component of the tension:
[ 1.2 \cos(40°) = μR ]
Substituting ( R \approx 1.7 , \text{N} ):
[ 1.2 \times 0.766 = μ \times 1.7 ]
Solving for μ gives:
[ μ = \frac{1.2 \times 0.766}{1.7} \approx 0.548 ]
Thus, the value of the coefficient of friction ( μ ) is approximately 0.55.