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A particle of weight 120 N is placed on a fixed rough plane which is inclined at an angle \( \alpha \) to the horizontal, where \( \tan \alpha = \frac{3}{4} \) - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1
Question 6
A particle of weight 120 N is placed on a fixed rough plane which is inclined at an angle \( \alpha \) to the horizontal, where \( \tan \alpha = \frac{3}{4} \). The... show full transcript
Worked Solution & Example Answer:A particle of weight 120 N is placed on a fixed rough plane which is inclined at an angle \( \alpha \) to the horizontal, where \( \tan \alpha = \frac{3}{4} \) - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1
Step 1
Show that the normal reaction between the particle and the plane has magnitude 114 N.
Answer
To find the normal reaction ( S ) between the particle and the plane, we resolve the forces acting on the particle perpendicular to the plane.
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The weight of the particle acting down the plane is ( 120 \sin \alpha ). Using ( \tan \alpha = \frac{3}{4} ), we can derive that: [ \sin \alpha = \frac{3}{5} \quad \text{and} \quad \cos \alpha = \frac{4}{5} ]
Thus, ( 120 \sin \alpha = 72 \text{ N} ). -
Also, the horizontal force of 30 N contributes to the normal force: it acts at an angle to the plane, contributing ( 30 \cos \alpha ) perpendicular to the plane, which results in: [ 30 \cos \alpha = 30 \times \frac{4}{5} = 24 ext{ N} ]
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Therefore, we can set up the equation for the normal force as: [ S = 120 \cos \alpha + 30 \cos \alpha = 120 \times \frac{4}{5} + 24 = 96 + 24 = 120 ext{ N} ]
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The normal force considering the weight is: [ S = 120 \cos \alpha + 30 \sin \alpha = 120 \times \frac{4}{5} + 30 \times \frac{3}{5} = 96 + 18 = 114 ext{ N} ]
Hence, the normal reaction is confirmed to be 114 N.
Step 2
Find the greatest possible value of P.
Answer
For the case when the horizontal force is replaced with ( P ), we will analyze the equilibrium of forces parallel and perpendicular to the plane.
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The forces acting perpendicular to the plane remain: [ R = 120 \cos \alpha = 120 \times \frac{4}{5} = 96 ext{ N} ]
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The maximum static friction acting on the particle is given by: [ F_{max} = \frac{1}{2} R = \frac{1}{2} \times 96 = 48 ext{ N} ]
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For the forces acting parallel to the plane: [ P_{max} = F_{max} + 120 \sin \alpha = 48 + 72 = 120 ext{ N} ]
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Therefore, the greatest possible value of ( P ) is 120 N.
Step 3
Find the magnitude and direction of the frictional force acting on the particle when P = 30.
Answer
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When ( P = 30 \text{ N} ), we again analyze the forces within the equilibrium state. First, we resolve the forces parallel to the plane: [ 30 + F - 120 \sin \alpha = 0 ]
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Plugging in the value of ( \sin \alpha ): [ 120 \times \frac{3}{5} = 72 ]
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Thus, we can find ( F ) as: [ 30 + F - 72 = 0 ] [ F = 72 - 30 = 42 ext{ N } ]
This value indicates that the frictional force acts up the plane, opposing the motion initiated by the force P.