A rough plane is inclined to the horizontal at an angle $ \alpha$, where \tan \alpha = \frac{3}{4}$. A small block $B$ of mass $5kg$ is held in equilibrium on the plane by a horizontal force of magnitude $X$ newtons, as shown in Figure 1. The force acts in a vertical plane which contains a line of greatest slope of the inclined plane. The block $B$ is modelled as a particle. The magnitude of the normal reaction of the plane on $B$ is $68.6N$. Using the model, (a) (i) find the magnitude of the frictional force acting on $B$, (ii) state the direction of the frictional force acting on $B$. The horizontal force of magnitude $X$ newtons is now removed and $B$ moves down the plane. Given that the coefficient of friction between $B$ and the plane is $0.5$ (b) find the acceleration of $B$ down the plane.

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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 2 - 2022 - Paper 1

Question 2

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A rough plane is inclined to the horizontal at an angle $ \alpha$, where \tan \alpha = \frac{3}{4}$. A small block $B$ of mass $5kg$ is held in equilibrium on the ... 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 2 - 2022 - Paper 1

Step 1

find the magnitude of the frictional force acting on B

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Answer

To find the magnitude of the frictional force acting on the block $B$ , we need to resolve the forces acting on it.

Identify Forces: The forces acting on block $B$ include:
- The normal reaction force ( $R = 68.6 N$ ).
- The weight of the block acting vertically downward ( $W = mg = 5 kg imes 9.8 m/s^2 = 49 N$ ).
- The horizontal force $X$ .
Resolve Forces: Using the inclined plane, we resolve the weight into components:
- Perpendicular to the plane: $W \cos(\alpha)$
- Parallel to the plane: $W \sin(\alpha)$
Apply Equilibrium Conditions: In equilibrium under the horizontal force,
- Perpendicular to the plane: $R + X \sin(\alpha) = 49 N$
- Parallel to the plane: $R - 68.6 = 5g \cos(\alpha)$
We can find the values of these components and then ultimately calculate the frictional force to be equal to the normal force multiplied by the coefficient of friction: $F_f = \mu R$ where \mu = 0.5$.

Step 2

state the direction of the frictional force acting on B

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Answer

The direction of the frictional force acting on block $B$ is up the plane. This is because the block is attempting to move down the plane due to gravity, and friction will always oppose the direction of motion.

Step 3

find the acceleration of B down the plane

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Answer

In this part, we need to apply Newton's second law, $F = ma$ .

Identify Forces Acting After Removal of $X$ : Now that the force $X$ is removed, the forces acting on $B$ are:
- The component of weight down the incline: $W_{down} = mg \sin(\alpha)$ .
- The frictional force $F_f$ acting up the incline, which we previously determined.
Set Up the Equation: The net force acting down the incline can be calculated as: $F_{net} = W_{down} - F_f = ma$ where $a$ is the acceleration of the block down the plane.
Substituting Values: Substitute the values derived: $5g \sin(\alpha) - \mu R = 5a$ After substituting for $\sin(\alpha)$ and $R$ , solve for $a$ .

This will yield the desired acceleration of block $B$ 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 2 - 2022 - 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 2 - 2022 - Paper 1

find the magnitude of the frictional force acting on B

state the direction of the frictional force acting on B

find the acceleration of B down the plane

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