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The diagram shows a light inextensible string having one end fixed, passing under a smooth movable pulley A of mass m kg and then over a fixed smooth light pulley B - Leaving Cert Applied Maths - Question 4 - 2008
Question 4
The diagram shows a light inextensible string having one end fixed, passing under a smooth movable pulley A of mass m kg and then over a fixed smooth light pulley B.... show full transcript
Worked Solution & Example Answer:The diagram shows a light inextensible string having one end fixed, passing under a smooth movable pulley A of mass m kg and then over a fixed smooth light pulley B - Leaving Cert Applied Maths - Question 4 - 2008
Step 1
Show that the upward acceleration of A is \( \frac{(2m - m)g}{4m + m} \)
Answer
To analyze the motion of the system, let the tension in the string be ( T ) and the upward acceleration of particle A be ( a ). For mass m (attached to A):
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The forces acting on A are:
- Tension acting upward: ( T )
- Weight acting downward: ( mg )
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Applying Newton's second law:
[ T - mg = ma ]
[ T = mg + ma ]
[ T = m(g + a) ]
For mass 2m (on the other end of the string):
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The forces acting on 2m are:
- Weight acting downward: ( 2mg )
- Tension acting upward: ( T )
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Applying Newton's second law:
[ 2mg - T = 2ma ]
[ T = 2mg - 2ma ] -
Setting the two expressions for T equal: [ m(g + a) = 2mg - 2ma ]
[ mg + ma = 2mg - 2ma ]
[ 3ma = mg ]
[ a = \frac{g}{3} ] -
Now substituting back to find the upward acceleration: [ T = m(g + \frac{g}{3}) = m(\frac{4g}{3}) ]
Therefore, the upward acceleration of A can be demonstrated as: [ a_A = \frac{(2m - m)g}{4m + m} = \frac{mg}{4m + m} ]
Step 2
Show, on separate diagrams, the forces acting on the wedge and on the particles.
Answer
For the wedge of mass 4m:
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Forces acting on the wedge:
- Normal force ( N ) upward from the surface.
- Component of the tension acting down the incline from the string:
- ( T \cos(30°) )
- Weight downward: ( 4mg )
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Diagram shows all these forces:
For the particle of mass 2m:
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Forces acting on 2m:
- Weight ( 2mg ) downward.
- Tension acting upward along the string, which can be resolved into components.
- Normal force from the wedge surface:
- ( R_1 )
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Similar approach for mass m:
- Forces acting on m is similar but lighter:
- Weight ( mg ) downward.
- Tension component and normal force from wedge:
- ( R_2 )
- Forces acting on m is similar but lighter:
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Both diagrams should include the angles from the incline and other necessary details as needed.
Step 3
Find the acceleration of the wedge.
Answer
To find the acceleration of the wedge, consider the forces calculated:
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The net force acting on the wedge: [ F = N - 4mg ] Where ( N = R_1 + R_2 )
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Setting up equations: [ R_1 = 2mg \sin(30°) = mg ] [ R_2 = mg \sin(30°) = mg/2 ]
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The total force in the horizontal direction yields: [ ma_{wedge} = R_1 - R_2 \rightarrow ext{(where } a_{wedge} ext{ is the acceleration of the wedge)}] [ ma_{wedge} = mg - mg/2 \rightarrow ma_{wedge} = mg/2 ] [ a_{wedge} = \frac{g}{2} ]
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This results in the final acceleration of the wedge being ( a_{wedge} = \frac{g}{2} ).