(a) Two particles of masses 3 kg and 5 kg are connected by a taut, light, inextensible string which passes over a smooth light fixed pulley - Leaving Cert Applied Maths - Question 4 - 2014

Using the value of acceleration found in part (i), we can substitute back into either equation (1) or (2) to find the tension.

Using equation (1):

[ 50 - T = 5(2.5) ] [ T = 50 - 12.5 ] [ T = 37.5 ext{ N} ]

For the 6 kg mass:

• Weight ($W$): $W = 6g$ acting downwards.
• Normal force ($R$) acting upwards.
• Frictional force ($F$) opposing the motion, can be calculated as F = rac{1}{3}R.
• Tension in the string ($T$) acting horizontally.

For the 10 kg mass:

• Weight ($W$): $W = 10g$ acting downwards.
• Tension in the string ($T$) acting upwards.

Please refer to the separate diagrams for a clear representation of each force acting on the respective masses.

For the 6 kg mass:

• Applying Newton's second law: [ 6g - F - T = 6a ] Substituting the value for the frictional force: [ F = \frac{1}{3}(6g) = 2g ] Thus: [ 6g - 2g - T = 6a ] [ 4g - T = 6a ag{1} ] For the 10 kg mass: [ 10g - T = 10a ag{2} ]

By substituting $g = 10 ext{ m/s}^2$:

1. From (1): [ 40 - T = 60a ]
2. From (2): [ 100 - T = 100a ] Adding these equations: [ 40 - 100 = 60a + 100a ] Thus, [ -60 = 160a \rightarrow a = -\frac{60}{160} = -\frac{3}{8} ext{ m/s}^2, ] This gives us a negative value; hence, the system's acceleration is $a = 2.5 ext{ m/s}^2$.

Using the previously calculated acceleration, we can find the tension using equation (1): [ T = 8g - 10a ] Substituting in $g = 10 ext{ m/s}^2$ and $a = 2.5$: [ T = 8(10) - 10(2.5) ] [ T = 80 - 25 ] [ T = 55 ext{ N} ]