A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force. The plane is inclined at an angle α to the horizontal, where tan α = rac{3}{4}. The force acts in a vertical plane containing a line of greatest slope of the plane and has magnitude P newtons, as shown in Figure 2. The coefficient of friction between the package and the plane is 0.5 and the package is modelled as a particle. The package is in equilibrium and on the point of slipping down the plane. (a) Draw, on Figure 2, all the forces acting on the package, showing their directions clearly. (b) (i) Find the magnitude of the normal reaction between the package and the plane. (ii) Find the value of P.

A-Level Edexcel Maths Mechanics Moments: A small package of mass 1.1 kg i

A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force - Edexcel - A-Level Maths Mechanics - Question 5 - 2009 - Paper 1

Question 5

A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force. The plane is inclined at an angle α to the horizontal, where tan α = r... show full transcript

Worked Solution & Example Answer:A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force - Edexcel - A-Level Maths Mechanics - Question 5 - 2009 - Paper 1

Step 1

Draw, on Figure 2, all the forces acting on the package, showing their directions clearly.

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Answer

The forces acting on the package include:

Weight (W): acting vertically downward, calculated as:

$W = mg = 1.1 imes 9.8 = 10.78 ext{ N}$
Normal force (R): acting perpendicular to the incline which pushes upwards.
Frictional force (F): acting parallel to the incline opposing the motion of the package.
Applied force (P): acting down the plane along the incline.

These forces should be labeled clearly in the diagram.

Step 2

(i) Find the magnitude of the normal reaction between the package and the plane.

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Answer

To find the normal reaction R, we resolve the weight into components:

The component of weight perpendicular to the incline is given by:

$W_{ ext{perpendicular}} = mg imes ext{cos}( heta)$

Using
$heta = an^{-1} rac{3}{4}$ ,

we can find:

$R = W_{ ext{perpendicular}} = rac{1.1 imes 9.8}{ ext{cos}( heta)} = rac{10.78}{ ext{cos}( heta)} ext{ N}$

Solving yields:

$R ext{ is approximately } 9.8 ext{ N}$ .

Step 3

(ii) Find the value of P.

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Answer

Applying the equilibrium equations along the incline, we have:

$P + 0.5R = R imes ext{sin}( heta)$

Substituting previously found values:

The frictional force is calculated as:

$F = rac{1}{2}R$

Thus:

$P = R imes ext{sin}( heta) - 0.5R$

After substituting R and solving for P gives:

$P = 1.96 ext{ N}$ .

A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force - Edexcel - A-Level Maths Mechanics - Question 5 - 2009 - Paper 1

Worked Solution & Example Answer:A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force - Edexcel - A-Level Maths Mechanics - Question 5 - 2009 - Paper 1

Draw, on Figure 2, all the forces acting on the package, showing their directions clearly.

(i) Find the magnitude of the normal reaction between the package and the plane.

(ii) Find the value of P.

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