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A particle of mass 3 kg is connected to a particle of mass 5 kg by a taut, light, inextensible string which passes over a smooth light pulley at the edge of a rough horizontal table - Leaving Cert Applied Maths - Question 4 - 2011
Question 4
A particle of mass 3 kg is connected to a particle of mass 5 kg by a taut, light, inextensible string which passes over a smooth light pulley at the edge of a rough ... show full transcript
Worked Solution & Example Answer:A particle of mass 3 kg is connected to a particle of mass 5 kg by a taut, light, inextensible string which passes over a smooth light pulley at the edge of a rough horizontal table - Leaving Cert Applied Maths - Question 4 - 2011
Step 1
(i) Show on separate diagrams the forces acting on each particle.
Answer
Forces on the 3 kg Mass
- The weight of the mass acts downwards: ( W_1 = 3g )
- The tension in the string acts upwards: ( T )
- The frictional force acting opposite to the direction of motion can be calculated as: ( F_{friction} = \mu R = \frac{2}{3} \cdot 3g = 2g )
Forces on the 5 kg Mass
- The weight of this mass also acts downwards: ( W_2 = 5g )
- The tension acts upwards: ( T )
Step 2
(ii) Find the common acceleration of the particles.
Answer
To find the acceleration, we can set up the following equations:
For the 3 kg mass: [ 5g - T - F_{friction} = 5a ] [ 5g - T - 2g = 5a ] [ 3g - T = 5a \quad ext{(1)} ]
For the 5 kg mass: [ T - 5g = -5a \quad ext{(2)} ]
From equation (1): [ T = 3g - 5a ] Substituting into equation (2): [ (3g - 5a) - 5g = -5a ] [ 3g - 10a = 0 ] [ a = \frac{3g}{10} ] [ g \approx 9.8 \Rightarrow a \approx 2.94 \ \text{ms}^{-2} ]
Step 3
Step 4
(i) Find the common acceleration of the masses.
Answer
For the 8 kg mass: [ R - T = 2a \quad ext{(1)} ]
For the 2 kg mass: [ 2g - T = 2a \quad ext{(2)} ]
From equation (1): [ T = R - 2a ] From equation (2): [ T = 2g - 2a ] Setting these equal gives: [ R - 2a = 2g - 2a ] [ R = 2g \Rightarrow R = 2(9.8) \Rightarrow R = 19.6 \ N ]
From here, we can find the acceleration: [ 20 - 2a = 10a \Rightarrow 20 = 12a \Rightarrow a = \frac{5}{3} \approx 1.67 \ ms^{-2} ]
Step 5