A particle of mass 2 kg is connected to another particle of mass 3 kg by a taut light inelastic string which passes over a smooth light pulley at the edge of a rough horizontal table - Leaving Cert Applied Maths - Question 4 - 2013

Using Newton’s second law for the 3 kg mass:

1. $3g - T = 3a$

For the 2 kg mass:

2. $T - \frac{1}{2} R - 2g = 2a$

Since $R = 2g - f$ and $f = \frac{1}{2}R$, substituting gives:

3. $T + \frac{1}{4}R = 2a$

Combining and solving these equations leads to:

4. $a = \frac{2g}{5} = 4 \, m/s^2$.

Substituting the acceleration back into the equations gives:

Using $T = 3g - 3a$ leads to:

5. $T = 3(10) - 3(4) = 18 \, N$.

For the 6 kg mass:

1. $6g \cos(30^{\circ}) - T = 6a$

For the 2 kg mass:

2. $T - 2g = 2a$

Substituting and simplifying gives:

3. $g = 8$ leads to $a = \frac{10}{8} = 1.25 \, m/s^2$.

Using the calculated acceleration in:

1. $T = 2g + 2a$ results in:

2. $T = 22.5 \, N$.