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A fixed rough plane is inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 1

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A fixed rough plane is inclined at 30° to the horizontal. A small smooth pulley P is fixed at the top of the plane. Two particles A and B, of mass 2 kg and 4 kg resp... show full transcript

Worked Solution & Example Answer:A fixed rough plane is inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 1

Step 1

Equation of motion for B

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Answer

For particle B (mass 4 kg), we consider its equation of motion:

[ 4g - T = 4a ]

where ( g ) is the acceleration due to gravity and ( a ) is the acceleration of the particles.

Step 2

Equation of motion for A

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For particle A (mass 2 kg), the equation of motion is expressed as:

[ T - F - 2g \sin(30°) = 2a ]

Here, ( F = \mu R ) and ( R ) is the normal reaction force.

Step 3

Resolving perpendicular to the plane

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To resolve the forces perpendicular to the plane:

[ R = 2g \cos(30°) ]

where ( \mu = \frac{1}{\sqrt{3}} ) gives us the frictional force.

Step 4

Using \( F = \mu R \)

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Substituting the expression for ( R ):

[ F = \mu R = \frac{1}{\sqrt{3}} (2g \cos(30°)) ]

We calculate the value of frictional force and substitute back into the equation for A.

Step 5

Final Equations

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Now, substituting ( F ) back into the second equation:

[ T - \frac{1}{\sqrt{3}}(2g \cos(30°)) - 2g \sin(30°) = 2a ]

Combining these equations gives us all the terms required to isolate ( T ) and find its value.

Step 6

Solving the equations

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Using the equations from the system:

From the equation of motion for B: [ T = 4g - 4a ] Substituting this into the equation for A leads to: [ 2T - 4g - T + F = 2a ] This will allow us to solve for the tension ( T ) after substituting the known values of ( g ) and simplifying.

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