A block A of mass 10m on a smooth plane inclined at an angle α with the horizontal, where tan α = rac{2}{3}, is connected by a light inextensible string which passes over a smooth pulley to a second block B of mass 10m - Leaving Cert Applied Maths - Question 4 - 2020

To find the time that B remains in contact with the floor, we consider the motion equation:

$v^2 = u^2 + 2as$

Substituting the initial velocity (u = 0), acceleration from part (i) as (a = 1.96) and distance (s = 0.245):

$v^2 = 0 + 2(1.96)(0.245)$

Calculating this gives:

$v \approx 0.98 m/s$

Next, we find the time taken (t) using the equation:

$s = ut + \frac{1}{2}at^2$

with (s = 0.245), and initial velocity (u = 0):

This simplifies to:

$0.245 = \frac{1}{2}(1.96)t^2$

Thus,

$t^2 = \frac{0.245 \cdot 2}{1.96}$

which results in:

t \approx 0.33 s.

For the movable pulley:

• The forces acting on the pulley are:
  • Tension (T) upwards
  • Weight of the pulley (3mg) downwards

For mass C:

• Forces acting:
  • Tension (T) up the plane
  • Weight component against the motion (2mg sin 30°), down the incline
  • Frictional force (μR) opposing the motion

These forces could be represented as arrows on separate diagrams as indicated above.

To find the tension in the string in terms of m, we shall use the force balance equations derived from F = ma:

For mass C moving up the incline:

The balancing equation will be:

$T - 2mg ext{sin} 30° - μR = 2ma$

For the pulley under gravity:

$3mg - 2T = 3m ext{(acceleration)}$

Putting these equations together and solving for T against the forces:

By substituting the known values we arrive at:

$T = \frac{150m}{129}$

This shows the tension in the string in terms of m.