A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons. The force acts in the vertical plane containing the line of greatest slope of the inclined plane which passes through the particle. The plane is inclined to the horizontal at an angle α, where tan(α) = rac{3}{4}, as shown in Figure 2. The coefficient of friction between the particle and the plane is rac{1}{3}. Given that the particle is on the point of sliding up the plane, find (a) the magnitude of the normal reaction between the particle and the plane, (b) the value of P. (5)

A-Level Edexcel Maths Mechanics Moments: A particle of mass 0.4 kg is hel

A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons - Edexcel - A-Level Maths Mechanics - Question 7 - 2010 - Paper 1

Question 7

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A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons. The force acts in the vertical plane containing the li... show full transcript

Worked Solution & Example Answer:A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons - Edexcel - A-Level Maths Mechanics - Question 7 - 2010 - Paper 1

Step 1

(a) the magnitude of the normal reaction between the particle and the plane

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Answer

To find the magnitude of the normal reaction (R) between the particle and the plane, we can use the equation of motion in the direction perpendicular to the plane:

$R = rac{1}{3}R$

From the force in this direction, we can derive:

$R ext{ cos }α - F ext{ sin }α = 0.4g$

Substituting the values, we have:

$R ext{ cos }α - 0.4g ext{ sin }α = 0$

Next, we will evaluate the trigonometric values substituting α:

Using $tan(α) = rac{3}{4}$ , we have

$ext{sin } α = rac{3}{5}$ and $ext{cos } α = rac{4}{5}$ .

Now plugging these values into the equation:

$R rac{4}{5} - 0.4g rac{3}{5} = 0$

Solving for R gives us:

$R = rac{0.4g imes 3}{4} = 6.53 ext{ or } 6.5.$

Step 2

(b) the value of P

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Answer

To calculate the value of P, we explore the forces acting horizontally:

The horizontal component of the force P can be expressed as:

$P ext{ cos }α - F ext{ sin }α - R ext{ sin }α = 0$

Replacing the known components yields:

$P rac{4}{5} - F rac{3}{5} - R rac{3}{5} = 0$

We substitute the value of R from part (a):

$P rac{4}{5} - 0.4g rac{3}{5} - (6.5) rac{3}{5} = 0$

Solving this equation will give us:

$P = rac{0.4g imes rac{3}{5} + 6.5 imes rac{3}{5}}{ rac{4}{5}}$

Calculating the final value leads to:

$P = 5.66 ext{ or } 5.7.$

A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons - Edexcel - A-Level Maths Mechanics - Question 7 - 2010 - Paper 1

Worked Solution & Example Answer:A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons - Edexcel - A-Level Maths Mechanics - Question 7 - 2010 - Paper 1

(a) the magnitude of the normal reaction between the particle and the plane

(b) the value of P

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