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A particle P of mass 2 kg is held at rest in equilibrium on a rough plane by a constant force of magnitude 40 N - Edexcel - A-Level Maths Mechanics - Question 5 - 2016 - Paper 1
Question 5
A particle P of mass 2 kg is held at rest in equilibrium on a rough plane by a constant force of magnitude 40 N. The direction of the force is inclined to the plane ... show full transcript
Worked Solution & Example Answer:A particle P of mass 2 kg is held at rest in equilibrium on a rough plane by a constant force of magnitude 40 N - Edexcel - A-Level Maths Mechanics - Question 5 - 2016 - Paper 1
Step 1
Resolve the forces acting on particle P
Answer
Using the given values, we resolve the forces acting on particle P along the direction parallel and perpendicular to the plane.
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Weight of Particle (W):
W = mg = 2 ext{ kg} imes 9.81 ext{ m/s}^2 = 19.62 ext{ N}
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Resolve forces parallel to the slope:
F_{ ext{parallel}} = W imes ext{sin}(20^ ext{o}) - R imes ext{sin}(30^ ext{o}) = 19.62 imes ext{sin}(20^ ext{o}) - 40 imes ext{sin}(30^ ext{o})
= 19.62 imes 0.342 - 40 imes 0.5
= 6.71 - 20 = -13.29 ext{ N}
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Resolve forces perpendicular to the slope:
R = mg imes ext{cos}(20^ ext{o}) - F imes ext{cos}(30^ ext{o})
R + 19.62 imes ext{cos}(20^ ext{o}) = 40 imes ext{cos}(30^ ext{o})
Step 2
Step 3
Equation of motion for friction
Answer
Using the coefficient of friction μ:
F_{ ext{friction}} = μR ext{ , where } F_{ ext{friction}} = W imes ext{cos}(20^ ext{o})
imes 0.9397
ext{Thus, with } F_{ ext{parallel}} = 0 ext{, we get: }
0 = W imes ext{ sin}(20^ ext{o}) - μR
μ = rac{19.62 imes 0.342}{R}
= rac{6.71}{16.21} = 0.414