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 Further Forces & Newtons Laws: 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 1

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 1

Step 1

Equation of motion of B:

96%

114 rated

Answer

For particle B (mass 4 kg), according to Newton's second law: $4g - T = 4a$ Where:

( g \approx 9.81 ,\text{m/s}^2 ) is the acceleration due to gravity,
( T ) is the tension in the string,
( a ) is the acceleration of the system.

Step 2

Equation of motion of A:

99%

104 rated

Answer

For particle A (mass 2 kg): $T - F - 2g \sin(30^{\circ}) = 2a$ Where:

( F = \mu R ) (frictional force),
( \mu = \frac{1}{\sqrt{3}} ) is the coefficient of friction,
( R = 2g \cos(30^{\circ}) ) is the normal reaction.

Thus, $F = \frac{1}{\sqrt{3}} \cdot 2g \cos(30^{\circ})$

Step 3

Resolving perpendicularly to the plane:

96%

101 rated

Answer

The normal reaction for particle A is: $R = 2g \cos(30^{\circ})$

Step 4

Eliminating a:

98%

120 rated

Answer

From the two equations:

From particle B: ( T = 4g - 4a )
Substitute ( T ) in the equation for A: $4g - 4a - 2g \sin(30^{\circ}) - F = 2a$ This allows us to find a in terms of known quantities.

Step 5

Final expression for T:

97%

117 rated

Answer

Using the earlier equations: $T - F = 2a + 2g \sin(30^{\circ})$ With the known values and solving yields the tension ( T ) in the string 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 1

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

Equation of motion of B:

Equation of motion of A:

Resolving perpendicularly to the plane:

Eliminating a:

Final expression for T:

Join the A-Level students using SimpleStudy...