A particle of mass 4 kg is connected to another particle of mass 6 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 - 2019

Using the equations of motion:

For the hanging mass: $6g - T = 6a$

For the mass on the table: $T - \mu R = 4a$

Substituting $R = 4g$: $T - \mu (4g) = 4a$

Adding both equations gives: $6g - \mu (4g) = 10a$

Plugging in values, we find: $a = 3 \, \text{m/s}^2$

Substituting $a = 3 \, \text{m/s}^2$ into the equation for the hanging mass:

$6g - T = 6(3)$

This simplifies to: $T = 6g - 18$

Given $g \approx 10 \, \text{m/s}^2$, we find: $T = 60 - 18 = 42 \, \text{N}$

From the equation for the mass on the table:

$T - \mu (4g) = 4(3)$

Substituting $T = 42$ into the equation: $42 - \mu (40) = 12$

So, $\mu (40) = 42 - 12$

Thus, $\mu = \frac{30}{40} = \frac{3}{4}$

For the 10 kg mass, the forces can be expressed as: $10g \sin \alpha - T = 10a$

And for the 6 kg mass: $6g - T = 6a$

Using the relation for acceleration: $a = \frac{g}{8} = 1.25 \, \text{m/s}^2$

Thus, the common acceleration is $1.25 \, \text{m/s}^2$.

From the equation for the mass on the inclined plane: $T = 10g \sin \alpha - 10a$

Using earlier equations, we can find: $T = 10(10)\left(\frac{4}{5}\right) - 10\left(\frac{g}{8}\right)$

Calculating this results in: $T = 67.5 \, \text{N}$