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 at an angle of 30°. The plane is inclined to the horizontal at an angle of 20°, as shown in Figure 2. The line of action of the force lies in the vertical plane containing P and a line of greatest slope of the plane. The coefficient of friction between P and the plane is μ. Given that P is on the point of sliding up the plane, find the value of μ.

A-Level Edexcel Maths Mechanics Moments: A particle P of mass 2 kg is hel

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

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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. 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 Forces Perpendicular to the Plane

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Answer

To find the normal reaction (R) acting on the particle P, we resolve the forces acting perpendicular to the inclined plane:

The weight of P acting downwards perpendicular to the plane is given by the component: $W = mg = 2 imes 9.81 ext{ N}$ where $g$ is the acceleration due to gravity.
Components of the 40 N force perpendicular to the plane: $F_{perpendicular} = 40 imes ext{cos}(30^ ext{o})$ This results in: $R = 2g ext{cos}(20^ ext{o}) + 40 ext{cos}(30^ ext{o})$

Step 2

Resolve Forces Along the Plane

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Answer

Next, we resolve the forces acting along the incline:

The component of weight parallel to the incline: $W_{parallel} = mg ext{sin}(20^ ext{o})$
The component of the force (40 N) parallel to the incline: $F_{parallel} = 40 imes ext{sin}(30^ ext{o})$
The frictional force can be expressed as: $F_{friction} = ext{μ} R$ Therefore, at the point of sliding, we have: $40 ext{sin}(30^ ext{o}) - mg ext{sin}(20^ ext{o}) - ext{μ} R = 0$

Step 3

Combine Equations to Find μ

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Answer

Substituting the expressions for R and the forces into the equation:

Substitute values for R: $R = 2g ext{cos}(20^ ext{o}) + 40 ext{cos}(30^ ext{o})$
Rearranging the equation will give: $ext{μ} = \frac{(40 ext{sin}(30^ ext{o}) - mg ext{sin}(20^ ext{o}))}{R}$
Solve for μ: After substituting the actual values, we calculate μ to find: $ext{μ} = 0.727$

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

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

Resolve Forces Perpendicular to the Plane

Resolve Forces Along the Plane

Combine Equations to Find μ

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