4. (a) Two particles of masses 7 kg and 3 kg are connected by a taut, light, inelastic string which passes over a smooth light pulley - Leaving Cert Applied Maths - Question 4 - 2007

By applying Newton's second law, we can write:

For the 3 kg mass: $T - 3g = -3a$

For the 7 kg mass: $7g - T = 7a$

Adding both equations:

$7g - 3g = 7a + 3a$
$4g = 10a$

Thus, solving for acceleration (a):
$a = \frac{4g}{10} = \frac{2g}{5}$
If we take g = 10 m/s²: $a = 4 m/s²$.

The common acceleration is ( 4 \text{ m/s}^2 ).

Using the equation for the 3 kg mass: $T - 3g = -3a$

Substituting the values: $T - 30 = -12$

Thus, solving for T: $T = 30 - 12 = 42 ext{ N}$.

The tension in the string is ( 42 \text{ N} ).

For the 2 kg mass on the inclined plane:

• Weight (W) = 2g downwards
• Normal force (R) perpendicular to the plane
• Frictional force (F_f) down the slope = ( \mu R )
• Tension (T) upwards along the plane

For the 3 kg mass:

• Weight (W') = 3g downwards
• Tension (T') upwards

The forces can be diagrammatically represented as:

For 2 kg Mass:
(Diagram)
W → 2g
↑ N
↘ F_f
↘ T

For 3 kg Mass:
(Diagram)
W' → 3g
↑ T'

From the force equations: For the 3 kg mass: $3g - T = 3a$
For the 2 kg mass: $T - 2g \sin 30° - \mu R = 2a$
With ( R = 2g \cos 30° ) substituting we get: $T - 2g \cdot \frac{1}{2} - \frac{1}{\sqrt{3}} R = 2a$ Solving these equations simultaneously, we find: $a = 2 ext{ m/s}^2$.

The common acceleration is ( 2 \text{ m/s}^2 ).

From the equation of motion for the 3 kg mass: $T - 3a = 3g$
Substituting the known values into the equation: $T - 6 = 30$
Thus, solving for T: $T = 30 + 6 = 24 ext{ N}$.

The tension in the string is ( 24 \text{ N} ).