A particle P of mass 2.7 kg lies on a rough plane inclined at 40° to the horizontal. The particle is held in equilibrium by a force of magnitude 15 N acting at an angle of 50° to the plane, as shown in Figure 4. The force acts in a vertical plane containing a line of greatest slope of the plane. The particle is in equilibrium and is on the point of sliding down the plane. Find a) the magnitude of the normal reaction of the plane on P, b) the coefficient of friction between P and the plane. The force of magnitude 15 N is removed. c) Determine whether P moves, justifying your answer.

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A particle P of mass 2.7 kg lies on a rough plane inclined at 40° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2014 - Paper 1

Question 7

A particle P of mass 2.7 kg lies on a rough plane inclined at 40° to the horizontal. The particle is held in equilibrium by a force of magnitude 15 N acting at an an... show full transcript

Worked Solution & Example Answer:A particle P of mass 2.7 kg lies on a rough plane inclined at 40° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2014 - Paper 1

Step 1

a) the magnitude of the normal reaction of the plane on P

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Answer

To find the normal reaction force (N), we consider the forces acting perpendicular to the inclined plane. The weight of the particle can be resolved into two components: one perpendicular to the plane and one parallel.

The weight, W, acting on the particle is: $W = mg = 2.7 ext{ kg} \times 9.81 \text{ m/s}^2 = 26.487 \text{ N}.$
The component of weight perpendicular to the plane: $W_{perpendicular} = W \cos(40°) = 26.487 \cos(40°),$

which approximately equals 20.32 N.
The vertical component of the applied force acting perpendicular to the plane is: $F_{perpendicular} = 15 \sin(50°) = 11.49 \text{ N}.$
Therefore, the normal reaction can be calculated as: $N = W_{perpendicular} - F_{perpendicular}$ $N = 20.32 - 11.49 = 8.83 \text{ N}.$

Thus, the magnitude of the normal reaction is approximately 8.83 N.

Step 2

b) the coefficient of friction between P and the plane

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Answer

The frictional force (F_f) can be expressed as: $F_f = ext{coefficient of friction} \times N$

We need to find the component of forces parallel to the plane:

The component of the weight acting down the slope is: $W_{parallel} = W \sin(40°) = 26.487 \sin(40°) \approx 17.01 \text{ N}.$
Since the particle is in equilibrium, the friction force must equal the parallel component of the weight: $F_f = W_{parallel} = 17.01 \text{ N}.$
Substituting in the formula for the coefficient of friction, we get: $F_f = \mu N \Rightarrow \mu = \frac{F_f}{N} = \frac{17.01}{8.83} \approx 1.93.$

Thus, the coefficient of friction between P and the plane is approximately 1.93.

Step 3

c) Determine whether P moves, justifying your answer

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Answer

After the removal of the 15 N force, we need to reconsider the forces acting on P:

The frictional force that was previously acting on P will still act, opposing motion. Using the coefficient from part b, the maximum static frictional force is: $F_{f(max)} = \mu N = 1.93 \times 8.83 \approx 17.01 \text{ N}.$
The parallel component of the weight acting down the slope remains: $W_{parallel} \approx 17.01 \text{ N}.$

Since the frictional force equals the weight component (17.01 N), the particle P is on the verge of sliding. Given that the applied force was removed, P will move down the plane as the static friction is not sufficient to hold it in equilibrium.

Thus, P will begin to slide down the plane.

A particle P of mass 2.7 kg lies on a rough plane inclined at 40° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2014 - Paper 1

Worked Solution & Example Answer:A particle P of mass 2.7 kg lies on a rough plane inclined at 40° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2014 - Paper 1

a) the magnitude of the normal reaction of the plane on P

b) the coefficient of friction between P and the plane

c) Determine whether P moves, justifying your answer

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