A parcel of mass 5 kg lies on a rough plane inclined at an angle $ \alpha $ to the horizontal, where $ \tan \alpha = \frac{1}{4} $. The parcel is held in equilibrium by the action of a horizontal force of magnitude 20 N, as shown in Fig. 2. The force acts in a vertical plane through a line of greatest slope of the plane. The parcel is on the point of sliding down the plane. Find the coefficient of friction between the parcel and the plane.

A-Level Edexcel Maths Mechanics Moments: A parcel of mass 5 kg lies on a

A parcel of mass 5 kg lies on a rough plane inclined at an angle $ \alpha $ to the horizontal, where $ \tan \alpha = \frac{1}{4} $ - Edexcel - A-Level Maths Mechanics - Question 4 - 2003 - Paper 1

Question 4

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A parcel of mass 5 kg lies on a rough plane inclined at an angle $ \alpha $ to the horizontal, where $ \tan \alpha = \frac{1}{4} $. The parcel is held in equilib... show full transcript

Worked Solution & Example Answer:A parcel of mass 5 kg lies on a rough plane inclined at an angle $ \alpha $ to the horizontal, where $ \tan \alpha = \frac{1}{4} $ - Edexcel - A-Level Maths Mechanics - Question 4 - 2003 - Paper 1

Step 1

Find the Normal Force $ R $

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Answer

To find the normal force, we use the components of forces acting on the parcel. The resultant normal force ( R ) can be expressed as:

$R = 5g \cos \alpha + 20 \sin \alpha$

Substituting ( g = 9.8 , \text{m/s}^2 ) gives us:

$R = 5(9.8) \cos \alpha + 20 \sin \alpha$ .

Step 2

Establish the Vertical Force Equation

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Answer

We set up the vertical force equation using the information provided:

$F + 20 \cos \alpha = 5g \sin \alpha$

This can help us find the forces acting vertically.

Step 3

Determine Tension and Force

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Using the previously calculated value of ( R ), we find:

( R = 51.2 , \text{N} ) and ( F = 13.4 , \text{N} ).

Step 4

Apply the Coefficient of Friction Formula

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Answer

We can establish the relationship for frictional force using:

$F = \mu R$

Substituting the value of ( F ) gives:

$13.4 = \mu \cdot 51.2$

Thus, solving for ( \mu ) we find:

$\mu = \frac{13.4}{51.2} \approx 0.262$

A parcel of mass 5 kg lies on a rough plane inclined at an angle \( \alpha \) to the horizontal, where \( \tan \alpha = \frac{1}{4} \) - Edexcel - A-Level Maths Mechanics - Question 4 - 2003 - Paper 1

Worked Solution & Example Answer:A parcel of mass 5 kg lies on a rough plane inclined at an angle \( \alpha \) to the horizontal, where \( \tan \alpha = \frac{1}{4} \) - Edexcel - A-Level Maths Mechanics - Question 4 - 2003 - Paper 1

Find the Normal Force \( R \)

Establish the Vertical Force Equation

Determine Tension and Force

Apply the Coefficient of Friction Formula

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