A uniform beam AB of length 30 m and mass 200 kg is held in limiting equilibrium by a light inextensible cable attached to B as shown in the diagram - Leaving Cert Applied Maths - Question 7 - 2016

Next, we apply the vertical force balance. The vertical forces must also sum to zero:

$T \cdot \sin(65^\circ) + Y = 200g$

Substituting $g \approx 9.81 \, m/s^2$:

$Y = 200 \cdot 9.81 - T \cdot \sin(65)$

After substituting T from earlier, we find:

$Y \approx 707.6956 \, N$

For the first rod, using equilibrium conditions:

When resolving forces vertically, we have:

$R_2 = W$ (weight of rod YZ)

Now for the tension in the string as both rods are inclined at angle $\theta$:

$R_1 + R_2 = W + T\sin(\theta)$

From the geometric relationship, we find:

$T = \frac{W}{\tan(\theta)}$

After adding a weight of $2W$ 25 cm from X:

Let T1 be the new tension after adding $2W$:

The new vertical force balance gives:

$T_1 = \frac{W}{4} + \frac{5W}{8}$

This results in:

$T_1 = \frac{9W}{8}$

Thus the tension increase is:

$T_1 - T = 0.25 T$

And hence it shows that the tension of the string is indeed increased by 25%.