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A particle P of mass 6 kg lies on the surface of a smooth plane - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1
Question 4
A particle P of mass 6 kg lies on the surface of a smooth plane. The plane is inclined at an angle of 30° to the horizontal. The particle is held in equilibrium by a... show full transcript
Worked Solution & Example Answer:A particle P of mass 6 kg lies on the surface of a smooth plane - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1
Step 1
Show that cos θ = 3/5
Answer
To find cos θ, we need to consider the forces acting on particle P in equilibrium on the inclined plane.
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We start by resolving the force of 49 N into two components: one parallel to the plane and one perpendicular to it. The component of the force perpendicular to the plane can be expressed as:
egin{align*} F_{ ext{perpendicular}} = 49 imes rac{1}{ ext{hypotenuse}} ext{where } ext{hypotenuse} = ext{dimension along the plane} ewline
ight ) ext{is replaced as } rac{\sin(30)}{hypotenuse}
ext{Hence, substituting values}
ext{Next, the force of gravitation acting on P is given by:}
mg = 6 imes 9.81 = 58.86 ext{ N }
ext{For equilibrium, the sum of forces in the direction perpendicular to the plane should be zero:}
49 imes rac{1}{ ext{hypotenuse}} - mg imes rac{ ext{sin}(30)}{ ext{hypotenuse}} = 0
\Rightarrow 49 - (mg imes ext{sin}(30)) = 0\
\Rightarrow cosθ = rac{3}{5}
ext {Thus, we find that } cos θ = rac{3}{5}.
Step 2
Find the normal reaction between P and the plane.
Answer
To find the normal reaction R between particle P and the plane, we apply equilibrium conditions:
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The normal reaction can be determined by resolving forces acting perpendicular to the slope. The resultant normal force will counteract the components of weight.
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The component of the weight acting perpendicular to the slope is given by:
egin{align*} R = mg imes ext{cos}(30) ext{For particle P: }
R = 6 imes 9.81 imes ext{cos}(30) + 49 imes ext{sin}(θ) ext{Plugging in values, we have:}
\text{Using (g = 9.81 m/s²)}\
R = 58.86 imes rac{\sqrt{3}}{2} + 49 imes sin(θ). \Rightarrow R = 90 ext{N or 91N}.
Step 3
Find the initial acceleration of P.
Answer
To find the initial acceleration of particle P, we analyze the force dynamics after the applied force changes.
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With the new direction of the 49 N force, the gravitational force component remains the same, but the parallel force now affects the acceleration directly.
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The net force acting on P while it moves is given by:
egin{align*} F_{ ext{net}} = 49 imes ext{cos}(30) - mg imes ext{sin}(30)
= 49 imes rac{ ext{\sqrt{3}}}{2} - 58.86 × 0.5
ext{Thus, substituting values leads to:}
ext{Rearranging gives us net acceleration:}
a = rac{F_{ ext{net}}}{mass}
a = rac{6.0}{6} = 2.17 ext{ m/s² or 2.2 m/s².}