A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$. A small block B of mass $5\,kg$ 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 $68.6\,N$. 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 $5\,kg$ is held in equilibrium on the p... 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, we first need to resolve the forces acting on the block B along the inclined plane. The forces include the weight component down the slope and the applied horizontal force X.

Using geometry, we can determine that the weight $W = mg$ can be resolved into two components:

Down the slope: $W\sin(\alpha) = 5g\sin(\alpha)$
Perpendicular to the slope: $W\cos(\alpha) = 5g\cos(\alpha)$

Substituting the normal reaction: $N = 68.6\,N$ .

Using $an(\alpha) = \frac{3}{4}$ , we can derive $rac{\sin(\alpha)}{\cos(\alpha)} = \frac{3}{4}$ . Therefore, $\sin(\alpha)$ and $\cos(\alpha)$ can be found using the triangle identity:

$\sin(\alpha) = \frac{3}{5} \text{ and } \cos(\alpha) = \frac{4}{5}$

Thus, the equation of motion perpendicular to the plane gives: $68.6 = 5g\cos(\alpha)$ We can use the above equation to find $g=9.8 \text{ m/s}^2$ .

The frictional force $f$ can then be expressed as: f = \mu N = 0.5 \cdot 68.6 = 34.3 N.

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 will oppose the motion, which is down the incline. Thus, the frictional force acts up the plane.

Step 3

Find the acceleration of B down the plane

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Answer

Once the horizontal force X has been removed, the only forces acting on B will be its weight component down the slope, the frictional force, and the normal force.

The effective force down the slope can be expressed as:

$F_{net} = W\sin(\alpha) - f$

Substituting the values we found: $F_{net} = (5g\sin(\alpha)) - 34.3 = (5 \cdot 9.8 \cdot \frac{3}{5}) - 34.3$

Calculating gives: $F_{net} = 29.4 - 34.3 = -4.9 \text{ N}$ This means that B is still moving down but with a net force of 4.9 N against the direction. Hence, this negative sign indicates the friction is greater than the component of the weight trying to pull B down the plane.

To find the acceleration, we can use Newton's second law:

$F = ma$ So, $a = \frac{F_{net}}{m} = \frac{-4.9}{5} = -0.98 \text{ m/s}^2$

The acceleration of B down the plane is thus $\approx -0.98 \text{ m/s}^2$ , indicating deceleration.

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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