A $$1 \text{ kg}$$ mass is suspended from the ceiling by a rope of length $$4 \text{ m}$$. A horizontal force '$$F$$' is applied at the mid point of the rope so that the rope makes an angle of $$45°$$ with respect to the vertical axis as shown in figure.

The magnitude of $$F$$ is : (Assume that the system is in equilibrium and $$g = 10 \text{ m/s}^2$$)

First lets write the Forces acting in Vertical and Horizontal direction

In the Vertical direction,

Forces are $$F_g=mg\ due\ to\ gravity\ and\ T\sin\theta\ \ \left(\ tension\ of\ the\ rope\right)$$

and in Horizontal , The forces are $$F\left(appplied\ force\right)\ and\ T\cos\theta\ \left(tension\ of\ the\ rope\right)$$

First lets equate the Vertical Forces

$$mg\ =\ T\sin\ \theta\ $$

$$10=\ T\times\ \frac{1}{\sqrt{\ 2}}$$

Now substitute this and equal Horizontal forces

$$F=\ T\cos\theta$$

$$F=\ \sqrt{\ 2}\times\ 10\times\ \cos45^{\circ\ }\ $$

$$F=\ \frac{10}{\sqrt{\ 2}}\times\ \sqrt{\ 2}$$

$$F=\ 10N$$