A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$. A brick P of mass $m$ is placed on the plane. The coefficient of friction between P and the plane is $\mu$. Brick P is in equilibrium and on the point of sliding down the plane. Brick P is modelled as a particle. Using the model, (a) find, in terms of $m$ and $g$, the magnitude of the normal reaction of the plane on brick P. (b) show that $\mu = \frac{3}{4}$. For parts (c) and (d), you are not required to do any further calculations. Brick P is now removed from the plane and a much heavier brick Q is placed on the plane. The coefficient of friction between Q and the plane is also $\frac{3}{4}$. (c) Explain briefly why brick Q will remain at rest on the plane. Brick Q is now projected with speed $0.5 \text{ m s}^{-1}$ down a line of greatest slope of the plane. Brick Q is modelled as a particle. (d) describe the motion of brick Q, giving a reason for your answer.

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A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$ - Edexcel - A-Level Maths Mechanics - Question 1 - 2020 - Paper 1

Question 1

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A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$. A brick P of mass $m$ is placed on the plane. The coefficien... show full transcript

Worked Solution & Example Answer:A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$ - Edexcel - A-Level Maths Mechanics - Question 1 - 2020 - Paper 1

Step 1

find, in terms of m and g, the magnitude of the normal reaction of the plane on brick P.

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Answer

To find the normal reaction $R$ on brick P, we can resolve the weight of the brick perpendicular to the inclined plane. The weight $W = mg$ , where $g$ represents the acceleration due to gravity. The component of the weight acting perpendicular to the plane is given by:

$R = mg \cos \alpha$

Since $\tan \alpha = \frac{3}{4}$ , we can use the relationship between sine and cosine. Therefore, we find $\cos \alpha = \frac{4}{5}$ :

$R = mg \cos \alpha = mg \cdot \frac{4}{5} = \frac{4}{5} mg.$

Step 2

show that \mu = \frac{3}{4}.

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Answer

For brick P to be on the verge of sliding, the force of friction $F$ must be equal to the component of the weight acting down the plane. Therefore:

The component of the weight down the plane is: $F = mg \sin \alpha$ Using $\sin \alpha = \frac{3}{5}$ , we have: $F = mg \cdot \frac{3}{5}.$
The maximum frictional force is given by: $F = \mu R$ Substituting for $R$ , we obtain: $\mu \cdot \frac{4}{5} mg = \frac{3}{5} mg.$
Dividing both sides by $mg$ and simplifying gives: $\mu \cdot \frac{4}{5} = \frac{3}{5} \Rightarrow \mu = \frac{3}{4}.$

Step 3

Explain briefly why brick Q will remain at rest on the plane.

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Answer

Brick Q will remain at rest because the forces acting on it (the gravitational force and the frictional force) will balance out. The frictional force opposing the motion down the slope will increase proportionally to the component of its weight acting down the inclined plane. Hence, as long as the static friction is greater than or equal to this component, brick Q will not slide.

Step 4

describe the motion of brick Q, giving a reason for your answer.

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Answer

Brick Q will slide down the plane with constant speed. This happens because it is projected down the slope with an initial speed of $0.5 \text{ m s}^{-1}$ . As it slides, the frictional force will work against its motion. However, because the system reaches a balance where the friction equals the force acting down the slope, there is no net acceleration. Therefore, brick Q will move at a constant speed down the plane.

A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$ - Edexcel - A-Level Maths Mechanics - Question 1 - 2020 - Paper 1

Worked Solution & Example Answer:A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$ - Edexcel - A-Level Maths Mechanics - Question 1 - 2020 - Paper 1

find, in terms of m and g, the magnitude of the normal reaction of the plane on brick P.

show that \mu = \frac{3}{4}.

Explain briefly why brick Q will remain at rest on the plane.

describe the motion of brick Q, giving a reason for your answer.

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