4. (a) Two particles of masses 5 kg and 7 kg are connected by a taut, light, inextensible string which passes over a smooth light pulley - Leaving Cert Applied Maths - Question 4 - 2010

To find the tension in the string, we substitute ( a ) into one of the original equations. Using the equation for the 5 kg particle:

$T - 5g = 5a$

Substituting the known values:

$T - 5 \times 10 = 5 \times \frac{5}{3}$

This simplifies to:

$T - 50 = \frac{25}{3}$

Adding 50 to both sides:

$T = \frac{25}{3} + 50 = \frac{25}{3} + \frac{150}{3} = \frac{175}{3} \approx 58.3 \, \text{N}$

For the 8 kg mass on a rough horizontal plane:

• The forces acting on the mass are:
  • Weight (8g downward)
  • Normal reaction force (R upwards)
  • Tension (T to the right)
  • Frictional force (( \mu R ) to the left)

The force diagram will depict these forces.

For the 10 kg mass on the inclined plane:

• The forces acting on the mass are:
  • Weight (10g downward)
  • Normal reaction force (R upwards, at 30 degrees to the plane)
  • Tension (T upwards along the plane)

The angles and direction of these forces should also be noted in the force diagram.

For the 8 kg mass on the horizontal plane:

Using the equation of motion:

1. The forces acting on the mass can be summarized as: $T - \mu R = 8a$ with ( R = 8g ) because it is on a horizontal surface: Therefore, $T - \mu (8g) = 8a$

For the 10 kg mass on the inclined plane:

2. Setting up the equation for the 10 kg mass: $10g \sin(30^\circ) - T = 10a$ where ( \sin(30^\circ) = \frac{1}{2} ): Thus, $5 - T = 10a$

Now solve the two equations simultaneously. From the first equation, substituting for T gives:

$T = 10g \sin(30^\circ) - 10a = 5 - 10a$

Combining this with the equation from the 8 kg mass leads to:

Solving will yield the common acceleration ( a ).

Using the values determined for ( a ) in the second part, substitute back into the equation for T:

From the mass on the horizontal plane: $T = 8a + \mu R$

Calculating ( R ) gives you: $R = 8g = 80 \, \text{N}$ Thus: $T = 8 \times a + \mu \times 80$

Substituting known values for ( a ) and ( \mu ) will give the result for the tension in the string.