Two blocks, A and B, of masses 2m and 3m respectively, are attached to the ends of a light string - Edexcel - A-Level Maths Mechanics - Question 3 - 2019 - Paper 1

To find the tension T, we start by analyzing the forces acting on both blocks A and B.

For block A, the forces acting on it are:

• The tension T acting upwards along the slope.
• The gravitational force component acting down the slope, which is (2mg \sin(\alpha)).
• The frictional force, which can be calculated as (F = \frac{2}{3} R), where (R) is the normal reaction force. The normal reaction can be found using:

$R = 2mg \cos(\alpha)$

Then the frictional force becomes:

$F = \frac{2}{3} (2mg \cos(\alpha))$

For block B, the only force acting on it is the gravitational force:

• The weight, which is (3mg) acting downwards.

Considering the equations of motion for both blocks:

• For A, we have:

$T - F - 2mg \sin(\alpha) = 2ma$

• For B:

$3mg - T = 3ma$

Next, since both blocks are connected, we can express acceleration a in terms of the tension. If we equate the two (since they are connected by the string):

1. From block A:

$T = F + 2mg \sin(\alpha) + 2ma$

2. From block B:

$T = 3mg - 3ma$

Now substituting the value of (a) derived from one equation into the other, we simplify to find:

Putting (\sin(\alpha) = \frac{5}{13}) (since, from the triangle, (\tan(\alpha) = \frac{5}{12})), we can derive the equations for T. Simplifying all these leads us to:

After plug-in values and balancing both equations properly, we arrive at:

$T = \frac{12mg}{5}$

Thus we have shown that the tension T can be expressed as (T = \frac{12mg}{5} ).