4. (a) A particle of mass 3 kg is connected to another particle of mass 4 kg by a taut light inelastic string which passes over a smooth light pulley, as shown in the diagram - Leaving Cert Applied Maths - Question 4 - 2018

To find the tension in the string using the acceleration calculated above:

Using the expression from the first mass: $T = 4g - 4a$ Substituting a = rac{g}{7} into the equation: $T = 4g - 4\left(\frac{g}{7}\right)$ $T = 4g - \frac{4g}{7}$ $T = \frac{28g}{7} - \frac{4g}{7} = \frac{24g}{7}$ Therefore, the tension in the string is: $T = \frac{24g}{7} \text{ or approximately } 34.3 ext{ N}.$

For the 6 kg mass on the inclined plane (inclined at 30°):

• The forces acting are the gravitational force down the slope and the tension in the string.
• The weight component acting down the slope is: $6g \sin(30°) = 3g$.

For the 5 kg mass on the rough horizontal surface:

• The forces include the weight, the normal force, and friction, which is: rac{1}{5}(5g) = g.
• Tension also acts towards the pulley.

This results in:

• For 6 kg mass: $T - 6g \sin(30°) = 6a$.
• For 5 kg mass: $T - 5g - g = 5a.$

Using the equations derived: For the 6 kg mass: $T - 3g = 6a$. For the 5 kg mass: $T - 6 - g = 5a$. Substituting $T$ from one equation into the other: From the second mass: $T = g + 5a$. Setting both expressions for tension equal gives: $g + 5a - 3g = 6a$. Solving for $a$: $2g = a(1) \rightarrow a = \frac{2g}{11}.$

Using the tension expression from before: From the 5 kg mass: $T = 5a + g$. Substituting $a = \frac{2g}{11}$ into the equation: $T = 5\left(\frac{2g}{11}\right) + g = \frac{10g}{11} + g = g\left(\frac{10+11}{11}\right) = g\left(\frac{21}{11}\right).$ Thus, the tension in the string is: $T = \frac{21g}{11}.$