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 respectively, are attached to the ends of a light inextensible string which passes over the pulley P. The part of the string from A to P is parallel to a line of greatest slope of the plane and B hangs freely below P, as shown in Figure 2. The coefficient of friction between A and the plane is $ \frac{1}{\sqrt{3}} $. Initially A is held at rest on the plane. The particles are released from rest with the string taut and A moves up the plane. Find the tension in the string immediately after the particles are released.

A-Level Edexcel Maths Mechanics Newtons Second Law: A fixed rough plane is inclined

A fixed rough plane is inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 2

Question 3

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 2

Step 1

Equation of motion of B

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Answer

For particle B of mass 4 kg, the forces acting on it include its weight downward and the tension T in the string upward. The equation of motion is given by:

$4g - T = 4a$

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

Step 2

Equation of motion of A

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Answer

For particle A of mass 2 kg, the forces acting include the tension T in the string upward, the normal reaction R perpendicular to the surface, and the frictional force F acting down the incline. The equation of motion is:

$T - R - F - 2g \sin(30°) = 2a$

The frictional force F can be expressed as ( F = \mu R ) where ( \mu = \frac{1}{\sqrt{3}} ). Therefore,

$T - R - \frac{1}{\sqrt{3}}R - 2g \sin(30°) = 2a$

Step 3

Resolve forces perpendicular to the plane

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Answer

Resolving forces perpendicular to the plane, we have:

$R = 2g \cos(30°)$

Substituting this into the equation of motion for A allows us to express everything in terms of g and a.

Step 4

Final calculations for tension

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Answer

Substituting for T from the equation of motion of B and resolving gives us:

$2T - 4g = - 4a$ This can be rearranged to find T:

$2T = 4g - 4a$ Now substituting the calculated value of a in terms of g allows us to find the tension T immediately after the particles are released.

A fixed rough plane is inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 2

Worked Solution & Example Answer:A fixed rough plane is inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 2

Equation of motion of B

Equation of motion of A

Resolve forces perpendicular to the plane

Final calculations for tension

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