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A rough plane is inclined to the horizontal at an angle α, where tan α = \frac{3}{4} - Edexcel - A-Level Maths Mechanics - Question 2 - 2022 - Paper 1
Question 2
A rough plane is inclined to the horizontal at an angle α, where tan α = \frac{3}{4}. A small block B of mass 5 kg is held in equilibrium on the plane by a horizont... show full transcript
Worked Solution & Example Answer:A rough plane is inclined to the horizontal at an angle α, where tan α = \frac{3}{4} - Edexcel - A-Level Maths Mechanics - Question 2 - 2022 - Paper 1
Step 1
(i) find the magnitude of the frictional force acting on B
Answer
To determine the magnitude of the frictional force acting on B, we will use the equilibrium of forces along the inclined plane. The forces acting on the block can be resolved into two components: along the plane and perpendicular to the plane.
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Resolve forces vertically:
R - 68.6 = 0 \ (Where R is the normal reaction, which is already given as 68.6N)
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Resolve forces horizontally (perpendicular to the plane):
The horizontal force acting on the block is X in the positive direction along the plane, and we can express the weight component acting down the slope as:
W = mg = 5g = 5 \cdot 9.8 = 49N \ (where g = 9.8 m/s²)
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Net Force in the direction of the plane:
68.6 \cdot \sin(\alpha) + F = X \
Since tan(α) = \frac{3}{4} -> we can find α using:
\sin(α) = \frac{3}{5}, \cos(α) = \frac{4}{5}
68.6 \cdot \frac{3}{5} + F = X \
F = X - 41.16
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Frictional Force:
The frictional force F = \mu R = 0.5 \cdot 68.6 = 34.3N.
Step 2
(ii) state the direction of the frictional force acting on B
Answer
The frictional force always acts in the opposite direction to the motion of the block. When the horizontal force X is removed and the block starts moving down the plane, the frictional force will act up the inclined plane to oppose the downward motion of the block. Thus, the direction of the frictional force acting on B is up the inclined plane.
Step 3
find the acceleration of B down the plane
Answer
To find the acceleration of B down the plane after the horizontal force is removed:
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Net forces acting on B:
Using Newton’s second law, F = ma, where m is the mass of the block and a is the acceleration:
F_{net} = mg \sin(α) - F \text{ (frictional force)}
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Substituting the values:
5g \sin(α) - 34.3 = 5a\
Using \sin(α) = \frac{3}{5}
5 \cdot 9.8 \cdot \frac{3}{5} - 34.3 = 5a\
29.4 - 34.3 = 5a
-4.9 = 5a\ -
Solve for a:
a = \frac{-4.9}{5} = -0.98 m/s²
Since the acceleration is downward, we express it as positive in the direction of motion, therefore, the acceleration of B down the plane is 0.98 m/s².