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A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2008 - Paper 1
Question 7
A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal. The package is held in equilibrium by a force of magnitude 45 N acting at an angle of... show full transcript
Worked Solution & Example Answer:A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2008 - Paper 1
Step 1
a) the magnitude of the normal reaction of the plane on the package
Answer
To find the normal reaction force (R), we analyze the forces acting on the package along the direction perpendicular to the surface of the inclined plane. The vertical forces include the component of the weight acting perpendicular to the incline and the vertical component of the applied force.
We can compute R as:
R = 45 ext{ N} imes rac{4}{5} ext{ (cos 40°)} + 4g imes rac{ ext{cos 30°}}{1}
Substituting values:
- The weight of the package is .
- Then calculating:
R = 45 imes rac{4}{5} imes ext{cos}(40°) + 39.24 imes ext{cos}(30°)
Solving this gives approximately R ≈ 68 N.
Step 2
b) the coefficient of friction between the plane and the package
Answer
To calculate the coefficient of friction (μ), we use the equilibrium condition along the plane. The forces acting along the incline must sum to zero since the package is in equilibrium:
Where F is the component of the applied force that acts along the slope. As the applied force (45 N) has a component down the slope given by:
Calculating each component, we find:
- The gravitational component down the slope:
4g imes ext{sin}(30°) = 4 imes 9.81 imes rac{1}{2} = 19.62 ext{ N}
Equating it:
Finally, solving for μ: μ = rac{45 + 19.62}{R}
This results in a coefficient of friction approximately μ ≈ 0.14.